# Newtonian Mechanics: Pully/Two Block System

1. Apr 23, 2016

### crichardson17

1. The problem statement, all variables and given/known data
The blocks of mass 20.0 kg and 10.0 kg are initially at rest on the floor and are connected by a massless string passing over a massless and frictionless pulley. An upward force F is applied to the pulley.

Find the accelerations a_A of block A and a_B for block B when F is
i) 124 N
ii) 294 N
iii) 424 N

2. Relevant equations
∑F = ma
m_A = mass of block A = 20 kg
m_B = mass of block B = 10 kg
a_A = acceleration of block A
a_B = acceleration of block B
W_A = weight of block A = (20 * 9.8) = 198 N
W_B = weight of block B = (10 * 9.8) = 98 N
T = Tension

3. The attempt at a solution
Taking up to be the positive y direction and right to be the positive x direction, I've drawn free body diagrams for the pulley (F_pull in the +y direction; (W_A + W_B) in the -y direction), block A (T in the +y direction, W_A in the -y direction), and block B (T in the +y direction, W_B in the -y direction).

Also, I've stated that since there are no forces in the x direction, the net force is equal to the net force of the y-components of the system, and that since the string is massless, T is the same for both block A and block B.

I have set up the following equations:

∑F_A = T - W_A = T - 198 = m_A * a_A
∑F_B = T - W_B = T - 98 = m_B * a_A

T = (m_A * a_A) + 198 = (m_B * a_B) + 98

but I'm not sure where to go from here. I realize that, with respect to the floor, the system itself will be accelerating at different rates with the different given forces of i, ii, and iii. I also realize that blocks A and B will have the same acceleration with respect to the pulley once they are lifted from the floor, regardless of what force is pulling upwards on the pulley.

Any guidance as to what I need to do next would be very much appreciated.

2. Apr 23, 2016

### drvrm

what is T ?
what happens if F through which the pulley is being lifted is less than or equal to , or greater than the downward forces on pulley.

3. Apr 23, 2016

### crichardson17

T is the tension in the string, so...

If F is less than the downward forces, the system won't move, and the tension in the string will be equal to the force pulling upwards.
If F is equal to the downward forces, the forces will cancel each other out and the system will stay at rest. So again, the tension in the string will be equal to the force pulling upwards? I'm not sure about this one.
If F is greater than the downward forces, then the tension in the string will be equal to (m_A * a_A) + 198 = (m_B * a_B) + 98.

4. Apr 23, 2016

### drvrm

The pulley does not know what is happening to the hanging masses-- so one should draw a free body diagram of pulley
and F can be related to T...... if F is equal to down ward forces.

similarly one should draw free body diagram for the masses hanging from the pulley and write Newton's equations of motion
i.e massxacceleration = net force in the direction of motion.

think over it .....
what will happen if the person raising the pulley accelerates the pulley upwards with say some net value a(p) ?

5. Apr 23, 2016

### crichardson17

So, for the second case, where the upward force is equal to the downward force, the FBD would look like this:

F_pull = W_A + W_B

T = W_A + W_B
so then F_pull would be equal to the tension in the string, wouldn't it?

FBD for block A:
T

W_A = 196 N (using g=9.81 m/s^2)
Since A is heavier than B, A will move in the downward direction, so ∑F_A = m_A * -a_A

FBD for block B:
T

W_B = 98 N
Again, since block B is lighter than block a, it will move in the upwards direction, so ∑F_B = m_B * a_B

Finally, if the person raising the pulley accelerates it with a(p) upwards, then the blocks will be undergoing that acceleration on top of the acceleration that they are undergoing within the system, as a result of their varying weights.

6. Apr 23, 2016

### drvrm

i differ to say that F(pull on the pulley) should be 2 times T as two string sections are oulling in downward direction.

7. Apr 24, 2016

### crichardson17

Okay. That makes sense. So, following that, the downward forces (T) on the pulley will be 2(W_A + W_B)?

8. Apr 25, 2016

### Anjum S Khan

Decompose the problem, then try to add complexity one by one.

Initial decomposition reveals the problem given as attachment.

How current problem differs from attachment ?

For pulley, taking upward as $+Y\uparrow$
$F \hat j + T(-\hat j) + T(-\hat j) = m_{pulley} . \vec{a_{pulley}} = 0.\vec{a_{pulley}}$
$~~\Rightarrow~~ T = \frac{F}{2}$

Now, draw FBD of each masses separately, and find the magnitude and direction of accelerations.

Sign of acceleration is important.

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• ###### Force-to-move-weights-up.png
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Last edited: Apr 25, 2016