What is the Coefficient of Kinetic Friction and Acceleration in a Pulley System?

AI Thread Summary
The discussion revolves around calculating the coefficient of kinetic friction and the acceleration in a pulley system involving two blocks, A and B. Block A rests on a table while Block B hangs and descends at a constant speed, prompting questions about the forces involved. The coefficient of kinetic friction (mu_k) is expressed as F_fr/w_A, but there is confusion about determining F_fr since Block B's constant speed suggests no net force acting on the system. The equations derived for acceleration incorporate both blocks' weights and friction, but the participants struggle with defining F_fr accurately. Overall, the conversation highlights the complexities of applying Newton's laws in this scenario.
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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.

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? :confused:

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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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.

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?
One may assume block A has zero velocity.

To find F_fr, what is F_fr opposing? What is the significance of Block B descending at constant speed?
 
F_fr = w_B, but when I plug this value into the expression for a that I found, acceleration would be 0. According to Newton's First Law, a body that has constant velocity and, therefore, 0 acceleration is acted on by no net forces. Where do I go from here?
 
Soaring -- didn't your last post just answer your own question? In your first post, you wanted to know what F_fr was. So, what is it?
 
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