6.1 Solving Problems with Newton’s Laws - University Physics Volume 1 | OpenStax (2023)

Applying Newton’s Laws of Motion

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

Different Tensions at Different Angles

Consider the traffic light (mass of 15.0 kg) suspended from two wires as shown in Figure 6.3. Find the tension in each wire, neglecting the masses of the wires.

Figure 6.3 A traffic light is suspended from two wires. (b) Some of the forces involved. (c) Only forces acting on the system are shown here. The free-body diagram for the traffic light is also shown. (d) The forces projected onto vertical (y) and horizontal (x) axes. The horizontal components of the tensions must cancel, and the sum of the vertical components of the tensions must equal the weight of the traffic light. (e) The free-body diagram shows the vertical and horizontal forces acting on the traffic light.

Strategy

The system of interest is the traffic light, and its free-body diagram is shown in Figure 6.3(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 6.3(d). There are two unknowns in this problem ($T_1$ and $T_2$), 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.

Solution

First consider the horizontal or x-axis:

$$F_{\text{net}x}=T_{2x}+T_{1x}=0.$$

Thus, as you might expect,

$$\left|T_{1x}\right|=\left|T_{2x}\right|.$$

This gives us the following relationship:

$$T_1\text{cos}30\text{°}=T_2\text{cos}45\text{°}.$$

Thus,

$$T_2=1.225T_1.$$

Note that $T_1$ and $T_2$ are not equal in this case because the angles on either side are not equal. It is reasonable that $T_2$ ends up being greater than $T_1$ because it is exerted more vertically than $T_1.$

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

$$F_{\text{net}y}=T_{1y}+T_{2y}-w=0.$$

This implies

$$T_{1y}+T_{2y}=w.$$

Substituting the expressions for the vertical components gives

$$T_1\text{sin}30\text{°}+T_2\text{sin}45\text{°}=w.$$

There are two unknowns in this equation, but substituting the expression for $T_2$ in terms of $T_1$ reduces this to one equation with one unknown:

$$T_1(0.500)+(1.225T_1)(0.707)=w=mg,$$

which yields

$$1.366T_1=(15.0\text{kg})(9.80{\text{m/s}}^2).$$

Solving this last equation gives the magnitude of $T_1$ to be

$$T_1=108\text{N}.$$

Finally, we find the magnitude of $T_2$ by using the relationship between them, $T_2=1.225T_1$, found above. Thus we obtain

$$T_2=132\text{N}.$$

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.

Drag Force on a Barge

Two tugboats push on a barge at different angles (Figure 6.4). The first tugboat exerts a force of $2.7\times {10}^5\text{N}$ in the x-direction, and the second tugboat exerts a force of $3.6\times {10}^5\text{N}$ in the y-direction. The mass of the barge is $5.0\times {10}^6\text{kg}$ and its acceleration is observed to be $7.5\times {10}^{\mathrm{-2}}{\text{m/s}}^2$ 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.)

Figure 6.4 (a) A view from above of two tugboats pushing on a barge. (b) The free-body diagram for the ship contains only forces acting in the plane of the water. It omits the two vertical forces—the weight of the barge and the buoyant force of the water supporting it cancel and are not shown. Note that ${\overrightarrow{F}}_{\text{app}}$ is the total applied force of the tugboats.

See Also

6.3 Centripetal Force - University Physics Volume 1 | OpenStax

Strategy

The directions and magnitudes of acceleration and the applied forces are given in Figure 6.4(a). We define the total force of the tugboats on the barge as ${\overrightarrow{F}}_{\text{app}}$ so that

$${\overrightarrow{F}}_{\text{app}}={\overrightarrow{F}}_1+{\overrightarrow{F}}_2.$$

