problem solving law of inertia

6 Applications of Newton’s Laws

6.1 Solving Problems with Newton’s Laws

Learning objectives.

By the end of the section, you will be able to:

• Apply problem-solving techniques to solve for quantities in more complex systems of forces
• Use concepts from kinematics to solve problems using Newton’s laws of motion
• Solve more complex equilibrium problems
• Solve more complex acceleration problems
• Apply calculus to more advanced dynamics problems

Success in problem solving is necessary to understand and apply physical principles. We developed a pattern of analyzing and setting up the solutions to problems involving Newton’s laws in Newton’s Laws of Motion ; in this chapter, we continue to discuss these strategies and apply a step-by-step process.

Problem-Solving Strategies

We follow here the basics of problem solving presented earlier in this text, but we emphasize specific strategies that are useful in applying Newton’s laws of motion . Once you identify the physical principles involved in the problem and determine that they include Newton’s laws of motion, you can apply these steps to find a solution. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, so the following techniques should reinforce skills you have already begun to develop.

Problem-Solving Strategy: Applying Newton’s Laws of Motion

• Identify the physical principles involved by listing the givens and the quantities to be calculated.
• Sketch the situation, using arrows to represent all forces.
• Determine the system of interest. The result is a free-body diagram that is essential to solving the problem.
• Apply Newton’s second law to solve the problem. If necessary, apply appropriate kinematic equations from the chapter on motion along a straight line.
• Check the solution to see whether it is reasonable.

Let’s apply this problem-solving strategy to the challenge of lifting a grand piano into a second-story apartment. Once we have determined that Newton’s laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in Figure (a). Then, as in Figure (b), we can represent all forces with arrows. Whenever sufficient information exists, it is best to label these arrows carefully and make the length and direction of each correspond to the represented force.

As with most problems, we next need to identify what needs to be determined and what is known or can be inferred from the problem as stated, that is, make a list of knowns and unknowns. It is particularly crucial to identify the system of interest, since Newton’s second law involves only external forces. We can then determine which forces are external and which are internal, a necessary step to employ Newton’s second law. (See Figure (c).) Newton’s third law may be used to identify whether forces are exerted between components of a system (internal) or between the system and something outside (external). As illustrated in Newton’s Laws of Motion , the system of interest depends on the question we need to answer. Only forces are shown in free-body diagrams, not acceleration or velocity. We have drawn several free-body diagrams in previous worked examples. Figure (c) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram.

Once a free-body diagram is drawn, we apply Newton’s second law. This is done in Figure (d) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—that is, if all forces are parallel—then the forces can be handled algebraically. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. We do this by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known. Generally, just write Newton’s second law in components along the different directions. Then, you have the following equations:

(If, for example, the system is accelerating horizontally, then you can then set [latex]{a}_{y}=0.[/latex]) We need this information to determine unknown forces acting on a system.

As always, we must check the solution. In some cases, it is easy to tell whether the solution is reasonable. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving; with experience, it becomes progressively easier to judge whether an answer is reasonable. Another way to check a solution is to check the units. If we are solving for force and end up with units of millimeters per second, then we have made a mistake.

There are many interesting applications of Newton’s laws of motion, a few more of which are presented in this section. These serve also to illustrate some further subtleties of physics and to help build problem-solving skills. We look first at problems involving particle equilibrium, which make use of Newton’s first law, and then consider particle acceleration, which involves Newton’s second law.

Particle Equilibrium

Recall that a particle in equilibrium is one for which the external forces are balanced. Static equilibrium involves objects at rest, and dynamic equilibrium involves objects in motion without acceleration, but it is important to remember that these conditions are relative. For example, an object may be at rest when viewed from our frame of reference, but the same object would appear to be in motion when viewed by someone moving at a constant velocity. We now make use of the knowledge attained in Newton’s Laws of Motion , regarding the different types of forces and the use of free-body diagrams, to solve additional problems in particle equilibrium .

Different Tensions at Different Angles

Consider the traffic light (mass of 15.0 kg) suspended from two wires as shown in Figure . Find the tension in each wire, neglecting the masses of the wires.

The system of interest is the traffic light, and its free-body diagram is shown in Figure (c). The three forces involved are not parallel, and so they must be projected onto a coordinate system. The most convenient coordinate system has one axis vertical and one horizontal, and the vector projections on it are shown in Figure (d). There are two unknowns in this problem ([latex]{T}_{1}[/latex] and [latex]{T}_{2}[/latex]), so two equations are needed to find them. These two equations come from applying Newton’s second law along the vertical and horizontal axes, noting that the net external force is zero along each axis because acceleration is zero.

First consider the horizontal or x -axis:

Thus, as you might expect,

This gives us the following relationship:

Note that [latex]{T}_{1}[/latex] and [latex]{T}_{2}[/latex] are not equal in this case because the angles on either side are not equal. It is reasonable that [latex]{T}_{2}[/latex] ends up being greater than [latex]{T}_{1}[/latex] because it is exerted more vertically than [latex]{T}_{1}.[/latex]

Now consider the force components along the vertical or y -axis:

This implies

Substituting the expressions for the vertical components gives

There are two unknowns in this equation, but substituting the expression for [latex]{T}_{2}[/latex] in terms of [latex]{T}_{1}[/latex] reduces this to one equation with one unknown:

which yields

Solving this last equation gives the magnitude of [latex]{T}_{1}[/latex] to be

Finally, we find the magnitude of [latex]{T}_{2}[/latex] by using the relationship between them, [latex]{T}_{2}=1.225{T}_{1}[/latex], found above. Thus we obtain

Significance

Both tensions would be larger if both wires were more horizontal, and they will be equal if and only if the angles on either side are the same (as they were in the earlier example of a tightrope walker in Newton’s Laws of Motion .

Particle Acceleration

We have given a variety of examples of particles in equilibrium. We now turn our attention to particle acceleration problems, which are the result of a nonzero net force. Refer again to the steps given at the beginning of this section, and notice how they are applied to the following examples.

Drag Force on a Barge

Two tugboats push on a barge at different angles ( Figure ). The first tugboat exerts a force of [latex]2.7\times {10}^{5}\,\text{N}[/latex] in the x -direction, and the second tugboat exerts a force of [latex]3.6\times {10}^{5}\,\text{N}[/latex] in the y -direction. The mass of the barge is [latex]5.0\times {10}^{6}\,\text{kg}[/latex] and its acceleration is observed to be [latex]7.5\times {10}^{-2}\,{\text{m/s}}^{2}[/latex] in the direction shown. What is the drag force of the water on the barge resisting the motion? ( Note: Drag force is a frictional force exerted by fluids, such as air or water. The drag force opposes the motion of the object. Since the barge is flat bottomed, we can assume that the drag force is in the direction opposite of motion of the barge.)

The directions and magnitudes of acceleration and the applied forces are given in Figure (a). We define the total force of the tugboats on the barge as [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}}[/latex] so that

The drag of the water [latex]{\mathbf{\overset{\to }{F}}}_{\text{D}}[/latex] is in the direction opposite to the direction of motion of the boat; this force thus works against [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}},[/latex] as shown in the free-body diagram in Figure (b). The system of interest here is the barge, since the forces on it are given as well as its acceleration. Because the applied forces are perpendicular, the x – and y -axes are in the same direction as [latex]{\mathbf{\overset{\to }{F}}}_{1}[/latex] and [latex]{\mathbf{\overset{\to }{F}}}_{2}.[/latex] The problem quickly becomes a one-dimensional problem along the direction of [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}}[/latex], since friction is in the direction opposite to [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}}.[/latex] Our strategy is to find the magnitude and direction of the net applied force [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}}[/latex] and then apply Newton’s second law to solve for the drag force [latex]{\mathbf{\overset{\to }{F}}}_{\text{D}}.[/latex]

Since [latex]{F}_{x}[/latex] and [latex]{F}_{y}[/latex] are perpendicular, we can find the magnitude and direction of [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}}[/latex] directly. First, the resultant magnitude is given by the Pythagorean theorem:

The angle is given by

From Newton’s first law, we know this is the same direction as the acceleration. We also know that [latex]{\mathbf{\overset{\to }{F}}}_{\text{D}}[/latex] is in the opposite direction of [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}},[/latex] since it acts to slow down the acceleration. Therefore, the net external force is in the same direction as [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}},[/latex] but its magnitude is slightly less than [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}}.[/latex] The problem is now one-dimensional. From the free-body diagram, we can see that

However, Newton’s second law states that

This can be solved for the magnitude of the drag force of the water [latex]{F}_{\text{D}}[/latex] in terms of known quantities:

Substituting known values gives

The direction of [latex]{\mathbf{\overset{\to }{F}}}_{\text{D}}[/latex] has already been determined to be in the direction opposite to [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}},[/latex] or at an angle of [latex]53^\circ[/latex] south of west.

The numbers used in this example are reasonable for a moderately large barge. It is certainly difficult to obtain larger accelerations with tugboats, and small speeds are desirable to avoid running the barge into the docks. Drag is relatively small for a well-designed hull at low speeds, consistent with the answer to this example, where [latex]{F}_{\text{D}}[/latex] is less than 1/600th of the weight of the ship.

In Newton’s Laws of Motion , we discussed the normal force , which is a contact force that acts normal to the surface so that an object does not have an acceleration perpendicular to the surface. The bathroom scale is an excellent example of a normal force acting on a body. It provides a quantitative reading of how much it must push upward to support the weight of an object. But can you predict what you would see on the dial of a bathroom scale if you stood on it during an elevator ride? Will you see a value greater than your weight when the elevator starts up? What about when the elevator moves upward at a constant speed? Take a guess before reading the next example.

What Does the Bathroom Scale Read in an Elevator?

Figure shows a 75.0-kg man (weight of about 165 lb.) standing on a bathroom scale in an elevator. Calculate the scale reading: (a) if the elevator accelerates upward at a rate of [latex]1.20\,{\text{m/s}}^{2},[/latex] and (b) if the elevator moves upward at a constant speed of 1 m/s.

If the scale at rest is accurate, its reading equals [latex]{\mathbf{\overset{\to }{F}}}_{\text{p}}[/latex], the magnitude of the force the person exerts downward on it. Figure (a) shows the numerous forces acting on the elevator, scale, and person. It makes this one-dimensional problem look much more formidable than if the person is chosen to be the system of interest and a free-body diagram is drawn, as in Figure (b). Analysis of the free-body diagram using Newton’s laws can produce answers to both Figure (a) and (b) of this example, as well as some other questions that might arise. The only forces acting on the person are his weight [latex]\mathbf{\overset{\to }{w}}[/latex] and the upward force of the scale [latex]{\mathbf{\overset{\to }{F}}}_{\text{s}}.[/latex] According to Newton’s third law, [latex]{\mathbf{\overset{\to }{F}}}_{\text{p}}[/latex] and [latex]{\mathbf{\overset{\to }{F}}}_{\text{s}}[/latex] are equal in magnitude and opposite in direction, so that we need to find [latex]{F}_{\text{s}}[/latex] in order to find what the scale reads. We can do this, as usual, by applying Newton’s second law,

From the free-body diagram, we see that [latex]{\mathbf{\overset{\to }{F}}}_{\text{net}}={\mathbf{\overset{\to }{F}}}_{s}-\mathbf{\overset{\to }{w}},[/latex] so we have

Solving for [latex]{F}_{s}[/latex] gives us an equation with only one unknown:

or, because [latex]w=mg,[/latex] simply

No assumptions were made about the acceleration, so this solution should be valid for a variety of accelerations in addition to those in this situation. ( Note: We are considering the case when the elevator is accelerating upward. If the elevator is accelerating downward, Newton’s second law becomes [latex]{F}_{s}-w=\text{−}ma.[/latex])

which gives

The scale reading in Figure (a) is about 185 lb. What would the scale have read if he were stationary? Since his acceleration would be zero, the force of the scale would be equal to his weight:

Thus, the scale reading in the elevator is greater than his 735-N (165-lb.) weight. This means that the scale is pushing up on the person with a force greater than his weight, as it must in order to accelerate him upward. Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators. In Figure (b), the scale reading is 735 N, which equals the person’s weight. This is the case whenever the elevator has a constant velocity—moving up, moving down, or stationary.

Check Your Understanding

Now calculate the scale reading when the elevator accelerates downward at a rate of [latex]1.20\,{\text{m/s}}^{2}.[/latex]

[latex]{F}_{\text{s}}=645\,\text{N}[/latex]

The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevator accelerates downward, a is negative, and the scale reading is less than the weight of the person. If a constant downward velocity is reached, the scale reading again becomes equal to the person’s weight. If the elevator is in free fall and accelerating downward at g , then the scale reading is zero and the person appears to be weightless.

