Solving a Pulley Forces Problem: Tension in B and C

[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:

  _____
 |     |
 | 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?

 

[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.

 

[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 :)

 

[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

 

[Hootenanny]

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).

 

[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

 

[Doc Al]

...so the tension in the cord connecting mass B and mass C is the same as the tension pulling mass A to the right?

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.

 

[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?

 

[Hootenanny]

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?

Looks good to me