: Approach: Diagrammatic RepresentationWe begin with a sketch that represents the situation, and then create the appropriate free body diagram.  Image Added
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It is important to note that friction works to prevent movement along the interface between the box and the truck bed. The truck bed is moving forward, so friction will attempt to pull the box forward as well. If the box moves at the same rate as bed, then the interface is static. For this reason, "static" friction will actually cause motion of the box in this case! |
Mathematical RepresentationUsing the free-body diagram, we construct the equations of Newton's Second Law applied to the box: | Latex |
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\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 ay = 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: | Latex |
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\begin{large}\[ F_{f} = (15\:{\rm kg})(3.5\:{\rm m/s}^{2}) = 53\:{\rm N} \]\end{large} |
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We're not finished yet! |
It is important now to check that this result does not conflict with the requirement that | Latex |
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\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: | Latex |
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\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, Ff = 53 N, is compatible with the static friction limit. |