Two Attached Blocks

Figure shows a block of mass [latex]{m}_{1}[/latex] on a frictionless, horizontal surface. It is pulled by a light string that passes over a frictionless and massless pulley. The other end of the string is connected to a block of mass [latex]{m}_{2}.[/latex] Find the acceleration of the blocks and the tension in the string in terms of [latex]{m}_{1},{m}_{2},\,\text{and}\,g.[/latex]

We draw a free-body diagram for each mass separately, as shown in Figure . Then we analyze each one to find the required unknowns. The forces on block 1 are the gravitational force, the contact force of the surface, and the tension in the string. Block 2 is subjected to the gravitational force and the string tension. Newton’s second law applies to each, so we write two vector equations:

For block 1: [latex]\mathbf{\overset{\to }{T}}+{\mathbf{\overset{\to }{w}}}_{1}+\mathbf{\overset{\to }{N}}={m}_{1}{\mathbf{\overset{\to }{a}}}_{1}[/latex]

For block 2: [latex]\mathbf{\overset{\to }{T}}+{\mathbf{\overset{\to }{w}}}_{2}={m}_{2}{\mathbf{\overset{\to }{a}}}_{2}.[/latex]

Notice that [latex]\mathbf{\overset{\to }{T}}[/latex] is the same for both blocks. Since the string and the pulley have negligible mass, and since there is no friction in the pulley, the tension is the same throughout the string. We can now write component equations for each block. All forces are either horizontal or vertical, so we can use the same horizontal/vertical coordinate system for both objects

The component equations follow from the vector equations above. We see that block 1 has the vertical forces balanced, so we ignore them and write an equation relating the x -components. There are no horizontal forces on block 2, so only the y -equation is written. We obtain these results:

When block 1 moves to the right, block 2 travels an equal distance downward; thus, [latex]{a}_{1x}=\text{−}{a}_{2y}.[/latex] Writing the common acceleration of the blocks as [latex]a={a}_{1x}=\text{−}{a}_{2y},[/latex] we now have

From these two equations, we can express a and T in terms of the masses [latex]{m}_{1}\,\text{and}\,{m}_{2},\,\text{and}\,g:[/latex]

Notice that the tension in the string is less than the weight of the block hanging from the end of it. A common error in problems like this is to set [latex]T={m}_{2}g.[/latex] You can see from the free-body diagram of block 2 that cannot be correct if the block is accelerating.

Calculate the acceleration of the system, and the tension in the string, when the masses are [latex]{m}_{1}=5.00\,\text{kg}[/latex] and [latex]{m}_{2}=3.00\,\text{kg}.[/latex]

[latex]a=3.68\,{\text{m/s}}^{2},[/latex] [latex]T=18.4\,\text{N}[/latex]

Atwood Machine

A classic problem in physics, similar to the one we just solved, is that of the Atwood machine , which consists of a rope running over a pulley, with two objects of different mass attached. It is particularly useful in understanding the connection between force and motion. In Figure , [latex]{m}_{1}=2.00\,\text{kg}[/latex] and [latex]{m}_{2}=4.00\,\text{kg}\text{.}[/latex] Consider the pulley to be frictionless. (a) If [latex]{m}_{2}[/latex] is released, what will its acceleration be? (b) What is the tension in the string?

We draw a free-body diagram for each mass separately, as shown in the figure. Then we analyze each diagram to find the required unknowns. This may involve the solution of simultaneous equations. It is also important to note the similarity with the previous example. As block 2 accelerates with acceleration [latex]{a}_{2}[/latex] in the downward direction, block 1 accelerates upward with acceleration [latex]{a}_{1}[/latex]. Thus, [latex]a={a}_{1}=\text{−}{a}_{2}.[/latex]

(The negative sign in front of [latex]{m}_{2}a[/latex] indicates that [latex]{m}_{2}[/latex] accelerates downward; both blocks accelerate at the same rate, but in opposite directions.) Solve the two equations simultaneously (subtract them) and the result is

Solving for a :

• Observing the first block, we see that [latex]\begin{array}{c}T-{m}_{1}g={m}_{1}a\hfill \\ T={m}_{1}(g+a)=(2\,\text{kg})(9.8\,{\text{m/s}}^{2}+3.27\,{\text{m/s}}^{2})=26.1\,\text{N}\text{.}\hfill \end{array}[/latex]

The result for the acceleration given in the solution can be interpreted as the ratio of the unbalanced force on the system, [latex]({m}_{2}-{m}_{1})g[/latex], to the total mass of the system, [latex]{m}_{1}+{m}_{2}[/latex]. We can also use the Atwood machine to measure local gravitational field strength.

Determine a general formula in terms of [latex]{m}_{1},{m}_{2}[/latex] and g for calculating the tension in the string for the Atwood machine shown above.

[latex]T=\frac{2{m}_{1}{m}_{2}}{{m}_{1}+{m}_{2}}g[/latex] (This is found by substituting the equation for acceleration in Figure (a), into the equation for tension in Figure (b).)

Newton’s Laws of Motion and Kinematics

Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. Newton’s laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics , and hence the relevance of earlier chapters.

When approaching problems that involve various types of forces, acceleration, velocity, and/or position, listing the givens and the quantities to be calculated will allow you to identify the principles involved. Then, you can refer to the chapters that deal with a particular topic and solve the problem using strategies outlined in the text. The following worked example illustrates how the problem-solving strategy given earlier in this chapter, as well as strategies presented in other chapters, is applied to an integrated concept problem.

What Force Must a Soccer Player Exert to Reach Top Speed?

A soccer player starts at rest and accelerates forward, reaching a velocity of 8.00 m/s in 2.50 s. (a) What is her average acceleration? (b) What average force does the ground exert forward on the runner so that she achieves this acceleration? The player’s mass is 70.0 kg, and air resistance is negligible.

To find the answers to this problem, we use the problem-solving strategy given earlier in this chapter. The solutions to each part of the example illustrate how to apply specific problem-solving steps. In this case, we do not need to use all of the steps. We simply identify the physical principles, and thus the knowns and unknowns; apply Newton’s second law; and check to see whether the answer is reasonable.

Substituting the known values yields

Substituting the known values of m and a gives

This is a reasonable result: The acceleration is attainable for an athlete in good condition. The force is about 50 pounds, a reasonable average force.

This example illustrates how to apply problem-solving strategies to situations that include topics from different chapters. The first step is to identify the physical principles, the knowns, and the unknowns involved in the problem. The second step is to solve for the unknown, in this case using Newton’s second law. Finally, we check our answer to ensure it is reasonable. These techniques for integrated concept problems will be useful in applications of physics outside of a physics course, such as in your profession, in other science disciplines, and in everyday life.

The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the ball?

What Force Acts on a Model Helicopter?

A 1.50-kg model helicopter has a velocity of [latex]5.00\mathbf{\hat{j}}\,\text{m/s}[/latex] at [latex]t=0.[/latex] It is accelerated at a constant rate for two seconds (2.00 s) after which it has a velocity of [latex](6.00\mathbf{\hat{i}}+12.00\mathbf{\hat{j}})\text{m/s}\text{.}[/latex] What is the magnitude of the resultant force acting on the helicopter during this time interval?

We can easily set up a coordinate system in which the x -axis [latex](\mathbf{\hat{i}}[/latex] direction) is horizontal, and the y -axis [latex](\mathbf{\hat{j}}[/latex] direction) is vertical. We know that [latex]\Delta t=2.00s[/latex] and [latex](6.00\mathbf{\hat{i}}+12.00\mathbf{\hat{j}}\,\text{m/s})-(5.00\mathbf{\hat{j}}\,\text{m/s}).[/latex] From this, we can calculate the acceleration by the definition; we can then apply Newton’s second law.

The magnitude of the force is now easily found:

The original problem was stated in terms of [latex]\mathbf{\hat{i}}-\mathbf{\hat{j}}[/latex] vector components, so we used vector methods. Compare this example with the previous example.

Find the direction of the resultant for the 1.50-kg model helicopter.

49.4 degrees

Baggage Tractor

Figure (a) shows a baggage tractor pulling luggage carts from an airplane. The tractor has mass 650.0 kg, while cart A has mass 250.0 kg and cart B has mass 150.0 kg. The driving force acting for a brief period of time accelerates the system from rest and acts for 3.00 s. (a) If this driving force is given by [latex]F=(820.0t)\,\text{N,}[/latex] find the speed after 3.00 seconds. (b) What is the horizontal force acting on the connecting cable between the tractor and cart A at this instant?

A free-body diagram shows the driving force of the tractor, which gives the system its acceleration. We only need to consider motion in the horizontal direction. The vertical forces balance each other and it is not necessary to consider them. For part b, we make use of a free-body diagram of the tractor alone to determine the force between it and cart A. This exposes the coupling force [latex]\mathbf{\overset{\to }{T}},[/latex] which is our objective.

Since acceleration is a function of time, we can determine the velocity of the tractor by using [latex]a=\frac{dv}{dt}[/latex] with the initial condition that [latex]{v}_{0}=0[/latex] at [latex]t=0.[/latex] We integrate from [latex]t=0[/latex] to [latex]t=3\text{:}[/latex]

• Refer to the free-body diagram in Figure (b). [latex]\begin{array}{ccc}\hfill \sum {F}_{x}& =\hfill & {m}_{\text{tractor}}{a}_{x}\hfill \\ \hfill 820.0t-T& =\hfill & {m}_{\text{tractor}}(0.7805)t\hfill \\ \hfill (820.0)(3.00)-T& =\hfill & (650.0)(0.7805)(3.00)\hfill \\ \hfill T& =\hfill & 938\,\text{N}.\hfill \end{array}[/latex]

Since the force varies with time, we must use calculus to solve this problem. Notice how the total mass of the system was important in solving Figure (a), whereas only the mass of the truck (since it supplied the force) was of use in Figure (b).

Recall that [latex]v=\frac{ds}{dt}[/latex] and [latex]a=\frac{dv}{dt}[/latex]. If acceleration is a function of time, we can use the calculus forms developed in Motion Along a Straight Line , as shown in this example. However, sometimes acceleration is a function of displacement. In this case, we can derive an important result from these calculus relations. Solving for dt in each, we have [latex]dt=\frac{ds}{v}[/latex] and [latex]dt=\frac{dv}{a}.[/latex] Now, equating these expressions, we have [latex]\frac{ds}{v}=\frac{dv}{a}.[/latex] We can rearrange this to obtain [latex]{a}^{}ds={v}^{}dv.[/latex]

Motion of a Projectile Fired Vertically

A 10.0-kg mortar shell is fired vertically upward from the ground, with an initial velocity of 50.0 m/s (see Figure ). Determine the maximum height it will travel if atmospheric resistance is measured as [latex]{F}_{\text{D}}=(0.0100{v}^{2})\,\text{N,}[/latex] where v is the speed at any instant.

The known force on the mortar shell can be related to its acceleration using the equations of motion. Kinematics can then be used to relate the mortar shell’s acceleration to its position.

Initially, [latex]{y}_{0}=0[/latex] and [latex]{v}_{0}=50.0\,\text{m/s}\text{.}[/latex] At the maximum height [latex]y=h,v=0.[/latex] The free-body diagram shows [latex]{F}_{\text{D}}[/latex] to act downward, because it slows the upward motion of the mortar shell. Thus, we can write

The acceleration depends on v and is therefore variable. Since [latex]a=f(v)\text{,}[/latex] we can relate a to v using the rearrangement described above,

We replace ds with dy because we are dealing with the vertical direction,

We now separate the variables ( v ’s and dv ’s on one side; dy on the other):

Thus, [latex]h=114\,\text{m}\text{.}[/latex]

Notice the need to apply calculus since the force is not constant, which also means that acceleration is not constant. To make matters worse, the force depends on v (not t ), and so we must use the trick explained prior to the example. The answer for the height indicates a lower elevation if there were air resistance. We will deal with the effects of air resistance and other drag forces in greater detail in Drag Force and Terminal Speed .

If atmospheric resistance is neglected, find the maximum height for the mortar shell. Is calculus required for this solution?

Explore the forces at work in this simulation when you try to push a filing cabinet. Create an applied force and see the resulting frictional force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body diagram of all the forces (including gravitational and normal forces).

