MODEL This situation can be modeled as dynamics with constant force—one of the forces being friction. Notice that this is a two-part problem: first while Carol is pushing the box, then as it slides after she releases it.

VISUALIZE This is a fairly complex situation, one that calls for careful visualization. FIGURE 6.15 shows the pictorial representation both while Carol pushes, when $\vec{a}=\vec{0}$, and after she stops. We’ve placed $x=0$ at the point where she stops pushing because this is the point where the kinematics calculation for How far? will begin. Notice that each part of the motion needs its own free-body diagram. The box is moving until the very instant that the problem ends, so only kinetic friction is relevant.

FIGURE 6.15 Pictorial representation of a box sliding across a floor.

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SOLVE We’ll start by finding how hard Carol has to push to keep the box moving at a steady speed. The box is in equilibrium $(\text{constant}\text{velocity,}\vec{a}=\vec{0})$, and Newton’s second law is

where we’ve noted from the free-body diagram that $\vec{F}{}_{\text{push}}$ and $\vec{f}{}_{\text{k}}$ have only x-components, $\vec{n}$ and $\vec{F}{}_{\text{G}}$ have only y-components, and we’ve used $F_{\text{G}}=mg$ for the gravitational force. The negative sign occurs in the first equation because $\vec{f}{}_{\text{k}}$ points to the left and thus the component is negative: $(f_{\text{k}}){}_x=-f_{\text{k}}$. Similarly, $(F_{\text{G}}){}_y=-F_{\text{G}}$ because the gravitational force vector—with magnitude mg—points down. In addition to Newton’s laws, we also have our model of kinetic friction:

Altogether we have three simultaneous equations in the three unknowns $F_{\text{push}}$, $f_{\text{k}}$, and n. Fortunately, these equations are easy to solve. The y-component of Newton’s second law tells us that $n=mg$. We can then find the friction force to be

We substitute this into the x-component of the second law, giving

Carol pushes this hard to keep the box moving at a steady speed.

The box is not in equilibrium after Carol stops pushing it. Our strategy for the second half of the problem is to use Newton’s second law to find the acceleration, then use constant-acceleration kinematics to find how far the box moves before stopping. We know from the motion diagram that $a_y=0$. Newton’s second law, applied to the second free-body diagram of Figure 6.15, is

We also have our model of friction,

We see from the y-component equation that $n=mg$, and thus $f_{\text{k}}={\mu }_{\text{k}}mg$. Using this in the x-component equation gives

This is easily solved to find the box’s acceleration:

The acceleration component $a_x$ is negative because the acceleration vector $\vec{a}$ points to the left, as we see from the motion diagram.

Now we are left with a problem of constant-acceleration kinematics. We are interested in a distance, rather than a time interval, so the easiest way to proceed is

from which the distance that the box slides is

REVIEW Carol was pushing at $2\text{m}\text{/}\text{s}\approx 4\text{mph}$, which is fairly fast. The box slides 1.0 m, which is slightly over 3 feet. That sounds reasonable.

MODEL Model the file cabinet as a particle in equilibrium. We’ll also use the model of static friction. The file cabinet will slip when the static friction force reaches its maximum value $f_{\text{s max}}$.

VISUALIZE FIGURE 6.16 shows the pictorial representation when the truck bed is tilted at angle $\theta$. We can make the analysis easier if we tilt the coordinate system to match the bed of the truck.

FIGURE 6.16 The pictorial representation of a file cabinet in a tilted dump truck.

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SOLVE The file cabinet is in equilibrium. Newton’s second law is

From the free-body diagram we see that $f_{\text{s}}$ has only a negative x-component and that $n$ has only a positive y-component. The gravitational force vector can be written $\vec{F}{}_{\text{G}}=+F_{\text{G}}\text{s}\text{i}\text{n}\theta î-F_{\text{G}}\text{c}\text{o}\text{s}\theta ĵ$, so $\vec{F}{}_{\text{G}}$ has both x- and y-components in this coordinate system. Thus the second law becomes

where we’ve used $F_{\text{G}}=mg$.

You might be tempted to solve the y-component equation for $n$, then to use Equation 6.12 to calculate the static friction force as ${\mu }_{\text{s}}n$. But Equation 6.12 does not say $f_s={\mu }_{s}n.$ Equation 6.12 gives only the maximum possible static friction force $f_{\text{s max}}$, the point at which the object slips. In nearly all situations, the actual static friction force is less than $f_{\text{s max}}$. In this problem, we can use the x-component equation—which tells us that static friction has to exactly balance the component of the gravitational force along the incline—to find the size of the static friction force:

Slipping occurs when the static friction reaches its maximum value

From the y-component of Newton’s law we see that $n=mg\text{cos}\theta$. Consequently,

Substituting this into the x-component of the first law gives

The mg in both terms cancels, and we find

REVIEW Steel doesn’t slide all that well on unlubricated steel, so a fairly large angle is not surprising. The answer seems reasonable.