(a) What is the minimum force of friction required to hold thesystem of the figu

Question

(a) What is the minimum force of friction required to hold thesystem of the figure below in equilibrium? Letw1 = 140 N andw2 = 53.0 N.


(b) What coefficient of static friction between the 140 N block and the table ensures equilibrium?


(c) If the coefficient of kinetic friction between the 140 N block and the table is 0.227142857142857, what hanging weight should replacethe 53.0 N weight to allow the system tomove at a constant speed once it is set in motion?

I have tried using some of the equations in mybook but to be honest I'm not even quite sure of what equation touse. Help appreciated! Thanks!

Explanation / Answer

w1 = 140 N,w2 = 53.0 N,k = 0.227142857142857, a) friction force f = tension of the rope = w2 =53.0 N b) s = f/w1 = 53.0/140 =0.379 c) w2 = tension of the rope = kinetic frictionforce = kw1 = 31.8 N

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