# Multiple Pulleys

1. Oct 7, 2012

### Lolagoeslala

Multiple Pulleys!!

1. The problem statement, all variables and given/known data
A 60 kg mass and a 10 kg mass are attached together by massless inextensible strings which pass over light frictionless pulleys. One pulley is free to move while the other is fixed.

a) The system is initially held at rest by means of a force supporting the 60 kg mass. Calculate the magnitude of the tensions T1 and T2.

b)When the system is released. calculate the magnitude of
1)Tension T1 ans T2
2)Aceeleration of the 10 Kg
3) Acceleration of the 60 Kg

Here is the diagram if it helps:
http://i1353.photobucket.com/albums/q674/lolagoeslala1/Untitled_zpseb420164.jpg

2. Relevant equations

Depends.

3. The attempt at a solution
I tried solving this for part a:
For Ti i did:
Fg=T1
mg=T1
(10kg)(9.8 m/s^2)=T1
98 N=T1

for T2:
Fg=T1+T2
mg=T1+T2
(60kg)(9.8m/s^2)-98N=T2
490 N=T2

2. Oct 7, 2012

Re: Multiple Pulleys!!

T1=120 N
T2=240 N
10kg....0.5g
60kg....0.25g

3. Oct 7, 2012

### Lolagoeslala

Re: Multiple Pulleys!!

Im sorry but how are you getting this tensions? I mean how did you get the T1 to be 120 while the 240 as the Tensions 2 to be 240? It says that the pulley is fixed... doesn't that mean the tensions have to be the same?

4. Oct 7, 2012

Re: Multiple Pulleys!!

you are right but the attempt is wrong....see my fig post...sorry for bad writing..hope it makes sense
http://i.imgur.com/Fon1i.jpg

5. Oct 7, 2012

### Staff: Mentor

Re: Multiple Pulleys!!

Lola, in your solution to part (a) you forgot to include the force that is supporting the 60 kg mass. There's also a relationship between T1 and T2 that you will find helpful.

6. Oct 7, 2012

### Lolagoeslala

Re: Multiple Pulleys!!

So you have to break the tensions 2 and tensions 1 into tensions 1,2,3 and 4? :O WOW. wait is that for when there is no acceleration whatsoever?,,,

7. Oct 7, 2012

### Lolagoeslala

Re: Multiple Pulleys!!

My well what i did was
Fnet=Fg+T1+T2
which means it would be going up eventually making the Fg negative.
so
Fg=T2+T1
(60 Kg)(9.8m/s^2)=T2+98 N
490 N=T2 going up..

8. Oct 7, 2012

### Staff: Mentor

Re: Multiple Pulleys!!

Part (a) is asking you for the tensions while nothing is moving because the 60 kg mass is being supported by an external force. You've calculated t1 correctly, but not t2. One way of seeing that is to try your values of t1 and t2 out on the movable pulley; in part (a) that pulley isn't supposed to be moving so the forces on it must balance.

Once you have part (a) nailed, part (b) about what happens when the supporting force is removed will be a lot easier.