# Pulley Question

1. Oct 21, 2006

### ubiquinone

Hi, I have a forces problem involving a pulley, I think I'm almost there. I was wondering if anyone may please give me some suggestions on how to solve this. Thank you.
Diagram:
Code (Text):

_____
|     |
| A   |______________
_|_____|______________O\
/  |
|  |
| _|_
||   | B
||___|
|  |
| _|_
||   | C
||   |
||___|

Question: Three boxes are connected by cords, one of which wraps over a pulley having negligible friction on its axle and negligible mass. The masses are $$m_A=30.0kg$$, $$m_B=40.0kg$$, and $$m_C=10.0kg$$. When the assembly is released from rest, what is the tension in the cord connecting $$B$$ and $$C$$.
I've started the problem by treating mass B and C as one big mass and drawing free body diagrams.
For mass A: $$F_{net}=F_{T_{sys}}=m_Aa_{sys}=30a_{sys}$$
$$a_{sys}=\frac{F_{T_{sys}}}{30}$$ (1)
For the "big mass" (mass B + mass C):
$$F_{net}=F_g-F_{T_{sys}}=50a_{sys}$$
$$a_{sys}=\frac{50g-F_{T_{sys}}}{50}$$ (2)
Solving for $$F_{T_{sys}}=183.75N$$ and $$a_{sys}=6.125m/s^2$$

Now how can I used this information to find the tension between mass B and C?

2. Oct 21, 2006

### PhanthomJay

Draw the FBD for block C alone.

Note: in your calculations, you didn't solve for the system tension, you solved for the tension in the rope wrapped around the pulley. There is no 'system' tension as such.

Last edited: Oct 21, 2006
3. Oct 22, 2006

### kcirick

FBD for all the blocks. Not just C. Just remember that tension in the string between A and B is the same in the FBD for A and B, and the tension between B and C are the same in the FBD for B and C.

Also remember that the system moves uniformly (i.e. The block A, B anc D all have the same velocity).

P.S. Nice drawing using normal characters. I like it :)

4. Oct 22, 2006

### ubiquinone

Hi thanks for replying guys, so would this work?
The net force acting on mass B and mass C is $$F_{net}=(40kg+10kg)(6.125N/kg)=306.25N$$
By drawing a free body diagram for mass B and mass C, the forces acting on it is, force of tension, force between B and C acting upwards and weight acting down.
Therefore, $$F_{net}=F_T+F_{T_{BC}}-F_g\Rightarrow F_{net}+F_g-F_T=F_{T_{BC}}$$
$$F_{T_{BC}}=306.25N+50g-183.75N=612.5N$$

5. Oct 22, 2006

### Hootenanny

Staff Emeritus
No, not quite. Start by setting up to equations, one for block A block B&C, thus;

$$(m_{b}+m_{c})g - T = (m_{b}+m_{c})a$$

$$T = (m_{a})a$$

I am assuming here that the table is frictionless. Can you see where the above to equations come from?

Now you can solve for T (knowing that the acceleration is uniform).

6. Oct 22, 2006

### ubiquinone

Hi Hootenanny! Thanks for answering back to my question, so the tension in the cord connecting mass B and mass C is the same as the tension pulling mass A to the right?

Because I think I've set up the two equations that you have and found the tension force to be $$183.75N$$

Last edited: Oct 22, 2006
7. Oct 22, 2006

### Staff: Mentor

No. In Hoot's equation, T stands for the tension in the cord connecting A and B, which is different from the tension in the cord connecting B and C.

In your first post you correctly calculated the acceleration of all the masses. Now just apply Newton's 2nd law to mass C alone--using that acceleration--to find the tension that you need.

8. Oct 22, 2006

### ubiquinone

After finding the acceleration of all three masses, $$a=6.125m/s^2$$
The net force on mass C is given by, $$F_{net}=(10kg)(6.125N/kg)=61.25N$$
The forces acting on mass C are: (1)Force of tension between mass B and C (2)the weight of mass C
Thus, $$F_{T_{B,C}}-98N=61.25N\Rightarrow F_{T_{B,C}}=159.25N$$

Is it correct now?

9. Oct 22, 2006

### Hootenanny

Staff Emeritus
Looks good to me