• Newton’s laws of motion can be applied in numerous situations to solve motion problems.
• Some problems contain multiple force vectors acting in different directions on an object. Be sure to draw diagrams, resolve all force vectors into horizontal and vertical components, and draw a free-body diagram. Always analyze the direction in which an object accelerates so that you can determine whether [latex]{F}_{\text{net}}=ma[/latex] or [latex]{F}_{\text{net}}=0.[/latex]
• The normal force on an object is not always equal in magnitude to the weight of the object. If an object is accelerating vertically, the normal force is less than or greater than the weight of the object. Also, if the object is on an inclined plane, the normal force is always less than the full weight of the object.
• Some problems contain several physical quantities, such as forces, acceleration, velocity, or position. You can apply concepts from kinematics and dynamics to solve these problems.

Conceptual Questions

To simulate the apparent weightlessness of space orbit, astronauts are trained in the hold of a cargo aircraft that is accelerating downward at g . Why do they appear to be weightless, as measured by standing on a bathroom scale, in this accelerated frame of reference? Is there any difference between their apparent weightlessness in orbit and in the aircraft?

The scale is in free fall along with the astronauts, so the reading on the scale would be 0. There is no difference in the apparent weightlessness; in the aircraft and in orbit, free fall is occurring.

A 30.0-kg girl in a swing is pushed to one side and held at rest by a horizontal force [latex]\mathbf{\overset{\to }{F}}[/latex] so that the swing ropes are [latex]30.0^\circ[/latex] with respect to the vertical. (a) Calculate the tension in each of the two ropes supporting the swing under these conditions. (b) Calculate the magnitude of [latex]\mathbf{\overset{\to }{F}}.[/latex]

a. 170 N; b. 170 N

Find the tension in each of the three cables supporting the traffic light if it weighs 2.00 × 10 2 N.

Three forces act on an object, considered to be a particle, which moves with constant velocity [latex]v=(3\mathbf{\hat{i}}-2\mathbf{\hat{j}})\,\text{m/s}\text{.}[/latex] Two of the forces are [latex]{\mathbf{\overset{\to }{F}}}_{1}=(3\mathbf{\hat{i}}+5\mathbf{\hat{j}}-6\mathbf{\hat{k}})\,\text{N}[/latex] and [latex]{\mathbf{\overset{\to }{F}}}_{2}=(4\mathbf{\hat{i}}-7\mathbf{\hat{j}}+2\mathbf{\hat{k}})\,\text{N}\text{.}[/latex] Find the third force.

[latex]{\mathbf{\overset{\to }{F}}}_{3}=(-7\mathbf{\hat{i}}+2\mathbf{\hat{j}}+4\mathbf{\hat{k}})\,\text{N}[/latex]

A flea jumps by exerting a force of [latex]1.20\times {10}^{-5}\,\text{N}[/latex] straight down on the ground. A breeze blowing on the flea parallel to the ground exerts a force of [latex]0.500\times {10}^{-6}\,\text{N}[/latex] on the flea while the flea is still in contact with the ground. Find the direction and magnitude of the acceleration of the flea if its mass is [latex]6.00\times {10}^{-7}\,\text{kg}[/latex]. Do not neglect the gravitational force.

Two muscles in the back of the leg pull upward on the Achilles tendon, as shown below. (These muscles are called the medial and lateral heads of the gastrocnemius muscle.) Find the magnitude and direction of the total force on the Achilles tendon. What type of movement could be caused by this force?

After a mishap, a 76.0-kg circus performer clings to a trapeze, which is being pulled to the side by another circus artist, as shown here. Calculate the tension in the two ropes if the person is momentarily motionless. Include a free-body diagram in your solution.

A 35.0-kg dolphin decelerates from 12.0 to 7.50 m/s in 2.30 s to join another dolphin in play. What average force was exerted to slow the first dolphin if it was moving horizontally? (The gravitational force is balanced by the buoyant force of the water.)

When starting a foot race, a 70.0-kg sprinter exerts an average force of 650 N backward on the ground for 0.800 s. (a) What is his final speed? (b) How far does he travel?

A large rocket has a mass of [latex]2.00\times {10}^{6}\,\text{kg}[/latex] at takeoff, and its engines produce a thrust of [latex]3.50\times {10}^{7}\,\text{N}.[/latex] (a) Find its initial acceleration if it takes off vertically. (b) How long does it take to reach a velocity of 120 km/h straight up, assuming constant mass and thrust?

a. [latex]7.70\,{\text{m/s}}^{2}[/latex]; b. 4.33 s

A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.900 m above the floor. (a) Calculate his velocity when he leaves the floor. (b) Calculate his acceleration while he is straightening his legs. He goes from zero to the velocity found in (a) in a distance of 0.300 m. (c) Calculate the force he exerts on the floor to do this, given that his mass is 110.0 kg.

A 2.50-kg fireworks shell is fired straight up from a mortar and reaches a height of 110.0 m. (a) Neglecting air resistance (a poor assumption, but we will make it for this example), calculate the shell’s velocity when it leaves the mortar. (b) The mortar itself is a tube 0.450 m long. Calculate the average acceleration of the shell in the tube as it goes from zero to the velocity found in (a). (c) What is the average force on the shell in the mortar? Express your answer in newtons and as a ratio to the weight of the shell.

a. 46.4 m/s; b. [latex]2.40\times {10}^{3}\,{\text{m/s}}^{2}\text{;}[/latex] c. 5.99 × 10 3 N; ratio of 245

A 0.500-kg potato is fired at an angle of [latex]80.0^\circ[/latex] above the horizontal from a PVC pipe used as a “potato gun” and reaches a height of 110.0 m. (a) Neglecting air resistance, calculate the potato’s velocity when it leaves the gun. (b) The gun itself is a tube 0.450 m long. Calculate the average acceleration of the potato in the tube as it goes from zero to the velocity found in (a). (c) What is the average force on the potato in the gun? Express your answer in newtons and as a ratio to the weight of the potato.

An elevator filled with passengers has a mass of [latex]1.70\times {10}^{3}\,\text{kg}[/latex]. (a) The elevator accelerates upward from rest at a rate of [latex]1.20\,{\text{m/s}}^{2}[/latex] for 1.50 s. Calculate the tension in the cable supporting the elevator. (b) The elevator continues upward at constant velocity for 8.50 s. What is the tension in the cable during this time? (c) The elevator decelerates at a rate of [latex]0.600\,{\text{m/s}}^{2}[/latex] for 3.00 s. What is the tension in the cable during deceleration? (d) How high has the elevator moved above its original starting point, and what is its final velocity?

a. [latex]1.87\times {10}^{4}\,\text{N;}[/latex] b. [latex]1.67\times {10}^{4}\,\text{N;}[/latex] c. [latex]1.56\times {10}^{4}\,\text{N;}[/latex] d. 19.4 m, 0 m/s

A 20.0-g ball hangs from the roof of a freight car by a string. When the freight car begins to move, the string makes an angle of [latex]35.0^\circ[/latex] with the vertical. (a) What is the acceleration of the freight car? (b) What is the tension in the string?

A student’s backpack, full of textbooks, is hung from a spring scale attached to the ceiling of an elevator. When the elevator is accelerating downward at [latex]3.8\,{\text{m/s}}^{2}[/latex], the scale reads 60 N. (a) What is the mass of the backpack? (b) What does the scale read if the elevator moves upward while slowing down at a rate [latex]3.8\,{\text{m/s}}^{2}[/latex]? (c) What does the scale read if the elevator moves upward at constant velocity? (d) If the elevator had no brakes and the cable supporting it were to break loose so that the elevator could fall freely, what would the spring scale read?

a. 10 kg; b. 90 N; c. 98 N; d. 0

A service elevator takes a load of garbage, mass 10.0 kg, from a floor of a skyscraper under construction, down to ground level, accelerating downward at a rate of [latex]1.2\,{\text{m/s}}^{2}[/latex]. Find the magnitude of the force the garbage exerts on the floor of the service elevator?

A roller coaster car starts from rest at the top of a track 30.0 m long and inclined at [latex]20.0^\circ[/latex] to the horizontal. Assume that friction can be ignored. (a) What is the acceleration of the car? (b) How much time elapses before it reaches the bottom of the track?

a. [latex]3.35\,{\text{m/s}}^{2}[/latex]; b. 4.2 s

The device shown below is the Atwood’s machine considered in Figure . Assuming that the masses of the string and the frictionless pulley are negligible, (a) find an equation for the acceleration of the two blocks; (b) find an equation for the tension in the string; and (c) find both the acceleration and tension when block 1 has mass 2.00 kg and block 2 has mass 4.00 kg.

Two blocks are connected by a massless rope as shown below. The mass of the block on the table is 4.0 kg and the hanging mass is 1.0 kg. The table and the pulley are frictionless. (a) Find the acceleration of the system. (b) Find the tension in the rope. (c) Find the speed with which the hanging mass hits the floor if it starts from rest and is initially located 1.0 m from the floor.

Shown below are two carts connected by a cord that passes over a small frictionless pulley. Each cart rolls freely with negligible friction. Calculate the acceleration of the carts and the tension in the cord.

A 2.00 kg block (mass 1) and a 4.00 kg block (mass 2) are connected by a light string as shown; the inclination of the ramp is [latex]40.0^\circ[/latex]. Friction is negligible. What is (a) the acceleration of each block and (b) the tension in the string?

6.1 Solving Problems with Newton’s Laws Copyright © 2016 by OpenStax. All Rights Reserved.

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28 Newton’s First Law of Motion: Inertia

[latexpage]

Learning Objectives

• Define mass and inertia.
• Understand Newton’s first law of motion.

Experience suggests that an object at rest will remain at rest if left alone, and that an object in motion tends to slow down and stop unless some effort is made to keep it moving. What Newton’s first law of motion states, however, is the following:

A body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless acted on by a net external force.

Note the repeated use of the verb “remains.” We can think of this law as preserving the status quo of motion.

Rather than contradicting our experience, Newton’s first law of motion states that there must be a cause (which is a net external force) for there to be any change in velocity (either a change in magnitude or direction) . We will define net external force in the next section. An object sliding across a table or floor slows down due to the net force of friction acting on the object. If friction disappeared, would the object still slow down?

The idea of cause and effect is crucial in accurately describing what happens in various situations. For example, consider what happens to an object sliding along a rough horizontal surface. The object quickly grinds to a halt. If we spray the surface with talcum powder to make the surface smoother, the object slides farther. If we make the surface even smoother by rubbing lubricating oil on it, the object slides farther yet. Extrapolating to a frictionless surface, we can imagine the object sliding in a straight line indefinitely. Friction is thus the cause of the slowing (consistent with Newton’s first law). The object would not slow down at all if friction were completely eliminated. Consider an air hockey table. When the air is turned off, the puck slides only a short distance before friction slows it to a stop. However, when the air is turned on, it creates a nearly frictionless surface, and the puck glides long distances without slowing down. Additionally, if we know enough about the friction, we can accurately predict how quickly the object will slow down. Friction is an external force.

Newton’s first law is completely general and can be applied to anything from an object sliding on a table to a satellite in orbit to blood pumped from the heart. Experiments have thoroughly verified that any change in velocity (speed or direction) must be caused by an external force. The idea of generally applicable or universal laws is important not only here—it is a basic feature of all laws of physics. Identifying these laws is like recognizing patterns in nature from which further patterns can be discovered. The genius of Galileo, who first developed the idea for the first law, and Newton, who clarified it, was to ask the fundamental question, “What is the cause?” Thinking in terms of cause and effect is a worldview fundamentally different from the typical ancient Greek approach when questions such as “Why does a tiger have stripes?” would have been answered in Aristotelian fashion, “That is the nature of the beast.” True perhaps, but not a useful insight.

The property of a body to remain at rest or to remain in motion with constant velocity is called inertia . Newton’s first law is often called the law of inertia . As we know from experience, some objects have more inertia than others. It is obviously more difficult to change the motion of a large boulder than that of a basketball, for example. The inertia of an object is measured by its mass . Roughly speaking, mass is a measure of the amount of “stuff” (or matter) in something. The quantity or amount of matter in an object is determined by the numbers of atoms and molecules of various types it contains. Unlike weight, mass does not vary with location. The mass of an object is the same on Earth, in orbit, or on the surface of the Moon. In practice, it is very difficult to count and identify all of the atoms and molecules in an object, so masses are not often determined in this manner. Operationally, the masses of objects are determined by comparison with the standard kilogram.

Which has more mass: a kilogram of cotton balls or a kilogram of gold?

They are equal. A kilogram of one substance is equal in mass to a kilogram of another substance. The quantities that might differ between them are volume and density.

Section Summary

• Newton’s first law of motion states that a body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless acted on by a net external force. This is also known as the law of inertia .
• Inertia is the tendency of an object to remain at rest or remain in motion. Inertia is related to an object’s mass.
• Mass is the quantity of matter in a substance.

Conceptual Questions

How are inertia and mass related?

