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# Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slide forward, crushing the cab, if the truck stops suddenly in an accident or even in braking. Assume, for example, that a $$10 000kg$$ load sits on the flatbed of a $$20 000kg$$ truck moving at $$12.0 m/s$$. Assume that the load is not tied down to the truck, but has a coefficient of friction of $$0.500$$ with the flatbed of the truck. (a) Calculate the minimum stopping distance for which the load will not slide forward relative to the truck. (b) Is any piece of data unnecessary for the solution.

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#### If the load is on the point of sliding forward on the bed of the slowing truck, static friction acts backward on the load with its maximum value, to give it the same acceleration as the truck:$$\sum F_{x} = ma_{x} : − f_{s} = m_{load}a_{x}$$$$\sum F_{y} = ma_{y} : n − m_{load}g = 0$$Solving for the normal force and substituting into the $$x$$ equation gives:$$-\mu_{s}m_{load}g = m_{load}a_{x}$$ or $$a_{x} = −\mu_{s}g$$We can then use$$v_{xf}^{2}=v_{xi}^{2}+2a_{x}(x_{f}-x_{i})$$Which becomes$$0=v_{xi}^{2}+2(-\mu_{s}g)(x_{f}-0)$$(a) $$x_{f}=\frac{v_{xi}^{2}}{2\mu_{s}g}=\frac{(12.0m/s)^{2}}{2(0.500)(9.80m/s^{2})}=14.7m$$(b) From expression $$x_{f}=\frac{v_{xi}^{2}}{2\mu_{s}g}$$ neither mass affects the answer.

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