The drag of the water ${\overrightarrow{F}}_{\text{D}}$ is in the direction opposite to the direction of motion of the boat; this force thus works against ${\overrightarrow{F}}_{\text{app}},$ as shown in the free-body diagram in Figure 6.4(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 ${\overrightarrow{F}}_1$ and ${\overrightarrow{F}}_2.$ The problem quickly becomes a one-dimensional problem along the direction of ${\overrightarrow{F}}_{\text{app}}$, since friction is in the direction opposite to ${\overrightarrow{F}}_{\text{app}}.$ Our strategy is to find the magnitude and direction of the net applied force ${\overrightarrow{F}}_{\text{app}}$ and then apply Newton’s second law to solve for the drag force ${\overrightarrow{F}}_{\text{D}}.$

Solution

Since $F_x$ and $F_y$ are perpendicular, we can find the magnitude and direction of ${\overrightarrow{F}}_{\text{app}}$ directly. First, the resultant magnitude is given by the Pythagorean theorem:

$$F_{\text{app}}=\sqrt{F_1^2+F_2^2}=\sqrt{{(2.7\times {10}^5\text{N})}^2+{(3.6\times {10}^5\text{N})}^2}=4.5\times {10}^5\text{N}\text{.}$$

The angle is given by

$$\theta ={\text{tan}}^{\mathrm{-1}}\left(\frac{F_2}{F_1}\right)={\text{tan}}^{\mathrm{-1}}\left(\frac{3.6\times {10}^5\text{N}}{2.7\times {10}^5\text{N}}\right)=53.1\text{°}.$$

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

$$F_{\text{net}}=F_{\text{app}}-F_{\text{D}}.$$

However, Newton’s second law states that

$$F_{\text{net}}=ma.$$

Thus,

$$F_{\text{app}}-F_{\text{D}}=ma.$$

This can be solved for the magnitude of the drag force of the water $F_{\text{D}}$ in terms of known quantities:

$$F_{\text{D}}=F_{\text{app}}-ma.$$

Substituting known values gives

$$F_{\text{D}}=\left(4.5\times {10}^5\text{N}\right)-\left(5.0\times {10}^6\text{kg}\right)\left(7.5\times {10}^{-2}{\text{m/s}}^2\right)=7.5\times {10}^4\text{N}\text{.}$$

The direction of ${\overrightarrow{F}}_{\text{D}}$ has already been determined to be in the direction opposite to ${\overrightarrow{F}}_{\text{app}},$ or at an angle of $53\text{°}$ south of west.

Significance

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 $F_{\text{D}}$ is less than 1/600th of the weight of the ship.

What Does the Bathroom Scale Read in an Elevator?

Figure 6.5 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 $1.20{\text{m/s}}^2,$ and (b) if the elevator moves upward at a constant speed of 1 m/s.

Figure 6.5 (a) The various forces acting when a person stands on a bathroom scale in an elevator. The arrows are approximately correct for when the elevator is accelerating upward—broken arrows represent forces too large to be drawn to scale. $\overrightarrow{T}$ is the tension in the supporting cable, $\overrightarrow{w}$ is the weight of the person, ${\overrightarrow{w}}_{\text{s}}$ is the weight of the scale, ${\overrightarrow{w}}_{\text{e}}$ is the weight of the elevator, ${\overrightarrow{F}}_{\text{s}}$ is the force of the scale on the person, ${\overrightarrow{F}}_{\text{p}}$ is the force of the person on the scale, ${\overrightarrow{F}}_{\text{t}}$ is the force of the scale on the floor of the elevator, and $\overrightarrow{N}$ is the force of the floor upward on the scale. (b) The free-body diagram shows only the external forces acting on the designated system of interest—the person—and is the diagram we use for the solution of the problem.

