Max Force Applied to Sled Before Slipping: 70N

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SUMMARY

The maximum force that can be applied to a sled with a block weighing 70N on top, before the block slips, is determined by analyzing the frictional forces involved. The coefficient of friction between the block and the sled is 0.700, while the coefficient between the sled and the ground is 0.100. The total weight of the system is 130N (70N block + 60N sled). The maximum static friction force between the block and sled is 49N (70N * 0.700), and the maximum static friction force between the sled and the ground is 6N (60N * 0.100). Therefore, the maximum force that can be applied before slipping occurs is limited by the lower friction force, which is 6N.

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stevengates45
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Im lost on this one & any help whatsoever would be more than appreciated... Here is the problem...

You have a block that weighs 70N that rests on a sled that weighs 60N. The coefficient of friction between the box & the sled is .700 & the coefficient of friction between the sled & the ground is .100. So what can the max force applied be before the top box slips?
 
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Consider the sum of the forces acting on the combined system (sled and block) and then the sum of the forces acting on the block.
 
Thanks... Appreciate it...
 

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