What is the relationship between weight and mass? Which is an intrinsic, unchanging property of a body?

Intro to Physics for Non-Majors Copyright © 2012 by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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Newton's First Law

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Mission NL1: Inertia and Newton's First Law

Mission NL1 pertains to the concepts of inertia, mass, and Newton's First Law of motion. The mission consists of 38 questions organized into 10 Question Groups. You must answer one question from each Question Group to complete the mission. The learning outcomes for this mission are ...  

Learning Outcomes

• The student should understand the significance of Newton's law of inertia by identifying and refuting classic misconceptions concerning the causes of motion.
• The student should recognize inertia as a property of an object which depends solely upon mass.

Launch Mission NL1

Getting help.

If you are not familiar with this topic, then you should first learn about the topic using our written Tutorial or our Video Tutorial: The Physics Classroom, Newton's Laws Unit, Lesson 1, Part a The Physics Classroom, Newton's Laws Unit, Lesson 1, Part b  

• What is inertia?
• How would an object move in the absence of an unbalanced force?
• What is meant by  balanced forces and  unbalanced forces ?
• What is meant by  the state of motion ?
• What is the law of inertia?
• Why does a sliding object eventually stop?
• What variables effect the inertia of an object?
• How can one use mass and speed information about two objects to determine which object has the greatest inertia?

Video Tutorials:

Inertia and Mass

Question-Specific Help

Each Question Group has its own Help page with information specific to the question. You can access the Help page from within the mission by tapping on the Help Me! icon (textbook). For your convenience, links to those pages are provided below:

• Physics Concept Questions And Answers
• Law Of Inertia Questions

Law of Inertia Questions

Motion is a phenomenon through which a body changes its location over time. It is mathematically expressed in terms of displacement, distance, speed, velocity, time and acceleration. The movement of a body is analysed by fixing a reference frame to the observer and calculating the change in location of the body with respect to the frame with variation in time.

Physical systems are connected to motion: matter particles, radiation, matter fields, radiation, and space-time. In general, motion signifies a continuous variation in the configuration or positions of a physical body in space. Newton’s laws of motion describe the connection between forces exerting on a system and its motion. These laws are the basic principles that control classical mechanics. The Aristotelian theory was the widely accepted theory about motion in the Western world. His theory described the motion as follows: In the absence of outside power, all bodies would come to a stationary state (rest) where moving bodies stay only to move until force or power is inducing the bodies to do so. Even though it was widely accepted, this motion theory was opposed by many notable scholars.

Later, Galileo Galilei further developed the theory of inertia. He hypothesised that a falling body acquires an equal value of velocity in equal time intervals. This implies that the speed rises at a constant figure as it descends. However, there was a major problem in testing this theorem: It was not possible for him to observe the free-falling movement of the body. Also, the technology was too underdeveloped to record such extreme velocities. If a body is released from the state of rest and acquires speed at a consistent rate, then the entire distance covered by the body is proportional to the time squared required for that travel.

In classical mechanics, there is no clear distinction between consistent motion in a straight path. They can be regarded as the exact state of motion perceived by different persons, one travelling at the exact velocity of the body and the other travelling at a steady velocity relative to the body. This is called the law of inertia. It was first developed by Galileo Galilei for horizontal movement on the Earth. It was later generalised by René Descartes. Even though the law of inertia is the beginning point and the basic foundation of classical mechanics, it is not very intuitive to the untrained observer.

In Aristotelian conception and in normal experience, bodies that are not being thrust tend to approach the state of rest. The principle of inertia was developed by Galileo from his experiments with rolling balls with inclined planes. Inertia is the property of an object through which it resists any change in the nature of motion or state of rest. Inertia is a passive character and does not allow an object to indulge in anything except resist active agents like torques and forces. A moving object stays moving not due to its inertia. It is only due to the lack of force to damp it down, vary its trajectory, or speed it up.

The video explains the basic concept of motion

Important Law of Inertia Questions with Answers

1) What is motion?

Motion is a phenomenon through which a body changes its location over time. It is mathematically expressed in terms of displacement, distance, speed, velocity, time and acceleration.

2) The branch that deals with the motion of bodies without concerning its cause is called ______.

Answer: kinematics

Explanation: The branch that deals with the motion of bodies without concerning its cause is called kinematics.

3) The branch that examines forces and their impacts on motion is called ______.

Answer: dynamics

Explanation: The branch that examines forces and their impacts on motion is called dynamics.

4) Why is it impossible to determine absolute motion?

Suppose a body is not changing with respect to a given reference frame. In that case, the body is considered to be at rest, immobile, stationary, or possesses a time-invariant or constant position with respect to its surroundings. Since there is no absolute reference frame, it is not possible to determine the absolute motion. Thus, everything in this cosmos can be regarded to be in motion.

5) What is inertia?

Inertia is the property of an object through which it resists any change in the nature of motion or state of rest. Inertia is a passive character and does not allow an object to indulge in anything except resist active agents like torques and forces. A moving object stays moving not due to its inertia. It is only due to the lack of force to damp it down, vary its trajectory, or speed it up.

6) Who developed the theory of inertia?

Galileo Galilei developed the theory of inertia. He hypothesised that a falling body acquires an equal value of velocity in equal time intervals. This implies that the speed rises at a constant figure as it descends.

7) How did Galileo measure the speed of the falling object?

In order to counter the technological disadvantage he had, Galileo tried to slow down the body’s motion by replacing the descending body with a ball rolling along an inclined plane. As free-falling motion is fundamentally equivalent to a fully vertical ramp, Galileo thought that a ball rolling along a ramp would accelerate in the exact way as a descending ball would. Using a water clock, he calculated the time consumed for the ball to reach a particular distance down the slanted plane. After many trials, he observed that the time consumed by the ball to roll the complete length of the ramp was equivalent to double the time consumed by the same ball to roll a quarter of the length. Through the test, he concluded that if a body is once in a state of motion, it moves with a uniform velocity if no external force acts on it.

8) Define Newton’s first law of motion.

Newton’s first law of motion states if an object is at the state of rest or in motion at a constant rate in a straight line, it will stay at rest or continue moving in a straight path at sustained speed unless it is disturbed by an external force.

9) What are the two quantitative measures of the inertia of an object?

There are fundamentally two quantitative measures of the inertia of an object: mass and moment of inertia. Mass control its resistance to the exertion of a force on the object. Moment of inertia along a particular axis measures the object’s resistance to the exertion of torque along the same axis. In general usage, inertia may refer to a body’s measure of resistance to change in speed or velocity. In simple terms, it is the opposition to a change in the state of motion (sometimes to the body’s momentum).

The law of inertia is one of the basic principles in classical mechanics that are still employed to explain the motion of bodies and how they are impacted by the applied forces on them.

10) What misled Aristotle to conclude that bodies would be in motion only if continuous force was exerted on them?

On the Earth’s surface, inertia is usually masked by the force of gravity and the impacts of air resistance and friction, both of which tend to reduce the speed of moving bodies (typically to the state of rest).

11) What are the main types of inertia?

The inertia of direction, the inertia of rest, and the inertia of motion are the main types of inertia.

12) What is meant by the inertia of rest?

A body continues to stay where it is positioned, and it will continue to be there until an external force moves it.

13) What is meant by the inertia of motion?

A body will continue to move at the same velocity until an external force is applied to it. For example, an object in a moving car going forward when the car stops suddenly.

14) What is meant by the inertia of direction?

A body will continue to move in the same direction until an outside force is applied to it.

15) Explain some common scenarios of inertia of rest.

When a vehicle is suddenly accelerated, passengers might feel as though they are moving backwards. Inertia forces the passengers to continue in place as the vehicle travels forward.

If a stationary vehicle is collided with a moving vehicle from behind, the driver may experience whiplash. This is the result of the driver’s body travelling in the forward direction but the driver’s head lagging behind. Here, the inertia is experienced by the head.

A balloon in a vehicle will seem to move when the vehicle travels forward. However, in reality, the balloon is trying to stay in the position it was in earlier. It is only the vehicle that is moving forward.

A table cloth can be taken out from underneath the plates and utensils if it is pulled quickly. The plates and utensils have the inherent tendency to stay still as long as the motion of the tablecloth is not too overpowering.

Practice Questions

1) What is Aristotle’s fallacy?

2) Explain Galileo’s fall experiment.

3) Give one scenario where the law of inertia is proven.

4) Which Newton’s law is connected to the law of inertia?

5) Which type of clock was used by Galileo to measure the time during the fall experiment?

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4.2 Newton’s First Law of Motion: Inertia

Learning objectives.

By the end of this section, you will be able to:

• Define mass and inertia.
• Understand Newton's first law of motion.

Experience suggests that an object at rest will remain at rest if left alone, and that an object in motion tends to slow down and stop unless some effort is made to keep it moving. What Newton’s first law of motion states, however, is the following:

Newton’s First Law of Motion

A body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless acted on by a net external force.

Note the repeated use of the verb “remains.” We can think of this law as preserving the status quo of motion.

Rather than contradicting our experience, Newton’s first law of motion states that there must be a cause (which is a net external force) for there to be any change in velocity (either a change in magnitude or direction) . We will define net external force in the next section. An object sliding across a table or floor slows down due to the net force of friction acting on the object. If friction disappeared, would the object still slow down?

The idea of cause and effect is crucial in accurately describing what happens in various situations. For example, consider what happens to an object sliding along a rough horizontal surface. The object quickly grinds to a halt. If we spray the surface with talcum powder to make the surface smoother, the object slides farther. If we make the surface even smoother by rubbing lubricating oil on it, the object slides farther yet. Extrapolating to a frictionless surface, we can imagine the object sliding in a straight line indefinitely. Friction is thus the cause of the slowing (consistent with Newton’s first law). The object would not slow down at all if friction were completely eliminated. Consider an air hockey table. When the air is turned off, the puck slides only a short distance before friction slows it to a stop. However, when the air is turned on, it creates a nearly frictionless surface, and the puck glides long distances without slowing down. Additionally, if we know enough about the friction, we can accurately predict how quickly the object will slow down. Friction is an external force.

Newton’s first law is completely general and can be applied to anything from an object sliding on a table to a satellite in orbit to blood pumped from the heart. Experiments have thoroughly verified that any change in velocity (speed or direction) must be caused by an external force. The idea of generally applicable or universal laws is important not only here—it is a basic feature of all laws of physics. Identifying these laws is like recognizing patterns in nature from which further patterns can be discovered. The genius of Galileo, who first developed the idea for the first law, and Newton, who clarified it, was to ask the fundamental question, “What is the cause?” Thinking in terms of cause and effect is a worldview fundamentally different from the typical ancient Greek approach when questions such as “Why does a tiger have stripes?” would have been answered in Aristotelian fashion, “That is the nature of the beast.” True perhaps, but not a useful insight.

The property of a body to remain at rest or to remain in motion with constant velocity is called inertia . Newton’s first law is often called the law of inertia . As we know from experience, some objects have more inertia than others. It is obviously more difficult to change the motion of a large boulder than that of a basketball, for example. The inertia of an object is measured by its mass . Roughly speaking, mass is a measure of the amount of “stuff” (or matter) in something. The quantity or amount of matter in an object is determined by the numbers of atoms and molecules of various types it contains. Unlike weight, mass does not vary with location. The mass of an object is the same on Earth, in orbit, or on the surface of the Moon. In practice, it is very difficult to count and identify all of the atoms and molecules in an object, so masses are not often determined in this manner. Operationally, the masses of objects are determined by comparison with the standard kilogram.

Check Your Understanding

Which has more mass: a kilogram of cotton balls or a kilogram of gold?

They are equal. A kilogram of one substance is equal in mass to a kilogram of another substance. The quantities that might differ between them are volume and density.

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Applications of Newton’s Laws

Solving Problems with Newton’s Laws

Learning objectives.

By the end of the section, you will be able to:

• Apply problem-solving techniques to solve for quantities in more complex systems of forces
• Use concepts from kinematics to solve problems using Newton’s laws of motion
• Solve more complex equilibrium problems
• Solve more complex acceleration problems
• Apply calculus to more advanced dynamics problems

Success in problem solving is necessary to understand and apply physical principles. We developed a pattern of analyzing and setting up the solutions to problems involving Newton’s laws in Newton’s Laws of Motion ; in this chapter, we continue to discuss these strategies and apply a step-by-step process.

Problem-Solving Strategies

We follow here the basics of problem solving presented earlier in this text, but we emphasize specific strategies that are useful in applying Newton’s laws of motion . Once you identify the physical principles involved in the problem and determine that they include Newton’s laws of motion, you can apply these steps to find a solution. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, so the following techniques should reinforce skills you have already begun to develop.

