# Solving a Pulley Forces Problem: Tension in B and C

• ubiquinone
In summary, the tension in the cord connecting mass B and mass C is the same as the tension pulling mass A to the right.
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:
  _____
|     |
| 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?

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:
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 :)

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, $$F_{net}=F_T+F_{T_{BC}}-F_g\Rightarrow F_{net}+F_g-F_T=F_{T_{BC}}$$
$$F_{T_{BC}}=306.25N+50g-183.75N=612.5N$$

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).

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:
ubiquinone said:
...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.

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?

ubiquinone said:
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

## What is the purpose of solving a pulley forces problem?

The purpose of solving a pulley forces problem is to determine the tension in each segment of the rope or cable that is connected to the pulley. This information can then be used to analyze the mechanical forces at play and ensure that the system is functioning properly.

## How do you calculate the tension in segments B and C?

To calculate the tension in segments B and C, you need to use the formula T = m x a, where T is the tension, m is the mass of the object being lifted or moved, and a is the acceleration of the object. You also need to take into account the direction of the forces and the angle of the segments in relation to the pulley.

## What factors can affect the tension in segments B and C?

The tension in segments B and C can be affected by several factors, including the weight of the object being lifted or moved, the angle and direction of the segments, and the friction and resistance in the pulley system. Other external forces, such as wind or vibrations, can also impact the tension.

## How can you ensure accuracy when solving a pulley forces problem?

To ensure accuracy when solving a pulley forces problem, it is important to carefully identify and label all forces acting on the system. This includes the weight of the object, the tension in each segment, and any external forces. It is also crucial to use the correct formulas and equations and double-check all calculations.

## What are some real-life applications of solving pulley forces problems?

Solving pulley forces problems has many real-life applications, such as in engineering and construction, where pulley systems are often used to lift and move heavy objects. It is also used in physics and mechanics to analyze the forces at play in various systems. Other applications include rock climbing, sailing, and weightlifting.

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