# Pulley + ramp problem

1. Mar 24, 2014

### ashimashi

Problem:
Block A is suspended from a vertical string. The string passes over a pulley and is attached to block B. The mass of block B is 1.8 kg. The mass of block C is 0.75 kg. If the coefficient of kinetic friction is 0.100 between block B and the ramp and the coefficient of static friction between block B and C is 0.900, determine the maximum mass of block A, such that block C does not slip off.

2. Relevant equations
f = ma

3. The attempt at a solution

I drew a FBD for block C and was able to calculate its force of friction which is: 5.61 N (μmgcosθ)

I also found out that the force of gravity for Block B is (M)(G)(COSθ). But I don't know what the force of Block C on B would be. If I could find that, I will be able to find the Force of Normal for Block B and solve for the force of friction for block B as well.

After that, I am not too sure what I must do next. I am thinking of solving for acceleration but not too sure if that will help.

Any help or any push in the right direction would be very helpful.

Thanks

Last edited: Mar 24, 2014
2. Mar 24, 2014

### BvU

Hello Ashi, and welcome to PF.
Your $\mu\, mg\,\cos\theta$ is the maximum force block B can exercise on block C without block C slipping, right ?

Can you calculate the friction force block B exercises on block C in the case that block A is so light that block C and B do not move ?

3. Mar 24, 2014

### ashimashi

Thank you BvU!

Okay I think I got it but I am not too sure:

I made a mistake on the Force of friction between block B and C. It should have been (mgμcosθ)/m where m is the mass of block c.

So force of friction must be greater than acceleration for block c to slip off.

Therefore: a = fnet/total mass = force of friction of block c

force of friction of block c = 7.48 N

total mass = M + 2.55

fnet = Mg - (2.55)(g)(0.1)(cos32) - (2.55)(g)(sin32)

acceleration = [Mg - (2.55)(g)(0.1)(cos32) - (2.55)(g)(sin32)] / [M + 255]

[Mg - (2.55)(g)(0.1)(cos32) - (2.55)(g)(sin32)] / [M + 255] = 7.48

M = 14.0KG

I believe that is the right answer but I am not 100% sure. Can someone please check my work and let me know if they believe my answer is correct.

Last edited: Mar 24, 2014
4. Mar 25, 2014

### BvU

No: $\mu\, g\,\cos\theta$ has the dimension m/s^2, an acceleration. So it cannot be a force.

You mean smaller ?

An acceleration can not be a force.

Where does the 7.48 come from ? Before, you had 5.616 as the maximum force of friction, right ? (I did ask rather explicitly).
However, with $\mu\, g\,\cos\theta =$ 7.49 m/s2 (not a force, but an acceleration) you do have a value for the maximum acceleration of block C (and it is NOT this 7.49 m/s2! (*)). And thereby a value for the maximum acceleration of all three of A, B and C.

An earlier small mistake I need to point out:
No: just mg.

So back to the drawing board: free body diagrams for C and for B !

(*) Can you calculate the friction force block B exercises on block C in the case that block A is so light that block C and B do not move ? (Didn' t I ask for that already?)