• Identify the physical principles involved by listing the givens and the quantities to be calculated.
• Sketch the situation, using arrows to represent all forces.
• Determine the system of interest. The result is a free-body diagram that is essential to solving the problem.
• Apply Newton’s second law to solve the problem. If necessary, apply appropriate kinematic equations from the chapter on motion along a straight line.
• Check the solution to see whether it is reasonable.

Let’s apply this problem-solving strategy to the challenge of lifting a grand piano into a second-story apartment. Once we have determined that Newton’s laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in (Figure) (a). Then, as in (Figure) (b), we can represent all forces with arrows. Whenever sufficient information exists, it is best to label these arrows carefully and make the length and direction of each correspond to the represented force.

As with most problems, we next need to identify what needs to be determined and what is known or can be inferred from the problem as stated, that is, make a list of knowns and unknowns. It is particularly crucial to identify the system of interest, since Newton’s second law involves only external forces. We can then determine which forces are external and which are internal, a necessary step to employ Newton’s second law. (See (Figure) (c).) Newton’s third law may be used to identify whether forces are exerted between components of a system (internal) or between the system and something outside (external). As illustrated in Newton’s Laws of Motion , the system of interest depends on the question we need to answer. Only forces are shown in free-body diagrams, not acceleration or velocity. We have drawn several free-body diagrams in previous worked examples. (Figure) (c) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram.

Once a free-body diagram is drawn, we apply Newton’s second law. This is done in (Figure) (d) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—that is, if all forces are parallel—then the forces can be handled algebraically. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. We do this by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known. Generally, just write Newton’s second law in components along the different directions. Then, you have the following equations:

As always, we must check the solution. In some cases, it is easy to tell whether the solution is reasonable. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving; with experience, it becomes progressively easier to judge whether an answer is reasonable. Another way to check a solution is to check the units. If we are solving for force and end up with units of millimeters per second, then we have made a mistake.

There are many interesting applications of Newton’s laws of motion, a few more of which are presented in this section. These serve also to illustrate some further subtleties of physics and to help build problem-solving skills. We look first at problems involving particle equilibrium, which make use of Newton’s first law, and then consider particle acceleration, which involves Newton’s second law.

Particle Equilibrium

Recall that a particle in equilibrium is one for which the external forces are balanced. Static equilibrium involves objects at rest, and dynamic equilibrium involves objects in motion without acceleration, but it is important to remember that these conditions are relative. For example, an object may be at rest when viewed from our frame of reference, but the same object would appear to be in motion when viewed by someone moving at a constant velocity. We now make use of the knowledge attained in Newton’s Laws of Motion , regarding the different types of forces and the use of free-body diagrams, to solve additional problems in particle equilibrium .

Different Tensions at Different Angles Consider the traffic light (mass of 15.0 kg) suspended from two wires as shown in (Figure) . Find the tension in each wire, neglecting the masses of the wires.

Solution First consider the horizontal or x -axis:

Thus, as you might expect,

This gives us the following relationship:

Now consider the force components along the vertical or y -axis:

This implies

Substituting the expressions for the vertical components gives

which yields

Significance Both tensions would be larger if both wires were more horizontal, and they will be equal if and only if the angles on either side are the same (as they were in the earlier example of a tightrope walker in Newton’s Laws of Motion .

Particle Acceleration

We have given a variety of examples of particles in equilibrium. We now turn our attention to particle acceleration problems, which are the result of a nonzero net force. Refer again to the steps given at the beginning of this section, and notice how they are applied to the following examples.

The angle is given by

However, Newton’s second law states that

Substituting known values gives

In Newton’s Laws of Motion , we discussed the normal force , which is a contact force that acts normal to the surface so that an object does not have an acceleration perpendicular to the surface. The bathroom scale is an excellent example of a normal force acting on a body. It provides a quantitative reading of how much it must push upward to support the weight of an object. But can you predict what you would see on the dial of a bathroom scale if you stood on it during an elevator ride? Will you see a value greater than your weight when the elevator starts up? What about when the elevator moves upward at a constant speed? Take a guess before reading the next example.

which gives

Significance The scale reading in (Figure) (a) is about 185 lb. What would the scale have read if he were stationary? Since his acceleration would be zero, the force of the scale would be equal to his weight:

Thus, the scale reading in the elevator is greater than his 735-N (165-lb.) weight. This means that the scale is pushing up on the person with a force greater than his weight, as it must in order to accelerate him upward. Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators. In (Figure) (b), the scale reading is 735 N, which equals the person’s weight. This is the case whenever the elevator has a constant velocity—moving up, moving down, or stationary.

The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevator accelerates downward, a is negative, and the scale reading is less than the weight of the person. If a constant downward velocity is reached, the scale reading again becomes equal to the person’s weight. If the elevator is in free fall and accelerating downward at g , then the scale reading is zero and the person appears to be weightless.

Strategy We draw a free-body diagram for each mass separately, as shown in (Figure) . Then we analyze each one to find the required unknowns. The forces on block 1 are the gravitational force, the contact force of the surface, and the tension in the string. Block 2 is subjected to the gravitational force and the string tension. Newton’s second law applies to each, so we write two vector equations:

Solution The component equations follow from the vector equations above. We see that block 1 has the vertical forces balanced, so we ignore them and write an equation relating the x -components. There are no horizontal forces on block 2, so only the y -equation is written. We obtain these results:

Solving for a :

Newton’s Laws of Motion and Kinematics

Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. Newton’s laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics , and hence the relevance of earlier chapters.

When approaching problems that involve various types of forces, acceleration, velocity, and/or position, listing the givens and the quantities to be calculated will allow you to identify the principles involved. Then, you can refer to the chapters that deal with a particular topic and solve the problem using strategies outlined in the text. The following worked example illustrates how the problem-solving strategy given earlier in this chapter, as well as strategies presented in other chapters, is applied to an integrated concept problem.

What Force Must a Soccer Player Exert to Reach Top Speed? A soccer player starts at rest and accelerates forward, reaching a velocity of 8.00 m/s in 2.50 s. (a) What is her average acceleration? (b) What average force does the ground exert forward on the runner so that she achieves this acceleration? The player’s mass is 70.0 kg, and air resistance is negligible.

Strategy To find the answers to this problem, we use the problem-solving strategy given earlier in this chapter. The solutions to each part of the example illustrate how to apply specific problem-solving steps. In this case, we do not need to use all of the steps. We simply identify the physical principles, and thus the knowns and unknowns; apply Newton’s second law; and check to see whether the answer is reasonable.

Substituting the known values yields

Substituting the known values of m and a gives

This is a reasonable result: The acceleration is attainable for an athlete in good condition. The force is about 50 pounds, a reasonable average force.

Significance This example illustrates how to apply problem-solving strategies to situations that include topics from different chapters. The first step is to identify the physical principles, the knowns, and the unknowns involved in the problem. The second step is to solve for the unknown, in this case using Newton’s second law. Finally, we check our answer to ensure it is reasonable. These techniques for integrated concept problems will be useful in applications of physics outside of a physics course, such as in your profession, in other science disciplines, and in everyday life.

Check Your Understanding The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the ball?

Solution We have

The magnitude of the force is now easily found:

Check Your Understanding Find the direction of the resultant for the 1.50-kg model helicopter.

49.4 degrees

Significance Since the force varies with time, we must use calculus to solve this problem. Notice how the total mass of the system was important in solving (Figure) (a), whereas only the mass of the truck (since it supplied the force) was of use in (Figure) (b).

Strategy The known force on the mortar shell can be related to its acceleration using the equations of motion. Kinematics can then be used to relate the mortar shell’s acceleration to its position.

We replace ds with dy because we are dealing with the vertical direction,

We now separate the variables ( v ’s and dv ’s on one side; dy on the other):

Significance Notice the need to apply calculus since the force is not constant, which also means that acceleration is not constant. To make matters worse, the force depends on v (not t ), and so we must use the trick explained prior to the example. The answer for the height indicates a lower elevation if there were air resistance. We will deal with the effects of air resistance and other drag forces in greater detail in Drag Force and Terminal Speed .

Check Your Understanding If atmospheric resistance is neglected, find the maximum height for the mortar shell. Is calculus required for this solution?

Explore the forces at work in this simulation when you try to push a filing cabinet. Create an applied force and see the resulting frictional force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body diagram of all the forces (including gravitational and normal forces).

• Newton’s laws of motion can be applied in numerous situations to solve motion problems.

• The normal force on an object is not always equal in magnitude to the weight of the object. If an object is accelerating vertically, the normal force is less than or greater than the weight of the object. Also, if the object is on an inclined plane, the normal force is always less than the full weight of the object.
• Some problems contain several physical quantities, such as forces, acceleration, velocity, or position. You can apply concepts from kinematics and dynamics to solve these problems.

Conceptual Questions

To simulate the apparent weightlessness of space orbit, astronauts are trained in the hold of a cargo aircraft that is accelerating downward at g . Why do they appear to be weightless, as measured by standing on a bathroom scale, in this accelerated frame of reference? Is there any difference between their apparent weightlessness in orbit and in the aircraft?

The scale is in free fall along with the astronauts, so the reading on the scale would be 0. There is no difference in the apparent weightlessness; in the aircraft and in orbit, free fall is occurring.

a. 170 N; b. 170 N

Find the tension in each of the three cables supporting the traffic light if it weighs 2.00 × 10 2 N.

Two muscles in the back of the leg pull upward on the Achilles tendon, as shown below. (These muscles are called the medial and lateral heads of the gastrocnemius muscle.) Find the magnitude and direction of the total force on the Achilles tendon. What type of movement could be caused by this force?

376 N pointing up (along the dashed line in the figure); the force is used to raise the heel of the foot.

After a mishap, a 76.0-kg circus performer clings to a trapeze, which is being pulled to the side by another circus artist, as shown here. Calculate the tension in the two ropes if the person is momentarily motionless. Include a free-body diagram in your solution.

A 35.0-kg dolphin decelerates from 12.0 to 7.50 m/s in 2.30 s to join another dolphin in play. What average force was exerted to slow the first dolphin if it was moving horizontally? (The gravitational force is balanced by the buoyant force of the water.)

When starting a foot race, a 70.0-kg sprinter exerts an average force of 650 N backward on the ground for 0.800 s. (a) What is his final speed? (b) How far does he travel?

A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.900 m above the floor. (a) Calculate his velocity when he leaves the floor. (b) Calculate his acceleration while he is straightening his legs. He goes from zero to the velocity found in (a) in a distance of 0.300 m. (c) Calculate the force he exerts on the floor to do this, given that his mass is 110.0 kg.

A 2.50-kg fireworks shell is fired straight up from a mortar and reaches a height of 110.0 m. (a) Neglecting air resistance (a poor assumption, but we will make it for this example), calculate the shell’s velocity when it leaves the mortar. (b) The mortar itself is a tube 0.450 m long. Calculate the average acceleration of the shell in the tube as it goes from zero to the velocity found in (a). (c) What is the average force on the shell in the mortar? Express your answer in newtons and as a ratio to the weight of the shell.

a. 10 kg; b. 90 N; c. 98 N; d. 0

The device shown below is the Atwood’s machine considered in (Figure) . Assuming that the masses of the string and the frictionless pulley are negligible, (a) find an equation for the acceleration of the two blocks; (b) find an equation for the tension in the string; and (c) find both the acceleration and tension when block 1 has mass 2.00 kg and block 2 has mass 4.00 kg.

Two blocks are connected by a massless rope as shown below. The mass of the block on the table is 4.0 kg and the hanging mass is 1.0 kg. The table and the pulley are frictionless. (a) Find the acceleration of the system. (b) Find the tension in the rope. (c) Find the speed with which the hanging mass hits the floor if it starts from rest and is initially located 1.0 m from the floor.

Shown below are two carts connected by a cord that passes over a small frictionless pulley. Each cart rolls freely with negligible friction. Calculate the acceleration of the carts and the tension in the cord.

Solving Problems with Newton’s Laws Copyright © 2016 by cnxuniphysics. All Rights Reserved.

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Newton’s first law of motion – problems and solutions

1. A person is in an elevator that moving upward at a constant velocity . The weight of the person is 800 N. Immediately the elevator rope is broke, so the elevator falls. Determine the normal force acted by elevator’s floor to the person just before and after the elevator’s rope broke.

A. 800 N and 0

B. 800 N and 800 N

C. 1600 N and 0

D. 1600 N and 800 N

Weight (w) = 800 Newton

Wanted: The normal force (N)

Before the elevator’s rope broke

When the person stands on the floor of the elevator, weight acts on the person where the direction of the person is downward. That person at rest so that there must a normal force acts on the person, where the direction of the normal force is upward and the magnitude of the normal force same as the magnitude of the weight.