Strategy

If the scale at rest is accurate, its reading equals ${\overrightarrow{F}}_{\text{p}}$, the magnitude of the force the person exerts downward on it. Figure 6.5(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 6.5(b). Analysis of the free-body diagram using Newton’s laws can produce answers to both Figure 6.5(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 $\overrightarrow{w}$ and the upward force of the scale ${\overrightarrow{F}}_{\text{s}}.$ According to Newton’s third law, ${\overrightarrow{F}}_{\text{p}}$ and ${\overrightarrow{F}}_{\text{s}}$ are equal in magnitude and opposite in direction, so that we need to find $F_{\text{s}}$ in order to find what the scale reads. We can do this, as usual, by applying Newton’s second law,

$${\overrightarrow{F}}_{\text{net}}=m\overrightarrow{a}.$$

From the free-body diagram, we see that ${\overrightarrow{F}}_{\text{net}}={\overrightarrow{F}}_s-\overrightarrow{w},$ so we have

$$F_{\text{s}}-w=ma.$$

Solving for $F_s$ gives us an equation with only one unknown:

$$F_{\text{s}}=ma+w,$$

or, because $w=mg,$ simply

$$F_{\text{s}}=ma+mg.$$

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 $F_s-w=\text{−}ma.$)

Solution

1. We have $a=1.20{\text{m/s}}^2,$ so that

   $$F_{\text{s}}=(75.0\text{kg})(9.80{\text{m/s}}^2)+(75.0\text{kg})(1.20{\text{m/s}}^2)$$

   yielding

   $$F_{\text{s}}=825\text{N}\text{.}$$

2. Now, what happens when the elevator reaches a constant upward velocity? Will the scale still read more than his weight? For any constant velocity—up, down, or stationary—acceleration is zero because $a=\frac{\text{Δ}v}{\text{Δ}t}$ and $\text{Δ}v=0.$ Thus,

   $$F_{\text{s}}=ma+mg=0+mg$$

   or

   $$F_{\text{s}}=(75.0\text{kg})(9.80{\text{m/s}}^2),$$

   which gives

   $$F_{\text{s}}=735\text{N}\text{.}$$

Significance

The scale reading in Figure 6.5(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:

$$F_{\text{net}}=ma=0=F_{\text{s}}-w$$

$$F_{\text{s}}=w=mg$$

$$F_{\text{s}}=(75.0\text{kg})(9.80{\text{m/s}}^2)=735\text{N}\text{.}$$

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 6.5(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.

Two Attached Blocks

Figure 6.6 shows a block of mass $m_1$ 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 $m_2.$ Find the acceleration of the blocks and the tension in the string in terms of $m_1,m_2,\text{and}g.$

Figure 6.6 (a) Block 1 is connected by a light string to block 2. (b) The free-body diagrams of the blocks.

Strategy

We draw a free-body diagram for each mass separately, as shown in Figure 6.6. 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: $\overrightarrow{T}+{\overrightarrow{w}}_1+\overrightarrow{N}=m_1{\overrightarrow{a}}_1$

For block 2: $\overrightarrow{T}+{\overrightarrow{w}}_2=m_2{\overrightarrow{a}}_2.$

Notice that $\overrightarrow{T}$ 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

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:

$$\begin{array}{cccc}\mathbf{\text{Block 1}}\hfill & & & \mathbf{\text{Block 2}}\hfill \\ {\displaystyle \sum F_x=m{a}_x}\hfill & & & {\displaystyle \sum F_y=m{a}_y}\hfill \\ T_x=m_1{a}_{1x}\hfill & & & T_y-m_{2}g=m_2{a}_{2y}.\hfill \end{array}$$

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

$$T=m_{1}a$$

and

$$T-m_{2}g=\text{−}{m}_{2}a.$$

From these two equations, we can express a and T in terms of the masses $m_1\text{and}{m}_2,\text{and}g:$

$$a=\frac{m_2}{m_1+m_2}g$$

and

$$T=\frac{m_1{m}_2}{m_1+m_2}g.$$

Significance

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 $T=m_{2}g.$ You can see from the free-body diagram of block 2 that cannot be correct if the block is accelerating.

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 6.7, $m_1=2.00\text{kg}$ and $m_2=4.00\text{kg}\text{.}$ Consider the pulley to be frictionless. (a) If $m_2$ is released, what will its acceleration be? (b) What is the tension in the string?

