Block A has weight w_A and block B has weight w_B. Block A is positioned on the horizontal surface of a table. Block A is connected to a cord passing over an easily turned pulley that has Block B hanging down from the pulley. Once Block B is now set into downward motion, it descends at a constant speed.(adsbygoogle = window.adsbygoogle || []).push({});

a. Calculate the coefficient of kinetic friction between Block A and the tabletop.

I drew Block A's force diagram. For this part, I am assuming Block A is not in motion as yet; it is just sitting on the tabletop. (Is this correct?)

Well, F_fr = mu_K * F_N and the normal force = w_A.

F_fr = mu_k*w_A

mu_k = F_fr/w_A

But I think this is somewhat incorrect. Musn't I express mu_k with the known variables give. What is the Value for F_fr?

b. A cat, also of weight w_A, falls asleep on top of block A. If Block B is now set into downward motion, what is its acceleration (magnitude and direction)? Express a in w_A, w_B, and g.

Well, I am unsure of the F_fr expression. This is what I did for this part.

For Block A, the expression is now F_T - F_fr = [(2*w_A)/g]*a

For Block B, ---------------------w_B - F_T = (w_B/g)*a

Adding these two expressions, I get

w_B - F_fr = [(2*w_A)/g]*a + (w_B/g)*a

w_B - F_fr = a[(2*w_A + w_B)/g]

a = [w_B - F_fr]/[(2*w_A + w_B)/g]

a = g*[(w_B-F_fr)/(2*w_A + w_B)]

But I don't know what to put for F_fr as noted in the first part.

Are my reasonings and math that I did do correct?

Thanks.

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# Homework Help: Forces and Pulley System

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