N = 800 Newton

After the elevator’s rope broke

After the elevator’s rope broke, the elevator and the person free fall together, where the magnitude and the direction of their acceleration same as acceleration due to gravity. There is no normal force on the person.

The correct answer is A.

2. A block with a mass of 20 gram moves at a constant velocity on a rough horizontal floor at a constant velocity if there is an external force of 2 N acts on the block. Determine the magnitude of the friction force experienced by the block.

Mass (m) = 20 gram

Force (F) = 2 Newton

Wanted: Magnitude of friction force experienced by the block.

– The magnitude of friction force (F fric ) same as the magnitude of the external force (F)

– The friction force (F fric ) has opposite direction with the external force (F)

Apply Newton’s first law of motion :

F – F fric = 0

F fric = 2 Newton

The correct answer is C.

3. A smooth inclined plane with the length of 0.6 m and height of 0.4 m. A block with the weight of, 1350 N will move upward using the inclined plane. Determine the magnitude of force need to move the block.

Weight of block (w) = 1350 Newton

hyp = 0.6 m

opp = 0.4 m

Wanted : The minimum force

opp = bc = 0.4 m

Sin θ = bc / ac = 0.4 / 0.6 = 4/6 = 2/3

Based on Newton’s first law of motion, the block start to moves upward then the external force (F) minimal same as the horizontal component of weight (w x ).

F – w x = 0

If F = w x , then object start to moving upward at constant velocity.

w x = w sin θ = (1350)(2/3) = (2)(450) = 900 Newton

The correct answer is D.

4. Three forces, F 1 = 22 N, F 2 = 18 N and F 3 = 40 N act on a block. Which figure describes Newton’s first law.

Newton’s first law : Net force (ΣF) = 0.

A. F 1 + F 2 – F 3 = 22 N + 18 N – 40 N = 40 N – 40 N = 0

B. F 2 + F 3 – F 1 = 18 N + 40 N – 22 N = 58 N – 22 N = 36 N (rightward)

C. F 2 + F 3 – F 1 = 18 N + 40 N – 22 N = 58 N – 22 N = 36 N (rightward)

D. F 1 + F 3 – F 2 = 22 N + 40 N – 18 N = 62 N – 18 N = 44 N (leftward)

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• Solving Law Firm’s Biggest Problem: Outsourcing Legal Billing to Improve Collection and Profitability

Streamlining and improving the billing process in law firms is often perceived as a daunting task, fraught with complexities that can challenge even the most seasoned legal professionals. Billing is a time consuming, labor-intensive necessary evil that is often fraught with inaccuracies due to inefficiencies in process and extensive operational silos which can cost a law firm potentially millions of dollars each year.

According to a recent Thomson Reuters survey , the average law firm experiences an 18% realization loss due to billing challenges. This means that nearly one-fifth of billable work is not converted into revenue, negatively impacting the firm’s profitability. 81% of law firms report issues with a significant portion of invoices remaining unpaid or delayed, creating cash flow challenges. This is a huge revenue loss that is creating major problems for law firms, but there are solutions.

Why is legal billing so complex?

Most unique to attorneys is that their work is often time-consuming difficult to tangibly track. Tracking billable hours accurately requires meticulous attention to detail and can be particularly challenging when dealing with tasks that don't neatly fit into predefined categories. This creates problems because clients often have unique billing preferences and expectations ranging from traditional hourly billing to flat fees, contingency fees and/or blended rates. Accommodating these distinctive preferences for each client and integrating them into a billing system adds a layer of complexity to billing that makes it extremely labor intensive. Additionally, silos that exist within law firms in the form of practice groups or other operational units can make regularity in billing guidelines either difficult to understand or completely nonexistent.

Another complexity in legal billing is the regulatory compliance and ethical guidelines that must be adhered to. Some of these issues include fee agreements, billing transparency, client confidentiality, and conflicts of interest. Staying compliant with these regulations while meeting client expectations and maintaining profitability requires a thorough understanding of legal and professional standards.

While most law firms rely on billing software to streamline the process, integrating these systems with other practice management tools and ensuring data accuracy and security can present technical challenges that require ongoing maintenance and support.

Reasons to outsource billing

By streamlining the billing processes , law firms can reduce the risk of errors, and focus on core competencies. Technical expertise and best practices ensure process optimization and service quality. Outsourcing can be a valuable solution for law firms aiming to stabilize their cash flow, reduce costs, and improve overall efficiency.

• Access to Advanced Technology: Outsourcing billing often provides access to advanced billing technologies and software platforms that may be cost-prohibitive for smaller law firms to implement independently. These technologies can automate billing processes, provide real-time reporting and analytics, and offer insights into billing trends and patterns that allow firms to optimize their billing practices for greater efficiency and profitability Additionally, predictive AI can provide billing staff with powerful tools to wade through complex billing guidelines in a timely manner, increasing productivity within billing departments.
• Expertise and Accuracy: Billing processes can be complex, especially in law firms where billing requirements may vary from client to client and case to case. Outsourcing billing to specialized professionals ensures that billing is handled by experts who are well-versed in legal billing practices. This expertise leads to greater accuracy in invoicing, reducing the likelihood of errors or discrepancies that could lead to disputes or delays in payment. This also leads to more standardized output.
• Centralization: Often, billing departments in law firms are segregated either geographically or by practice area, leading to inefficiencies and lack of scalability. Centralizing billing can assist by allowing billing to be looked at holistically as a firm rather than piecemeal, leading to a more effective business model for financial processes.
• Efficiency and Timeliness: Outsourcing billing allows law firms to streamline their billing processes and improve efficiency. Billing professionals have the necessary tools and systems in place to generate invoices promptly and ensure timely submission to clients. This helps expedite the payment cycle, leading to improved cash flow for the firm.
• Cost Savings: While some may initially view outsourcing as an additional expense, it often proves to be a cost-effective solution in the long run. By outsourcing billing, law firms eliminate the need to hire and train in-house billing staff, invest in billing software, and bear the overhead costs associated with maintaining billing infrastructure, lowering both soft and hard operational costs and simplifying a billing budget. Additionally, outsourcing billing allows firms to scale their billing operations up or down as needed without incurring significant fixed costs.
• Compliance and Regulation: Legal billing is subject to various regulations and compliance requirements, including those related to client confidentiality, billing transparency, and fee agreements. Outsourcing billing to professionals who are experts in these regulations helps ensure compliance and mitigates the risk of potential legal or ethical issues arising from billing practices.
• Focus on Core Competencies: Attorneys and legal staff excel in providing legal counsel and representation, but are not usually trained in managing billing and administrative tasks, and even if they are, legal staff turning their attention to these tasks from an administrative perspective harms the firm’s bottom line by taking time away from client-facing, billable projects. By outsourcing billing, law firms can free up valuable time and resources to concentrate on what they do best, which is serving clients legal needs.
• Enhanced Client Satisfaction: Timely and accurate billing contributes to a positive client experience. By outsourcing billing , law firms can ensure that clients receive clear, transparent invoices in a timely manner, enhancing client satisfaction and fostering long-term client relationships.

Legal billing is a challenge that costs firms millions of dollars every year. It has become such an issue that many of law firms have said this is their main priority to fix by 2025. Centralized teams face challenges due to the varied requirements of different attorneys and clients. The benefits to outsourcing billing not only will help realization rates but will foster a better relationship with clients and will help alleviate headaches of both lawyers and legal administrators as well.

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10.3: Dynamics of Rotational Motion - Rotational Inertia

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Learning Objectives

By the end of this section, you will be able to:

• Understand the relationship between force, mass and acceleration.
• Study the turning effect of force.
• Study the analogy between force and torque, mass and moment of inertia, and linear acceleration and angular acceleration.

If you have ever spun a bike wheel or pushed a merry-go-round, you know that force is needed to change angular velocity as seen in Figure 10.4.1. In fact, your intuition is reliable in predicting many of the factors that are involved. For example, we know that a door opens slowly if we push too close to its hinges. Furthermore, we know that the more massive the door, the more slowly it opens. The first example implies that the farther the force is applied from the pivot, the greater the angular acceleration; another implication is that angular acceleration is inversely proportional to mass. These relationships should seem very similar to the familiar relationships among force, mass, and acceleration embodied in Newton’s second law of motion. There are, in fact, precise rotational analogs to both force and mass.

To develop the precise relationship among force, mass, radius, and angular acceleration, consider what happens if we exert a force \(F\) on a point mass \(m\) that is at a distance \(r\) from a pivot point, as shown in Figure 10.4.2. Because the force is perpendicular to \(r\), an accelerationn\(a = frac{F}{m}\) is obtained in the direction of \(F\). We can rearrange this equation such that \(F = ma\) and then look for ways to relate this expression to expressions for rotational quantities. We note that \(a = r\alpha\), and we substitute this expression into \(F = ma\), yielding \[F = mr\alpha.\] Recall that torque is the turning effectiveness of a force. In this case, because \(F\) is perpendicular to \(r\), torque is simply \(\tau = Fr\). So, if we multiply both sides of the equation above by \(r\), we get torque on the left-hand side. That is, \[rF = mr^2\alpha\] or \[\tau = mr^2\alpha.\]

This last equation is the rotational analog of Newton’s second law \(F = ma\), where torque is analogous to force, angular acceleration is analogous to translational acceleration, and \(mr^2\) is analogous to mass (or inertia). The quantity \(mr^2\) is called the rotational inertia or moment of inertia of a point mass \(m\) a distance \(r\) from the center of rotation.

Making Connections: Rotational Motion Dynamics

Dynamics for rotational motion is completely analogous to linear or translational dynamics. Dynamics is concerned with force and mass and their effects on motion. For rotational motion, we will find direct analogs to force and mass that behave just as we would expect from our earlier experiences.

Rotational Inertia and Moment of Inertia

Before we can consider the rotation of anything other than a point mass like the one in Figure , we must extend the idea of rotational inertia to all types of objects. To expand our concept of rotational inertia, we define the moment of inertia \(I\) of an object to be the sum of \(mr^2\) for all the point masses of which it is composed. That is, \(I = \sum mr^2\). Here \(I\) is analogous to \(m\) in translational motion. Because of the distance \(r\), the moment of inertia for any object depends on the chosen axis. Actually, calculating \(I\) is beyond the scope of this text except for one simple case—that of a hoop, which has all its mass at the same distance from its axis. A hoop’s moment of inertia around its axis is therefore \(MR^2\), where \(M\) is its total mass and \(R\) its radius. (We use \(M\) and \(R\) for an entire object to distinguish them from \(m\) and \(r\) for point masses.) In all other cases, we must consult Figure 10.4.3 (note that the table is piece of artwork that has shapes as well as formulae) for formulas for \(I\) that have been derived from integration over the continuous body. Note that \(I\) has units of mass multiplied by distance squared \((kg \cdot m^2)\) as we might expect from its definition.

The general relationship among torque, moment of inertia, and angular acceleration is \[net \, \tau = I \alpha\] or \[\alpha = \dfrac{net \, \tau}{I},\] where net \(\tau\) is the total torque from all forces relative to a chosen axis. For simplicity, we will only consider torques exerted by forces in the plane of the rotation. Such torques are either positive or negative and add like ordinary numbers. The relationship in \(\tau = I\alpha\), \(\alpha = \frac{net \, \tau}{I}\) is the rotational analog to Newton’s second law and is very generally applicable. This equation is actually valid for any torque, applied to any object, relative to any axis.

As we might expect, the larger the torque is, the larger the angular acceleration is. For example, the harder a child pushes on a merry-go-round, the faster it accelerates. Furthermore, the more massive a merry-go-round, the slower it accelerates for the same torque. The basic relationship between moment of inertia and angular acceleration is that the larger the moment of inertia, the smaller is the angular acceleration. But there is an additional twist. The moment of inertia depends not only on the mass of an object, but also on its distribution of mass relative to the axis around which it rotates. For example, it will be much easier to accelerate a merry-go-round full of children if they stand close to its axis than if they all stand at the outer edge. The mass is the same in both cases; but the moment of inertia is much larger when the children are at the edge.

Take-Home Experiment

• Cut out a circle that has about a 10 cm radius from stiff cardboard. Near the edge of the circle, write numbers 1 to 12 like hours on a clock face. Position the circle so that it can rotate freely about a horizontal axis through its center, like a wheel. (You could loosely nail the circle to a wall.) Hold the circle stationary and with the number 12 positioned at the top, attach a lump of blue putty (sticky material used for fixing posters to walls) at the number 3. How large does the lump need to be to just rotate the circle? Describe how you can change the moment of inertia of the circle. How does this change affect the amount of blue putty needed at the number 3 to just rotate the circle? Change the circle’s moment of inertia and then try rotating the circle by using different amounts of blue putty. Repeat this process several times.

