A force F=50N is applied to a block of mass M=9kg from an angle =20°above the ho

Question

A force F=50N is applied to a block of mass M=9kg from an angle =20°above the horizontal direction; the block sits under a smaller block of mass m=3kg. the coefficient of friction between "M" and the surface is k=0.25 while the coefficient of friction between the two blocks is unknown. What minimun coefficient of static friction is required between two blocks in order forr "m" not to slide along the surface of "M"? Also what will be M's acceleration?

Explanation / Answer

Some of our 50N force is fighting friction. Only the leftover force accelerates the blocks. A small part of our force is pushing down, increasing the normal force. A large part is pushing horizontally. Cos(20) is larger than sin(20), so cos(20) is horizontal. F net = push - µkF norm = 50N*cos(20) - 0.25* [9.8m/s^2*(9+3)kg + 50N*sin(20)]= 13.31N Fnet = ma a = Fnet/m = 13.31N/(9kg) = 1.11m/s^2 for the 3 kg block to accelerate at 1.11m/s^2 without slipping, we need F fric >= 3kg*1.11m/s^2 µkF norm = ma µk = ma/mg = 1.11m/s^2/9.8m/s^2 = 0.11 this shouldn't be a problem, this is a low coefficient of friction.

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