# Homework Help: Tension between 2 Masses

1. Mar 15, 2009

### tascja

1. The problem statement, all variables and given/known data
Two objects of masses M and m are attached to one another with a string and pulley. Object M sits on the horizontal surface, which allows for frictionless motion. The second object m is hanging over the edge of the table. The string tied to the two objects is massless and passes over a massless pulley that rotates without friction. If M=4.53 kg and m=1.89 kg, what is the magnitude of the acceleration of the sliding object?

2. Relevant equations
Fnet = ma

3. The attempt at a solution
I know that the tension will be uniform across the entire length of the string. But im not sure how to write out the equations for everything. Im also confused about the one object moving in the y direction while the other moves in the x direction.. An explanation about how to set up the equations would be greatly appreciated!

2. Mar 15, 2009

### LowlyPion

What force is acting on the horizontally movable block?

And what forces are acting on the vertically mobile block.

Draw a diagram of the 2 in isolation and recall that the Tension in both pictures must be the same.

3. Mar 15, 2009

### tascja

So for the block sitting on the table it has the following forces acting on it:
Normal force
Tension
Gravity
* the normal force and gravity must equal 0 because it is not moving in the y direction, thus Fnet = T
For the object that is in the air, the following forces act on it:
Tension
Gravity
* no forces in the x direction, thus Fnet = T - mg

Newtons 2 law: Fnet = ma
so do i just isolate T in one of the equations to plug into the other?
and do i have to add both masses together when looking at the hanging block?

4. Mar 15, 2009

### LowlyPion

Good.
What is the acceleration of the system then?

What is the force acting on both masses.

5. Mar 15, 2009

### tascja

so gravity is the only force acting as an acceleration to the system...?

tension...?

6. Mar 15, 2009

### xxChrisxx

Why would you think you would add the masses together for the hanging block?

7. Mar 15, 2009

### tascja

because isnt it like the hanging object is pulling the object on the table? ...

8. Mar 15, 2009

### LowlyPion

The force of the hanging block is your only force on the system. Mv*g = (Mv + Mh)*a where Mv is vertical and Mh is horizontal.

Your tension is then Mh*a, but it is internal to the system.

9. Mar 15, 2009

### tascja

ok so to get that formula: Mv*g = (Mv + Mh)*a
did you just add the Fnet of each object together?

10. Mar 15, 2009

### LowlyPion

Not really. I just looked at it.

∑ Ext Forces = ∑ mass * acceleration

It's true for this case anyway.

11. Mar 15, 2009

### tascja

Will that generally be true for systems where there are 2 or more masses attached by a string?

12. Mar 15, 2009

### xxChrisxx

Ok this has got me confused now. Whats all this adding masses got to do with anything.

You want the acceleration for block M, F=ma for it. You find the T=mg from the other FBD. Plug T=mg into your FBD for the block M and get out a.

EDIT: If im missing something here, be kind and dont make me look all silly.

13. Mar 15, 2009

### tascja

well i think it is because both blocks will have the same acceleration.. and the only force acting to accelerate the 2 blocks is Fg for the hanging block. and i believe because this force is pulling both of the objects, both masses need to be added together when you find it...

14. Mar 15, 2009

### tascja

im not sure if that really makes sense.. lol

15. Mar 15, 2009

### xxChrisxx

They wont have the same acceleration as they have different masses. Fg is only accelerating the hanging block, this then provides a force (T) to the stationary block.

The frictionless pully allows you to draw two seperate free body diagrams with T being the same in both.

16. Mar 15, 2009

### LowlyPion

No. It's true here because the two masses are effectively joined as one without any other complicating factors, like the other mass having other forces acting on it like a ramp or a double pulley or any other hellish invention like an anchored or free-acting pulley that creates mechanical advantages. So no it is not a general equation.

17. Mar 15, 2009

### LowlyPion

Actually you do have the same acceleration. You must. The string doesn't change length. In this case then the weight of the hanging block is the motive force to be accelerating both blocks - together as one.

Making separate drawings is the correct method. I took a shortcut that I can see is not necessarily useful for understanding.

18. Mar 15, 2009

### xxChrisxx

Im not convinced.

Edit: god knows how a statics problem can have me confused. must be time for bed.

19. Mar 15, 2009

### LowlyPion

Work it out then. T eliminates readily enough.

20. Mar 15, 2009

### xxChrisxx

I forgot about the system acceleration, knew there was something utterly wrong with the way I was thinking. Looks like its time for me to refresh basic statics.

Definately time for bed then.

21. Mar 15, 2009

### LowlyPion

Good. Let's put the problem to bed then too.