Problem-Solving Strategy for Rotational Dynamics

• Examine the situation to determine that torque and mass are involved in the rotation . Draw a careful sketch of the situation.
• Determine the system of interest .
• Draw a free body diagram . That is, draw and label all external forces acting on the system of interest.
• Apply \(net \, \tau = \alpha\), \(\alpha = \frac{net \, \tau}{I},\)
• the rotational equivalent of Newton’s second law, to solve the problem . Care must be taken to use the correct moment of inertia and to consider the torque about the point of rotation.
• As always, check the solution to see if it is reasonable .

Making Connections: Statics vs. Kinetics

In statics, the net torque is zero, and there is no angular acceleration. In rotational motion, net torque is the cause of angular acceleration, exactly as in Newton’s second law of motion for rotation.

Example \(\PageIndex{1}\): Calculating the Effect of Mass Distribution

Consider the father pushing a playground merry-go-round in Figure . He exerts a force of 250 N at the edge of the 50.0-kg merry-go-round, which has a 1.50 m radius. Calculate the angular acceleration produced (a) when no one is on the merry-go-round and (b) when an 18.0-kg child sits 1.25 m away from the center. Consider the merry-go-round itself to be a uniform disk with negligible retarding friction.

Angular acceleration is given directly by the expression \(\alpha = \frac{net \, \tau}{I}\) \[\alpha = \dfrac{\tau}{I}.\] To solve for \(\alpha\), we must first calculate the torque (which is the same in both cases) and moment of inertia \(I\) (which is greater in the second case). To find the torque, we note that the applied force is perpendicular to the radius and friction is negligible, so that\[\tau = rF \, sin \, \theta = (1.50 \, m)(250 \, N) = 375 \, N \cdot m.\]

Solution for (a)

The moment of inertia of a solid disk about this axis is given in Figure to be \[\dfrac{1}{2}MR^2,\] where \(M = 50.0 \, kg\) and \(R = 1.50 \, m\), so that

\[I = (0.500)(50.0 \, kg)(1.50 \, m)^2 = 56.25 \, kg \cdot m^2\]

Now, after we substitute the known values, we find the angular acceleration to be

\[\alpha = \dfrac{\tau}{I} = \dfrac{375 \, N \cdot m}{56.25 \, kg \cdot m^2} = 6.67 \dfrac{rad}{s^2}.\]

Solution for (b)

We expect the angular acceleration for the system to be less in this part, because the moment of inertia is greater when the child is on the merry-go-round. To find the total moment of inertia \(I\), we first find the child’s moment of inertia \(I_c\) by considering the child to be equivalent to a point mass at a distance of 1.25 m from the axis. Then,

\[I_c = MR^2 = (18.0 \, kg)(1.25 \, m)^2 = 28.13 \, kg \cdot m^2.\]

The total moment of inertia is the sum of moments of inertia of the merry-go-round and the child (about the same axis). To justify this sum to yourself, examine the definition of \(I\):

\[I = 28.13 \, kg \cdot m^2 + 56.25 \, kg \cdot m^2 = 84.38 \, kg \cdot m^2.\]

Substituting known values into the equation for \(\alpha \) gives

\[\alpha = \dfrac{\tau}{I} = \dfrac{375 \, N \cdot m}{84.38 \, kg \cdot m^2} = 4.44 \dfrac{rad}{s^2}.\]

The angular acceleration is less when the child is on the merry-go-round than when the merry-go-round is empty, as expected. The angular accelerations found are quite large, partly due to the fact that friction was considered to be negligible. If, for example, the father kept pushing perpendicularly for 2.00 s, he would give the merry-go-round an angular velocity of 13.3 rad/s when it is empty but only 8.89 rad/s when the child is on it. In terms of revolutions per second, these angular velocities are 2.12 rev/s and 1.41 rev/s, respectively. The father would end up running at about 50 km/h in the first case. Summer Olympics, here he comes! Confirmation of these numbers is left as an exercise for the reader.

Exercise \(\PageIndex{1}\):Check Your Understanding

Torque is the analog of force and moment of inertia is the analog of mass. Force and mass are physical quantities that depend on only one factor. For example, mass is related solely to the numbers of atoms of various types in an object. Are torque and moment of inertia similarly simple?

No. Torque depends on three factors: force magnitude, force direction, and point of application. Moment of inertia depends on both mass and its distribution relative to the axis of rotation. So, while the analogies are precise, these rotational quantities depend on more factors

• The farther the force is applied from the pivot, the greater is the angular acceleration; angular acceleration is inversely proportional to mass.
• If we exert a force \(F\) on a point mass \(m\) that is a distance r from a pivot point and because the force is perpendicular to r and acceleration \(a = F/m\) is obtained in the direction of \(F\). We can rearrange this equation such that \[F = ma,\] and then look for ways to relate this expression to expressions for rotational quantities. We note that \(a = r \alpha\), and we substitute this expression into \(F = ma\), yielding \[F = mr\alpha\]
• Torque is the turning effectiveness of a force. In this case, because \(F\) is perpendicular to \(r\), torque is simply \(\tau = rF\). If we multiply both sides of the equation above by \(r\), we get torque on the left-hand side. That is, \[rF = mr^2\alpha\] or \[\tau = mr^2\alpha.\]
• The moment of inertia \(I\) of an object is the sum of \(MR^2\) for all the point masses of which it is composed. That is, \[I = \sum mr^2.\]
• The general relationship among torque, moment of inertia, and angular acceleration is \[\tau = I \alpha\] or \[\alpha = \dfrac{net \, \tau}{I}.\]

‘Law & Order: SVU’ Season 25 Finale Brings in Stabler for Important Conversation, Plus Who’s Shot? (RECAP)

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[Warning: The below contains MAJOR spoilers for the Law & Order: SVU Season 25 Finale “Duty to Hope.”]

Pressure is high to catch a rapist in the  SVU Season 25 finale.

There have been four victims in the past six months, and their unknown subject is escalating. He uses a coat hanger to bind his victims and sexually assaults them with a gun; the latest is still in the ICU. Not making it any easier is ADA Heidi Russell (Kate Loprest), the new chief of the Trial Division, isn’t exactly making any friends—with Benson ( Mariska Hargitay ) or Carisi ( Peter Scanavino ), the only one of the two who does work for her. In fact, she’s fine with charging a man, Hedges, who seems good for the attack that occurs near the beginning of this episode (his fingerprint at the scene)—and that man even pleads guilty—but then there’s another assault.

Russell still believes that Hedges is good for the crime he’s confessed to, even though the others think he was just pressured by his incompetent public defender. And the latest attack does give them a lead: A neighbor interrupted it, and the woman was able to kick the rapist’s engraved leather holster under the stove when he ran. The engraving points to a special ops branch of the Marines. Could this man and Hedges be working together? There’s nothing to suggest that.

'SVU’ Is Finally Solving Its Rollins Problem

Benson sends everyone home to live their lives… and that’s when Fin ( Ice-T ) is surprised by a kid holding him at gunpoint as he’s throwing out the trash. This kid is Hedges’ son, and he insists his father is innocent. In the ensuing scuffle, Fin is shot in the arm, and he tells the kid to get out of there and he won’t tell anyone. He then tries to treat the wound himself and turns in the gun claiming a neighbor gave it to him, but then he collapses and ends up in the hospital. He does, when she pushes, tell Benson what happened, and she sends him to talk to Hedges in Rikers with Carisi. It’s then that they realize how his fingerprint got on the inside of a glass door in the victim’s apartment: He worked in a home and hardware store that sold it.

Touch DNA on the leather holster IDs the rapist: Glenn Duncan, an ex-Marine who was dishonorably discharged in 2019. With IDs from two victims, Benson, the squad, and ESU go to his apartment, and he immediately begins shooting at them from his window. He hits an officer, and Benson puts herself in the line of fire to drive over and get him back to the paramedics. Benson tries to talk to Glenn over the phone, but he says either he goes down shooting or goes to prison for the rest of his life. He holds his girlfriend hostage, but when ESU moves in, they’re able to save her. And Glenn is good for all the assaults. Hedges will get out. Fin goes to tell his son and brings him to pick his dad up from Rikers—and make it clear he never wants to catch him touching a gun again.

While Benson says she would love to go with Fin to tell Hedges’ son, she says she has somewhere to be: Maddie’s 16th birthday. Maddie’s mom thanks Benson for taking her daughter to therapy; she’s on the road to recovery. She suggests she needs a compass like Benson has around her neck (the one from Stabler) to navigate what’s going on in her life—and Benson tells her she can borrow the necklace! “This has gotten me through some tough times, and you can return it whenever you want. It’s not like we’re going to lose touch,” the captain points out.

The episode ends with Benson calling Stabler ( Christopher Meloni ) to check in and tell him. “The necklace, the compass that you gave me last year. …  I haven’t taken it off since you gave it to me, and I wanted to let you know it has been incredibly meaningful to me and it has guided me in terms of my healing,” she says. “I’m glad,” he tells her. He also knows exactly who she means when she reveals she lent it to someone. “Sounds like she needed it a lot more than you do,” he remarks. “I knew you’d understand,” she tells him, and the season ends with them still talking.

What did you think of the  SVU finale and Benson giving away the compass necklace for now? Let us know in the comments section, below.

Law & Order: SVU , Season 26, Fall 2024, NBC

If you or someone you know is the victim of sexual assault, contact the Rape, Abuse & Incest National Network ‘s National Helpline at 1-800-656-HOPE (4673). If you or a loved one are in immediate danger, call 911.

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Laws of Motion Numericals

• Newton's First Law of Motion
• Newton's Third Law of Motion
• Motion in a Vertical Circle
• Laws of Logarithms
• Kepler's Laws of Planetary Motion
• Applications of Fluid Dynamics
• Uniform Circular Motion
• Applications of Charles Law
• Motion Under Gravity
• Equation of Tangents and Normals
• Motion of Center of Mass
• Applications of Kepler's Laws
• Law of Inertia
• What is Motion?
• Uniform and Non-Uniform Motion
• Oscillatory Motion Formula
• Kinematics of Rotational Motion
• Dynamics of Circular Motion
• Secant Method of Numerical analysis

Laws of Motion describe how objects move under the influence of different types of forces. These forces can be due to any physical phenomenon, but their effect is the same. All the forces change the momentum of the system on which they are acting. Newton gave three laws, these laws describe the interaction between two objects and the forces between them. These laws become the building block for the further theory of mechanics and motion. Let’s look at these concepts and some problems with them.

Table of Content

Newton’s Laws of Motion

• Newton’s First Law

Newton’s Second Law

• Newton’s Third Law
• Solved Examples

Numericals on Laws of Motion

Sir Isaac Newton was an English physicist and mathematician who gave the three laws of motion called Newton’s Law of Motion which formed the base for classical mechanics. These laws still keep giving accurate predictions except for bodies traveling with speeds comparable to light or the size of an electron. These were the first laws that described the forces acting on the bodies and the motion of the body which is governed by these forces. There are three laws of motion that are given by Sir Issac Newton that include,

• Newton’s First Law of Motion
• Newton’s Second Law of Motion
• Newton’s Third Law of Motion

Now let’s learn about these laws further in this article.

Newton’s First Law (Law of Inertia)

Newton’s First Law of motion also called the law of interia states that if a body is at rest or is moving in a straight line with constant speed. It will keep moving in a straight line at constant speed or will remain at rest until it is acted upon by an external force. This property of any object to resist a change in its state is called inertia and thus this law is known Law of Inertia.  

Newton’s Second Law is a quantitative description of the changes that take place when an external force acts on the body. The momentum of the body is defined as the product of the mass and velocity of that body. When a force acts on the body, it brings about changes in the momentum of the body or its direction, or both. It is one of the most important laws in the field of classical mechanics. Assuming the mass of the body is “m”, the law is given by,

Here, F is the force acting on the particle, and “a” denotes the acceleration produced in the body. The direction of acceleration is the same as the direction of motion.

Newton’s Third Law (Law of Action and Reaction)

Newton’s Third Law also called law of action and reaction states that when two bodies interact with each other, they apply forces to one another which are equal in magnitude and opposite in direction. This law is also known as action-reaction law. It allows us to explain phenonmenon such as static equilibrium, where all the forces are balanced, but it also applies to bodies in uniform or accelerated motion. If the net forces acting on the body are equal, the body is said to be in equilibrium.

Also, Check

• Acceleration

Examples on Newton Laws of Motion

Example 1: Calculate the momentum of a ball thrown at a speed of 10 m/s and weighing 800 g.

