The two blocks (m = 30 kg and M = 100 kg) in the figure below are not attached t

Question

The two blocks (m = 30 kg and M = 100 kg) in the figure below are not attached to each other. The coefficient of static friction between the blocks is ?s = 0.49, but the surface beneath the larger block is frictionless. What is the minimum magnitude of the horizontal force required to keep the smaller block from slipping down the larger block?

Explanation / Answer

Q: The two blocks (m = 18 kg and M = 91 kg) in Fig. 6-38 are not attached to each other. The coefficient of static friction between the blocks is µ_s = 0.47, but the surface beneath the larger block is frictionless. What is the minimum magnitude of the horizontal force F-> required to keep the smaller block from slipping down the larger block? ANS: The force F will accelerate the two blocks in the horizontal direction. From Newton's second law, the acceleration will be: a = F/(M+m). The small block, also accelerating at this value, experiences two horizontal forces: the force F and the normal force that the large block exerts on the small block (in the opposite direction). Let's denote this force by N. Then Newton's second law, applied to the horizontal motion of the small block only, gives: m a = F - N . Hence we have an expression for N: N = F - m a = F - m F/(m+M) = F ( 1 - m/(m+M) ) = F M/(m+M) For the small block not slipping down during the motion, we must have that µ_s N >= m g Hence N >= m g / µ_s Combining this with the earlier expression for N we have a lower bound on F: F M/(M+m) >= m g / µ_s F >= m M /(M+m) * (g / µ_s) With the given data this is F>= 18 kg * 91kg / 109 kg * 9.81 m/s^2 /0.47 F>= 313.7 N

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