MODEL The crate and the truck are separate objects that form the system. We’ll model them as particles. The earth and the road surface are part of the environment.

VISUALIZE The sketch in FIGURE 7.16 establishes a coordinate system, lists the known information, and—new to problems of interacting objects—identifies the acceleration constraint. As long as the crate doesn’t slip, it must accelerate with the truck. Both accelerations are in the positive x-direction, so the acceleration constraint in this problem is $a_{\text{C}x}=a_{\text{T}x}=a_x$.

FIGURE 7.16 Pictorial representation of the crate and truck in Example 7.4.

The interaction diagram of Figure 7.16 shows the crate interacting twice with the truck—a friction force parallel to the surface of the truck bed and a normal force perpendicular to this surface. The truck interacts similarly with the road surface, but notice that the crate does not interact with the ground; there’s no contact between them. The two interactions within the system are each an action/reaction pair, so this is a total of four forces. You can also see four external forces crossing the system boundary, so the free-body diagrams should show a total of eight forces.

Finally, the interaction information is transferred to the free-body diagrams, where we see friction between the crate and truck as an action/reaction pair and the normal forces (the truck pushes up on the crate, the crate pushes down on the truck) as another action/reaction pair. It’s easy to overlook forces such as $\vec{f}{}_{\text{C on T}}$, but you won’t make this mistake if you first identify action/reaction pairs on an interaction diagram. Note that $\vec{f}{}_{\text{C on T}}$ and $\vec{f}{}_{\text{T on C}}$ are static friction forces because they are forces that prevent slipping; force $\vec{f}{}_{\text{T on C}}$ must point forward to prevent the crate from sliding out the back of the truck.

SOLVE Now we’re ready to write Newton’s second law. For the crate:

For the truck:

Be sure you agree with all the signs, which are based on the free-body diagrams. The net force in the y-direction is zero because there’s no motion in the y-direction. It may seem like a lot of effort to write all the subscripts, but it is very important in problems with more than one object.

Notice that we’ve already used the acceleration constraint $a_{\text{C}x}=a_{\text{T}x}=a_x$. Another important piece of information is Newton’s third law, which tells us that $f_{\text{C on T}}=f_{\text{T on C}}$ and $n_{\text{C on T}}=n_{\text{T on C}}$. Finally, we know that the maximum value of $F_{\text{T}}$ will occur when the static friction on the crate reaches its maximum value:

The friction depends on the normal force on the crate, not the normal force on the truck.

Now we can assemble all the pieces. From the y-equation of the crate, $n_{\text{T on C}}=m_{\text{C}}g$. Thus

Using this in the x-equation of the crate, we find that the acceleration is

This is the crate’s maximum acceleration without slipping. Now use this acceleration and the fact that $f_{\text{C on T}}=f_{\text{T on C}}={\mu }_s{m}_{\text{C}}g$ in the x-equation of the truck to find

Solving for $F_{\text{T}}$, we find the maximum propulsion without the crate sliding is

REVIEW This is a hard result to assess. Few of us have any intuition about the size of forces that propel cars and trucks. Even so, the fact that the forward force on the truck is a significant fraction $(80\%)$ of the combined weight of the truck and the crate seems plausible. We might have been suspicious if $F_{\text{T}}$ had been only a tiny fraction of the weight or much greater than the weight.

As you can see, there are many equations and many pieces of information to keep track of when solving a problem of interacting objects. These problems are not inherently harder than the problems you learned to solve in Chapter 6, but they do require a high level of organization. Using the systematic approach of the problem-solving strategy will help you solve similar problems successfully.