# Pulley and blocks

1. Sep 30, 2007

### bsmith2000

A friend of mine gave me this problem, and it was somewhat hard:

A rectangular block of mass M1 rests on a horizontal table. Masses M2 and M3 are free to slide along the surfaces of M1 and are attached by a massless string over a frictionless pulley.

For the diagram, go to: http://img451.imageshack.us/img451/7456/28142628dx2.jpg[/PLAIN] [Broken]

Also, assume all surfaces are frictionless.

a) Draw free body diagrams for M1 and M2. Assume the pulley makes a 45 degree angle with the horizontal.

b) Find the external force F that should be applied to M1 so that M3 will stay at a fixed height above the table. (meaning M2 and M3 dont move relative to M1)

c) Suppose there is no external force. All masses are held fixed up until time t=0, and then they are released. Find the initial tension in the string T, and the initial acceleration a of M1 just after being released. Express the answer in terms of the masses given and g.

My thoughts so far:

For a), on M1, I drew the normal force it had with the table and the gravitational force. Do I need to draw some sort of normal force for the contact it has with M3? The M2 diagram has its normal force with M1, gravity, and the tension force to the right.

For b), I thought the force would be M3*g, because that is the tension force, so applying that force would somehow counter it? It is somewhat of a wild guess, and I am pretty sure it is incorrect, b/c that seems too easy.

For c), I don't see how how M1 would even accelerate, nor do I see how the "initial" tension would ever change after time.

Last edited by a moderator: May 3, 2017
2. Sep 30, 2007

### bsmith2000

Whoops - sorry, I didn't know we had to place questions in a separate forum.

Anyone have any insight to this problem?

3. Oct 1, 2007

### pgraves013

for part b, x: (M1+M2+M3)(a)=F_x. solve for a. Since you want it to remain stationary, a_y=0=T-M3g. (M2)a_x=T. So then you have the acceleration you solved for multiplied by M2 and that will equal M3g. Do algebra to get F= M3g(M3+M2+M1)/M2.

4. Oct 1, 2007

### pgraves013

for part a, you dont have to draw a force for the contact. it is assumed that friction is negligible and won't affect it.

for part c, the tension will remain the same seeing as though the length of the string and the masses do not change. Therefore, you are correct and the "initial T" will be the same until the block reaches the ground. So, T=M2M3g/(M2+M3)

5. Oct 1, 2007

### bsmith2000

Thank you so much pgraves013!! That makes so much sense!

Do you know how to find the initial acceleration of mass M1 just after the masses are released? (there is no external force during that part)

Last edited: Oct 1, 2007
6. Oct 1, 2007

### FedEx

I think that the application of pseudo force would make it even easier.

Are you familiar with pseudo forces?

7. Oct 1, 2007

### pgraves013

since the blocks are all connected in one way or another, i think that the acceleration of each would be the same. im really not sure though, but i think it may be a=M3g/(M1+M2+M3). i'm not 100% sure, but i think that is correct

8. Oct 31, 2010

### Kevii

Hi, I need help with this similar problem and to make things short, I also found that

a = m3g/m2 from part b which states: Find the force F by ONLY using the seperate free body diagrams of M2, M3, and M1+M2+M3.

so for M1+M2+m3 F=(M3g/M2)(M1+M2+M3)

but for my part c it says: Find the force F by ONLY using the seperate free body diagrams of m1, m2, and m3. (The force from part b and c should agree! Make sure your answer is in terms of m1, m2, m3, and g)

And when it comes to this I found that F=(M1+M3)(M3g/M2) . . . . :(