Given, M = 800 g V = 10 m/s Momentum is given by, p = MV Plugging in the values in the formula p = MV p = (800)(10)  p = 8000 gm/s p = 8 × 10 3 gm/s

Example 2: Calculate the momentum of a ball thrown at a speed of 10 m/s and weighing 20 g.

Given, M = 20 g V = 10 m/s Momentum is given by, p = MV Plugging in the values in the formula p = MV p = (20)(10) p = 200 gm/s p = 2 x 10 2 gm/s

Example 3: A force of 20N is acting on a body of mass 2Kg. Find the acceleration produced.

Given, m = 2 Kg F = 20 N Acceleration will be given by, F = ma Plugging in the values, F = ma 20 = (2)(a) 10 m/s 2 = a

Example 4: A force of 100N is acting on a body of mass 5Kg. Find the acceleration produced.

Given, m = 5 Kg F = 100 N Acceleration will be given by, F = ma Plugging in the values, F = ma 100 = (5)(a) 20 m/s 2 = a

Example 5: A body of 2 kg is moving at a velocity of 50m/s. A force starts acting on it and the velocity becomes 20m/s in a time of 5 seconds. Find the force applied to the body.

Given, m = 5 Kg v i = 50 m/s v f = 20 m/s t = 5 s Force is defined as the rate of change of momentum. F = m(v f  – v i )/t F = (5)(50 – 20)/(5) F = 30N

Example 6: A body of 10 kg is moving at a velocity of 100m/s. A force starts acting on it and the velocity becomes 20m/s in a time of 10 seconds. Find the force applied to the body.

Given, m = 10 Kg v i = 100 m/s v f = 20 m/s t = 10 s Force is defined as the rate of change of momentum F = m(v f  – v i )/t F = (10)(80 – 20)/(10) F = 80N

Example 7: The momentum of the body is given by the equation below,

p(t) = 3t 2 + 4t + 5

Find the force acting on the body at t = 5.

Force rate of change of momentum, F = dp/dt Given, p(t) = 3t 2 + 4t + 5 F = dp/dt = d/dt(3t 2 + 4t + 5) F = 6t + 4 At t = 5 F = 6(5) + 4 F = 34 N Thus, force acting on the body at t = 5 sec is 34 N.

Example 8: The momentum of the body is given by the equation below,

p(t) = e t + t 2 + 20

Find the force acting on the body at t = 0.

Force rate of change of momentum, F = dp/dt Given, p(t) = e t + t 2 + 20 F = dp/dt F = d/dt (e t + t 2 + 20) F = e t + 2t At t = 0 F = e 0 + 2×1    = 1 + 2    = 3 F = 3 N Thus, the force acting on the body at t = 0 is 3 N.

Lets learn Law of Motion Numericals for class 9 and class 11

1. If the momentum of any body is, p(t) = 3t 3 + 5t 2 + t. Find the force acting on the body at t = 2.

2. If the distance covered by an object is given by, d(t) = t 3 + t. Find the acceleration on the body at t = 3.

3. If the distance covered by an object is given by, d(t) = 4t 4 + 3t + 5. Find the velocity on the body at t = 0.

4. A body of 19 kg is moving at a velocity of 100 m/s. A force starts acting on it and the velocity becomes 120 m/s in a time of 10 seconds. Find the force applied to the body.

5. Force acting on body is 120 N, the mass of the body is 12 Kg. Find the acceleration produced produced by the body.

FAQs on Laws of Motion

1. who discovered the laws of motion.

The laws of motion were discovered by Sir Isaac Newton an English Mathematician.

2. What is Law of Inertia?

Law of Inertia is also called the Newton First Law of Motion. This law states that an object at the state of rest or at the state of motion stays in its state until an external force is applied.

3. What is Law of Action and Reaction?

Law of Action and Reaction is the other name of Newton Third Law of Motion. This law states that, “Every action has its equal and opposite reaction.”

4. What is Formula for Force Acting on a Body?

The force acting on a body is given using the formula, F = ma where, m is Mass of Object a is Acceleration of Object

5. What is Formula for Momentum of a Body?

The momentum of an object is given using the formula, F = mv where, m is Mass of Object v is Velocity of Object

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Solving Law Firm’s Biggest Problem: Outsourcing Legal Billing to Improve Collection and Profitability

• Legal Operations

Streamlining and improving the billing process in law firms is often perceived as a daunting task, fraught with complexities that can challenge even the most seasoned legal professionals. Billing is a time consuming, labor-intensive necessary evil that is often fraught with inaccuracies due to inefficiencies in process and extensive operational silos which can cost a law firm potentially millions of dollars each year.

According to a recent Thomson Reuters survey , the average law firm experiences an 18% realization loss due to billing challenges. This means that nearly one-fifth of billable work is not converted into revenue, negatively impacting the firm’s profitability. 81% of law firms report issues with a significant portion of invoices remaining unpaid or delayed, creating cash flow challenges. This is a huge revenue loss that is creating major problems for law firms, but there are solutions.

Why is legal billing so complex?

Most unique to attorneys is that their work is often time-consuming difficult to tangibly track. Tracking billable hours accurately requires meticulous attention to detail and can be particularly challenging when dealing with tasks that don't neatly fit into predefined categories. This creates problems because clients often have unique billing preferences and expectations ranging from traditional hourly billing to flat fees, contingency fees and/or blended rates. Accommodating these distinctive preferences for each client and integrating them into a billing system adds a layer of complexity to billing that makes it extremely labor intensive. Additionally, silos that exist within law firms in the form of practice groups or other operational units can make regularity in billing guidelines either difficult to understand or completely nonexistent.

Another complexity in legal billing is the regulatory compliance and ethical guidelines that must be adhered to. Some of these issues include fee agreements, billing transparency, client confidentiality, and conflicts of interest. Staying compliant with these regulations while meeting client expectations and maintaining profitability requires a thorough understanding of legal and professional standards.

While most law firms rely on billing software to streamline the process, integrating these systems with other practice management tools and ensuring data accuracy and security can present technical challenges that require ongoing maintenance and support.

Reasons to outsource billing

By streamlining the billing processes , law firms can reduce the risk of errors, and focus on core competencies. Technical expertise and best practices ensure process optimization and service quality. Outsourcing can be a valuable solution for law firms aiming to stabilize their cash flow, reduce costs, and improve overall efficiency.

• Access to Advanced Technology: Outsourcing billing often provides access to advanced billing technologies and software platforms that may be cost-prohibitive for smaller law firms to implement independently. These technologies can automate billing processes, provide real-time reporting and analytics, and offer insights into billing trends and patterns that allow firms to optimize their billing practices for greater efficiency and profitability Additionally, predictive AI can provide billing staff with powerful tools to wade through complex billing guidelines in a timely manner, increasing productivity within billing departments.
• Expertise and Accuracy: Billing processes can be complex, especially in law firms where billing requirements may vary from client to client and case to case. Outsourcing billing to specialized professionals ensures that billing is handled by experts who are well-versed in legal billing practices. This expertise leads to greater accuracy in invoicing, reducing the likelihood of errors or discrepancies that could lead to disputes or delays in payment. This also leads to more standardized output.
• Centralization: Often, billing departments in law firms are segregated either geographically or by practice area, leading to inefficiencies and lack of scalability. Centralizing billing can assist by allowing billing to be looked at holistically as a firm rather than piecemeal, leading to a more effective business model for financial processes.
• Efficiency and Timeliness: Outsourcing billing allows law firms to streamline their billing processes and improve efficiency. Billing professionals have the necessary tools and systems in place to generate invoices promptly and ensure timely submission to clients. This helps expedite the payment cycle, leading to improved cash flow for the firm.
• Cost Savings: While some may initially view outsourcing as an additional expense, it often proves to be a cost-effective solution in the long run. By outsourcing billing, law firms eliminate the need to hire and train in-house billing staff, invest in billing software, and bear the overhead costs associated with maintaining billing infrastructure, lowering both soft and hard operational costs and simplifying a billing budget. Additionally, outsourcing billing allows firms to scale their billing operations up or down as needed without incurring significant fixed costs.
• Compliance and Regulation: Legal billing is subject to various regulations and compliance requirements, including those related to client confidentiality, billing transparency, and fee agreements. Outsourcing billing to professionals who are experts in these regulations helps ensure compliance and mitigates the risk of potential legal or ethical issues arising from billing practices.
• Focus on Core Competencies: Attorneys and legal staff excel in providing legal counsel and representation, but are not usually trained in managing billing and administrative tasks, and even if they are, legal staff turning their attention to these tasks from an administrative perspective harms the firm’s bottom line by taking time away from client-facing, billable projects. By outsourcing billing, law firms can free up valuable time and resources to concentrate on what they do best, which is serving clients legal needs.
• Enhanced Client Satisfaction: Timely and accurate billing contributes to a positive client experience. By outsourcing billing , law firms can ensure that clients receive clear, transparent invoices in a timely manner, enhancing client satisfaction and fostering long-term client relationships.

Legal billing is a challenge that costs firms millions of dollars every year. It has become such an issue that many of law firms have said this is their main priority to fix by 2025. Centralized teams face challenges due to the varied requirements of different attorneys and clients. The benefits to outsourcing billing not only will help realization rates but will foster a better relationship with clients and will help alleviate headaches of both lawyers and legal administrators as well.

The contents of this article are intended to convey general information only and not to provide legal advice or opinions.

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10.5: Moments of Inertia: Problem Solving

Chapter 0: physics basics, chapter 1: an introduction to statics, chapter 2: force vectors, chapter 3: equilibrium of a particle, chapter 4: force system resultants, chapter 5: equilibrium of a rigid body, chapter 6: structural analysis, chapter 7: internal forces, chapter 8: friction, chapter 9: center of gravity and centroid, chapter 10: moment of inertia, chapter 11: virtual work, chapter 12: kinematics of a particle, chapter 13: kinetics of a particle: force and acceleration, chapter 14: kinetics of a particle: impulse and momentum, chapter 15: planar kinematics of a rigid body, chapter 16: 3-dimensional kinetics of a rigid body, chapter 17: concept of stress, chapter 18: stress and strain - axial loading.

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Consider airplanes, where the wing components are triangular plates. Approximating this component as a schematic triangle, what is the moment of inertia about the centroidal axis?

The centroid of a triangle is located at one-third of the triangle's height from its base.

Consider a differential strip at a certain distance from the base parallel to the centroidal axis. This strip's differential moment of inertia equals the square of the distance from the axis times the differential area.

The differential area is the product of the strip's length and width. The strip's length is estimated considering the law of similar triangles.

Integrating the differential moment of inertia along the entire height gives the moment of inertia about the base.

Using parallel axis theorem, the moment of inertia about the centroid equals the moment of inertia about the base minus the product of the traingle's area and the square of the distance from the centroidal axis.

Substituting the values, the moment of inertia of the triangle about the centroidal axis is obtained.

The second moment of an area, also known as the moment of inertia of an area, is a geometric property of a shape that reflects its resistance to change. The moment of inertia of an area can be calculated for both two-dimensional and three-dimensional shapes. The moment of inertia of an area is calculated by taking the sum of the product of the area and the square of its distance from a chosen axis of rotation. For two-dimensional shapes, the moment of inertia can be expressed as a single equation in terms of the shape's x and y coordinates.

To determine the moment of inertia for a schematic triangle along the centroidal axis, as shown above, we must begin by considering a differential strip at a certain distance from the base parallel to the centroidal axis. The differential moment of inertia equals the distance squared multiplied by the differential area.

The length of the strip is estimated using the law of similar triangles. The strip's length is proportional to its distance from the base.

Rewriting the differential area in terms of the length and width of the differential strip and using the expression for the strip length, the equation is modified. Now, integrating the differential moment of inertia along the entire height gives the total moment of inertia along the base.

The centroid of a triangle is located at one-third of the triangle's height from its base. Using the parallel axis theorem, the moment of inertia along the centroid equals the moment of inertia along the base minus the product of the triangle's area and the centroidal axis distance squared.

By substituting the relevant values, the moment of inertia of the triangle along the centroidal axis can be obtained.

Triangular plates are commonly used in the design of the tail section of aircraft. This section of the aircraft is responsible for providing stability and control during flight. Triangular plates are often used to form the vertical and horizontal stabilizers, which help keep the aircraft stable and pointing in the right direction.

• F.P. Beer, E.R. Johnston, D.F. Mazurek, P.J. Cornwell, B.P. Self, Vector Mechanics For Engineers Statics and Dynamics Engineering Mechanics Statics, Mc Graw-Hill Education. Pp. 491

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