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Part A

A 15 kg box is sitting in the bed of a pickup truck. The truck begins to accelerate at a constant rate of 3.5 m/s². Given that the friction between the box and the truck bed is characterized by a coefficient of kinetic friction of 0.25 and a coefficient of static friction equal to 0.40, what is the magnitude of the friction force acting on the box once the truck begins its acceleration?

Solution

We begin with a sketch that represents the situation, and then create the appropriate free body diagram.

Using the free-body diagram, we construct the equations of Newton's Second Law applied to the box:

\begin{large}\[\sum F_{x} = F_{f} = ma_{x} \]\[\sum F_{y} = N - mg = ma_{y} \] \end{large}

Since the truck is moving only in the x-direction, we expect a_y = 0. Thus, we know that the normal force acting on the box will equal its weight. If friction is adequate, we expect that the box will accelerate in the x-direction at the same rate as the truck does. In that case, we expect:

\begin{large}\[ F_{f} = (15\:{\rm kg})(3.5\:{\rm m/s}^{2}) = 53\:{\rm N} \]\end{large}

It is important now to check that this result does not conflict with the requirement that

\begin{large} \[ F_{f} \le \mu_{s} N\] \end{large}

Since we have already used the y-direction equation of Newton's Second Law to conclude that the normal force on the box is equal to mg, we find:

\begin{large} \[ F_{f} = 53\:{\rm N} < \mu_{s} N = 0.40(15\:{\rm kg})(9.8\:{\rm m/s}^{2}) = 59\:{\rm N} \]\end{large}

We therefore conclude that our answer, F_f = 53 N, is compatible with the static friction limit.

Part B

Consider the same basic situation as above, but now suppose that the truck accelerates at a rate of 4.0 m/s² rather than 3.5 m/s². If the center of mass of the box is located 2.0 m horizontally from the edge of the truck bed, how much time will elapse from the instant the truck begins to accelerate until the instant the box falls off the truck bed?

Solution