MODEL Model the car’s motion as dynamics with constant force and use the model of kinetic friction. We want to solve the problem only once, not three separate times, so we’ll leave the slope angle $\theta$ unspecified until the end.

VISUALIZE FIGURE 6.24 shows the pictorial representation. We’ve shown the car sliding uphill, but these representations work equally well for a level or downhill slide if we let $\theta$ be zero or negative, respectively. We’ve used a tilted coordinate system so that the motion is along one of the axes. We’ve assumed that the car is traveling to the right, although the problem didn’t state this. You could equally well make the opposite assumption, but you would have to be careful with negative values of x and $v_x$. The car skids to a halt, so we’ve taken the coefficient of kinetic friction for rubber on concrete from Table 6.1.

FIGURE 6.24 Pictorial representation of a skidding car.

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SOLVE Newton’s second law and the model of kinetic friction are

We’ve written these equations by “reading” the motion diagram and the free-body diagram. Notice that both components of the gravitational force vector $\vec{F}{}_{\text{G}}$ are negative. $a_y=0$ because the motion is entirely along the x-axis.

The second equation gives $n=mg\text{cos}\theta$. Using this in the friction model, we find $f_{\text{k}}={\mu }_{\text{k}}mg\text{cos}\theta$. Inserting this result back into the first equation then gives

This is a constant acceleration. Constant-acceleration kinematics gives

which we can solve for the stopping distance $x_1:$

Notice how the minus sign in the expression for $a_x$ canceled the minus sign in the expression for $x_1$. Evaluating our result at the three different angles gives the stopping distances:

The implications are clear about the danger of driving downhill too fast!

REVIEW $30\text{m}\text{/}\text{s}\approx 60\text{mph}$ and $57\text{m}\approx 180\text{feet}$ on a level surface. This is similar to the stopping distances you learned when you got your driver’s license, so the results seem reasonable. Additional confirmation comes from noting that the expression for $a_x$ becomes $-g\text{sin}\theta$ if ${\mu }_{\text{k}}=0$. This is what you learned in Chapter 2 for the acceleration on a frictionless inclined plane.

MODEL Model the cart as a particle acted on by constant forces.

VISUALIZE The cart changes velocity—it accelerates—when both pulled and rolled. Consequently, there must be a net force for both motions. For rolling, force identification finds that the only horizontal force is rolling friction, a force that opposes the motion and slows the cart. There is no “force of motion” or “force of the hand” because the hand is no longer in contact with the cart. (Recall Newton’s “zeroth law”: The cart responds only to forces applied at this instant.) Pulling adds a tension force in the direction of motion. The two free-body diagrams are shown in FIGURE 6.26.

FIGURE 6.26 Pictorial representations of the cart.

SOLVE The cart’s acceleration when pulled, which we can find from the velocity data, will allow us to find the net force. Isolating the tension force will require knowing the friction force, but we can determine that from the rolling motion. For the rolling motion, Newton’s second law can be written by “reading” the free-body diagram on the left:

Make sure you understand where the signs come from and how we used our knowledge that $\vec{a}$ has only an x-component, which we called $a_{\text{roll}}$. The magnitude of the friction force, which is all we’ll need to determine the tension, is found from the x-component equation:

But we’ll need to do a bit more analysis to get the coefficient of rolling friction. The y-component equation tells us that $n=mg$. Using this in the model of rolling friction, $f_{\text{r}}={\mu }_{\text{r}}n={\mu }_{\text{r}}mg$, we see that the coefficient of rolling friction is

The x-component equation of Newton’s second law when the cart is pulled is

Thus the tension that we seek is

FIGURE 6.27 shows the graphs of the velocity data. The accelerations are the slopes of these lines, and from the equations of the best-fit lines we find $a_{\text{roll}}=-0.124\text{m}\text{/}{\text{s}}^2$ and $a_{\text{pulled}}=1.55\text{m}\text{/}{\text{s}}^2$. Thus the friction force is

Knowing this, we find that the string tension pulling the cart is

and the coefficient of rolling friction is

FIGURE 6.27 The velocity graphs of the rolling and pulled motion. The slopes of these graphs are the cart’s acceleration.

REVIEW The coefficient of rolling friction is very small, but it’s similar to the values in Table 6.1 and thus believable. That gives us confidence that our value for the tension is also correct. It’s reasonable that the tension needed to accelerate the cart is small because the cart is light and there’s very little friction.

