Blocks, Tension, Acceleration, Friction Problem

[trc214]

Homework Statement

Block B, with mass 5.00 kg, rests on block A, with mass 8.00 kg, which in turn is on a horizontal tabletop. There is no friction between block A and the tabletop, but the coefficient of static friction between block A and block B is 0.750. A light string attached to block A passes over a frictionless, massless pulley, and block C is suspended from the other end of the string. What is the largest mass that block C can have so that blocks A and B still slide together when the system is released from rest?

Homework Equations

f=ma
coefficient of static friction = Fmax/normal force

The Attempt at a Solution

At first I took found the weights of blocks A and B.

[w Block A = 8.00kg(9.8m/s^2)] = 78.4 N
[w Block B = 5.00(9.8m/s^2)] = 49 N

w of A+B = 78.4+49= 127.4 N

I calculated that the normal force of Blocks A+B = 127.4 N. Then since the weight of block C which is unknown would equal the tension in the rope, I figured I would use this information using the coefficient of friction to find the mass of block C.

So then I set 0.750 which was the coefficient of friction = to fmax/normal force.

0.750 = Fmax/127.4

I found the Fmax=36.75 N

Then I divided 36.75 N by g to get the mass of block C. I got the mass of block C to be 3.75 kg. Apparently this is incorrect and the correct answer is 9.75 kg.

Can anyone pease help me? Thank you very much.

 

[PhanthomJay]

Homework Statement

Block B, with mass 5.00 kg, rests on block A, with mass 8.00 kg, which in turn is on a horizontal tabletop. There is no friction between block A and the tabletop, but the coefficient of static friction between block A and block B is 0.750. A light string attached to block A passes over a frictionless, massless pulley, and block C is suspended from the other end of the string. What is the largest mass that block C can have so that blocks A and B still slide together when the system is released from rest?

Homework Equations

f=ma
coefficient of static friction = Fmax/normal force

The Attempt at a Solution

At first I took found the weights of blocks A and B.

[w Block A = 8.00kg(9.8m/s^2)] = 78.4 N
[w Block B = 5.00(9.8m/s^2)] = 49 N

w of A+B = 78.4+49= 127.4 N

I calculated that the normal force of Blocks A+B = 127.4 N.

This is the normal force of the floor on Block A

Then since the weight of block C which is unknown would equal the tension in the rope, I figured I would use this information using the coefficient of friction to find the mass of block C.

It would only equal the rope tension if it was at rest or moving at constant velocity. Is it?

So then I set 0.750 which was the coefficient of friction = to fmax/normal force.

0.750 = Fmax/127.4

I found the Fmax=36.75 N

127.4 N is not the normal force of A on B

Then I divided 36.75 N by g to get the mass of block C. I got the mass of block C to be 3.75 kg. Apparently this is incorrect and the correct answer is 9.75 kg.

Can anyone pease help me? Thank you very much.

It may be late, but I don't get the book answer. Draw free body diagrams to identify the forces on each mass, and use Newton's laws. All blocks are accelearting at the same magnitude.