Figure 6.7 An Atwood machine and free-body diagrams for each of the two blocks.

Strategy

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 $a_2$ in the downward direction, block 1 accelerates upward with acceleration $a_1$. Thus, $a=a_1=\text{−}{a}_2.$

Solution

1. We have

   $$\begin{array}{cccc}\text{For}{m}_1,{\displaystyle \sum F_y=}T-m_{1}g=m_{1}a.\hfill & & & \text{For}{m}_2,{\displaystyle \sum F_y=}T-m_{2}g=\text{−}{m}_{2}a.\hfill \end{array}$$

   (The negative sign in front of $m_{2}a$ indicates that $m_2$ accelerates downward; both blocks accelerate at the same rate, but in opposite directions.) Solve the two equations simultaneously (subtract them) and the result is

   $$(m_2-m_1)g=(m_1+m_2)a.$$

   Solving for a:

   $$a=\frac{m_2-m_1}{m_1+m_2}g=\frac{4\text{kg}-2\text{kg}}{4\text{kg}+2\text{kg}}(9.8{\text{m/s}}^2)=3.27{\text{m/s}}^2.$$

2. Observing the first block, we see that

   $$\begin{array}{c}T-m_{1}g=m_{1}a\hfill \\ T=m_1(g+a)=(2\text{kg})(9.8{\text{m/s}}^2+3.27{\text{m/s}}^2)=26.1\text{N}\text{.}\hfill \end{array}$$

Significance

The result for the acceleration given in the solution can be interpreted as the ratio of the unbalanced force on the system, $(m_2-m_1)g$, to the total mass of the system, $m_1+m_2$. We can also use the Atwood machine to measure local gravitational field strength.

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.

Solution

1. We are given the initial and final velocities (zero and 8.00 m/s forward); thus, the change in velocity is $\text{Δ}v=8.00\text{m/s}$. We are given the elapsed time, so $\text{Δ}t=2.50\text{s}\text{.}$ The unknown is acceleration, which can be found from its definition:

   $$a=\frac{\text{Δ}v}{\text{Δ}t}.$$

   Substituting the known values yields

   $$a=\frac{8.00\text{m/s}}{2.50\text{s}}=3.20{\text{m/s}}^2.$$

2. Here we are asked to find the average force the ground exerts on the runner to produce this acceleration. (Remember that we are dealing with the force or forces acting on the object of interest.) This is the reaction force to that exerted by the player backward against the ground, by Newton’s third law. Neglecting air resistance, this would be equal in magnitude to the net external force on the player, since this force causes her acceleration. Since we now know the player’s acceleration and are given her mass, we can use Newton’s second law to find the force exerted. That is,

   $$F_{\text{net}}=ma.$$

   Substituting the known values of m and a gives

   $$F_{\text{net}}=(70.0\text{kg})(3.20{\text{m/s}}^2)=224\text{N}\text{.}$$

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.

What Force Acts on a Model Helicopter?

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

Strategy

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

Solution

We have

$$a=\frac{\text{Δ}v}{\text{Δ}t}=\frac{(6.00\widehat{i}+12.00\widehat{j}\text{m/s})-(5.00\widehat{j}\text{m/s})}{2.00\text{s}}=3.00\widehat{i}+3.50\widehat{j}{\text{m/s}}^2$$

$${\displaystyle \sum \overrightarrow{F}=m\overrightarrow{a}}=(1.50\text{kg})(3.00\widehat{i}+3.50\widehat{j}{\text{m/s}}^2)=4.50\widehat{i}+5.25\widehat{j}\text{N}\text{.}$$

The magnitude of the force is now easily found:

$$F=\sqrt{{(4.50\text{N})}^2+{(5.25\text{N})}^2}=6.91\text{N}\text{.}$$

Significance

The original problem was stated in terms of $\widehat{i}-\widehat{j}$ vector components, so we used vector methods. Compare this example with the previous example.

