Homework Help: 3 Blocks and a Pulley

1. Jun 15, 2013

postfan

1. The problem statement, all variables and given/known data

This is a 2-part question

A block of mass m1 is on top of a block of mass m2. Block 2 is connected by an ideal rope passing through a pulley to a block of unknown mass m3 as shown. The pulley is massless and frictionless. There is friction between block 1 and 2 and between the horizontal surface and block 2. Assume that the coefficient of kinetic friction between block 2 and the surface, mu, is equal to the coefficient of static friction between blocks 1 and 2.

(Question 1)The mass of block 3 is such that block 1 and block 2 are moving together with a given acceleration of magnitude a. What is the magnitude and the direction of the force of friction exerted by block 2 on block 1? Express your answer in terms of some or all of the variables a, m1, m2, mu , and g (acceleration due to gravity). To indicate the direction, use a minus sign if the force is pointing to the left.

(Question 2)What is the minimum value of m3 for which block 1 will start to move relative to block 2? Express your answer in terms of some or all of the variables m1, m2, \mu, and g.

2. Relevant equations

F=ma

3. The attempt at a solution

For part 1 I tried to find the force of friction by finding the force that opposes it (the force of motion) and putting a negative sign on it because friction opposes motion. I ended up getting F=mu*(m1+m2)*g

For part 2 I rearranged Newton's 2nd law to get m=F/a. I then found the force of the system to be mu*(m1+m2)*g and the acceleration to be g. Using the modified Newton's second, I got (m1+m2)*mu.

I know the answer to both of these is wrong because I am taking an online course (so I know whether I am right or wrong instantly).

If someone could take his/her time to help me get the right answer (and more importantly how to find the right answer), I would be grateful.

Last edited: Jun 15, 2013
2. Jun 15, 2013

haruspex

Pls post your working. In particular, what are the forces on m1, and what does F=ma (horizontal) give you for m1?

3. Jun 19, 2013

postfan

I calculated the normal force (m1+m2)*g then multiplied my mu then added a negative sign (friction opposes motion).

4. Jun 21, 2013

haruspex

Where exactly does that normal force act? And where does the force of friction you calculated act? (You previously said it was between blocks 1 and 2 - did you mean that?)

5. Jun 21, 2013

postfan

Yes I did mean that. The Normal force acts in opposition of gravity on the blocks and the force of fiction acts oppostite to the motion caused by the pulley. So would the answer then be -mu*m2*g?

6. Jun 21, 2013

haruspex

You seem very confused. Try to concentrate on block 1 first. What are all the forces on it? What is it's resultant acceleration? Bear in mind that the frictional force when not sliding is not in general equal to the Nμk; it's any value up to that.