MODEL This is a somewhat different problem from any we have looked at thus far. Let the box, which we’ll model as a particle, be the object of interest. It contacts other objects only where it touches the truck bed, so only the truck can exert contact forces on the box. If the box does not slip, then there is no motion of the box relative to the truck and the box must accelerate with the truck: $a_{\text{box}}=a_{\text{truck}}$. As the box accelerates, it must, according to Newton’s second law, have a net force acting on it. But from what?

Imagine, for a moment, that the truck bed is frictionless. The box would slide backward (as seen in the truck’s reference frame) as the truck accelerates. The force that prevents sliding is static friction, so the truck must exert a static friction force on the box to “pull” the box along with it and prevent the box from sliding relative to the truck.

VISUALIZE This situation is shown in FIGURE 6.28. There is only one horizontal force on the box, $\vec{f}{}_{\text{s}}$, and it points in the forward direction to accelerate the box. Notice that we’re solving the problem with the ground as our reference frame. Newton’s laws are not valid in the accelerating truck because it is not an inertial reference frame.

FIGURE 6.28 Pictorial representation for the box in a flatbed truck.

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SOLVE Newton’s second law, which we can “read” from the free-body diagram, is

Now, static friction, you will recall, can be any value between 0 and $f_{\text{s max}}$. If the truck accelerates slowly, so that the box doesn’t slip, then $f_{\text{s}}{<}_{}{f}_{\text{s max}}$. However, we’re interested in the acceleration $a_{\text{max}}$ at which the box begins to slip. This is the acceleration at which $f_{\text{s}}$ reaches its maximum possible value

The y-equation of the second law and the friction model combine to give $f_{\text{s max}}={\mu }_{\text{s}}mg$. Substituting this into the x-equation, and noting that $a_x$ is now $a_{\text{max}}$, we find

The truck must keep its acceleration less than $3.9\text{m}\text{/}{\text{s}}^2$ if slipping is to be avoided.

REVIEW $3.9\text{m}\text{/}{\text{s}}^2$ is about one-third of g. You may have noticed that items in a car or truck are likely to tip over when you start or stop, but they slide only if you really floor it and accelerate very quickly. So this answer seems reasonable. Notice that neither the dimensions of the crate nor ${\mu }_{\text{k}}$ was needed. Real-world situations rarely have exactly the information you need, no more and no less. Many problems in this textbook will require you to assess the information in the problem statement in order to learn which is relevant to the solution.

MODEL Model the object as a particle. But we cannot use the constant-force model or constant-acceleration kinematics.

VISUALIZE The sine function is 0 at $t=0$ and again at $t=T$, when the value of the argument is $\pi \text{rad}$. Over the interval $0\le t\le T$, the force grows from 0 to $c$ and then returns to 0, always pointing in the positive x-direction. FIGURE 6.29 shows a graph of the force and a pictorial representation.

FIGURE 6.29 Pictorial representation for a variable force.

SOLVE The object’s acceleration increases between 0 and $T\text{/}2$ as the force increases. You might expect the object to slow down between $T\text{/}2$ and $T$ as the force decreases. However, there’s still a net force in the positive x-direction, so there must be an acceleration in the positive x-direction. The object continues to speed up, only more slowly as the acceleration decreases. Maximum velocity is reached at $t=T$.

a. This is not constant-acceleration motion, so we cannot use the familiar equations of constant-acceleration kinematics. Instead, we must use the definition of acceleration as the rate of change—the time derivative—of velocity. With no friction, we need only the x-component equation of Newton’s second law:

First we rewrite this as

Then we integrate both sides from the initial conditions $(v_x=$ $v_{0x}=0\text{at t}=t_0=0)$ to the final conditions $(v_x$ at the later time $t)$:

The fraction $c\text{/}m$ is a constant that we could take outside the integral. The integral on the right side is of the form

Using this, and integrating both sides of the equation, we find

Simplifying, we find the object’s velocity at time $t$ is

This expression is graphed in FIGURE 6.30, where we see that, as predicted, maximum velocity is reached at $t=T$.

b. Maximum velocity, at $t=T$, is

FIGURE 6.30 The object’s velocity as a function of time.

REVIEW A steady $2.5\text{N}$ force would cause a 0.5 kg object to accelerate at $5\text{m}\text{/}{\text{s}}^2$ and reach a speed of $5\text{m}\text{/}\text{s}$ in $1\text{s}$. A variable force with a maximum of $2.5\text{N}$ will produce less acceleration, so a top speed of $3.2\text{m}\text{/}\text{s}$ seems reasonable.