Baggage Tractor

Figure 6.8(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 $F=(820.0t)\text{N,}$ 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?

Figure 6.8 (a) A free-body diagram is shown, which indicates all the external forces on the system consisting of the tractor and baggage carts for carrying airline luggage. (b) A free-body diagram of the tractor only is shown isolated in order to calculate the tension in the cable to the carts.

Strategy

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 $\overrightarrow{T},$ which is our objective.

Solution

1. ${\displaystyle \sum F_x}=m_{\text{system}}{a}_x$ and ${\displaystyle \sum F_x}=820.0t,$ so

   $$\begin{array}{ccc}\hfill 820.0t& =\hfill & (650.0+250.0+150.0)a\hfill \\ \hfill a& =\hfill & 0.7809t.\hfill \end{array}$$

   Since acceleration is a function of time, we can determine the velocity of the tractor by using $a=\frac{dv}{dt}$ with the initial condition that $v_0=0$ at $t=0.$ We integrate from $t=0$ to $t=3\text{:}$

   $$dv=adt,{\displaystyle {\int }_0^{3}dv=}{\displaystyle {\int }_0^{3.00}adt}={\displaystyle {\int }_0^{3.00}0.7809tdt},v=0.3905{t^2]}_0^{3.00}=3.51\text{m/s}\text{.}$$

2. Refer to the free-body diagram in Figure 6.8(b).

   $$\begin{array}{ccc}\hfill {\displaystyle \sum F_x}& =\hfill & m_{\text{tractor}}{a}_x\hfill \\ \hfill 820.0t-T& =\hfill & m_{\text{tractor}}(0.7805)t\hfill \\ \hfill (820.0)(3.00)-T& =\hfill & (650.0)(0.7805)(3.00)\hfill \\ \hfill T& =\hfill & 938\text{N}.\hfill \end{array}$$

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 6.8(a), whereas only the mass of the truck (since it supplied the force) was of use in Figure 6.8(b).

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 6.9). Determine the maximum height it will travel if atmospheric resistance is measured as $F_{\text{D}}=(0.0100v^2)\text{N,}$ where v is the speed at any instant.

Figure 6.9 (a) The mortar fires a shell straight up; we consider the friction force provided by the air. (b) A free-body diagram is shown which indicates all the forces on the mortar shell. (credit a: modification of work by OS541/DoD; The appearance of U.S. Department of Defense (DoD) visual information does not imply or constitute DoD endorsement.)

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.

Solution

Initially, $y_0=0$ and $v_0=50.0\text{m/s}\text{.}$ At the maximum height $y=h,v=0.$ The free-body diagram shows $F_{\text{D}}$ to act downward, because it slows the upward motion of the mortar shell. Thus, we can write

$${\displaystyle \sum }{F}_y=m{a}_y$$

$$\begin{array}{}\\ \hfill -F_{\text{D}}-w& =\hfill & m{a}_y\hfill \\ \hfill -0.0100v^2-98.0& =\hfill & 10.0a\hfill \\ \hfill a& =\hfill & \mathrm{-0.00100}{v}^2-9.80.\hfill \end{array}$$

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

$$a^{}ds=v^{}dv.$$

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

$$ady=vdv,(\mathrm{-0.00100}{v}^2-9.80)dy=vdv.$$

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

$$\begin{array}{}\\ {\displaystyle {\int }_0^{h}dy=}{\displaystyle {\int }_{50.0}^0\frac{vdv}{(\mathrm{-0.00100}{v}^2-9.80)}}\hfill \\ {\displaystyle {\int }_0^{h}dy=}-{\displaystyle {\int }_{50.0}^0\frac{vdv}{(0.00100v^2+9.80)}}=(\mathrm{-5}\times {10}^2)\text{ln}(0.00100v^2+9.80){|}_{50.0}^0.\hfill \end{array}$$

Thus, $h=114\text{m}\text{.}$

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.