# Homework Help: Two-Mass Pulley System

1. Jul 15, 2017

### Neolight

1. The problem statement, all variables and given/known data
In the figure shown if the system is released from rest, acceleration of block B will be
(a). g/2
(b). g/3
(c). 2g/3
(d). g/7
2. Relevant equations
ma= F[net]
taking the direction of motion as positive

3. The attempt at a solution

So my attempt was to first find tension in the string connected to Mass 3M from
3Ma= 3Mg-T
T= 3M(g-a)

now the tension in the second string connected to the M say T1 is half of T ( am i correct?)
so
T1= T/2= 3Mg/2 - 3Ma/2

now equation of motion for the mass M

Ma= T1-Mg
Ma= M(3g/2 -3a/2 -g)

a = 3g/2 - g - 3a/2
a + 3a/2 = 3g/2 -g
5a/2= g/2
∴ a= g/5

where did i go wrong , because i looked up the answer and it is g/7

2. Jul 15, 2017

### Vibhor

Acceleration of block B is not 'a' i.e it is not equal to that of block A .

But they have a simple relationship .

3. Jul 15, 2017

### Neolight

can you please tell me the relationship between the two accelerations?

4. Jul 15, 2017

### Vibhor

If the lower pulley goes down by a distance 'x' how much does block B go down ?

5. Jul 15, 2017

### Neolight

since the string isn't directly connected to the pulley so they won't travel the same distance, the string goes along the circumference of the pulley, this is as far as i can think sorry i'm really bad at this type of thinking

6. Jul 15, 2017

### Vibhor

No problem .

Suppose you hold block B i.e you do not allow it to move .Now let lower pulley go down by distance 'x' , how much string length of string whose one end is fixed at ground and other end connects block B , goes slack/loose ?

Hint: Carefully look at both the left and right parts of string going over lower pulley .

7. Jul 15, 2017

### Neolight

its half the circumference (πr) so if the pulley moves up say x distance the mass B will move πr +x ?

8. Jul 15, 2017

### Vibhor

No.

You do not need to worry about the part that lies on the pulley circumference . Why ?
Because before pulley moves down , πr length is over the pulley and after pulley moves down , same length πr is on the pulley circumference , so net change in the length of string as far as part that lies on the string is zero .

So forget about the part of string that lies on pulley circumference .

Just look at left and right parts of string .Pick a pen and paper and make a rough sketch . Without moving block B , move lower pulley down by a distance 'x' .

How much string length of left part gets loosened ?

How much string length of right part gets loosened ?

What is the total length of string that gets slack/loosened ?

9. Jul 15, 2017

### Neolight

hahah i get it now 2x

so acceleration of block B will be two times the accelaration of the pulley

i have done the calculation using this result and now the answer is g/7 ...... thanks for your help , i really appreciate it

10. Jul 15, 2017

### Vibhor

Well done !

11. Jul 15, 2017

### TSny

Is this what you get for the acceleration of block B, or is it the acceleration of block A?

12. Jul 15, 2017

### Neolight

It's for block B

13. Jul 15, 2017

### TSny

Maybe I'm making a mistake, but I get that block A has the acceleration of g/7.
Do you agree with the following equations?

3Mg - TA = 3MaA

TB - Mg = MaB

TA = 2TB

aA = aB / 2

14. Jul 15, 2017

### Neolight

Are you taking the acceleration of the block A and B as the same?

15. Jul 15, 2017

### TSny

No. See the last equation I wrote.

16. Jul 15, 2017

### Neolight

The acceleration of block B will be two times that of block A while tension of Block B will be half that of block A

17. Jul 15, 2017

### TSny

Yes, that agrees with the 3rd and 4th equations that I wrote.

18. Jul 15, 2017

### Neolight

I have done the math again ... And the result is

Acceleration for block A is g/7 and for B is 2g/7

Correct?

I made a mistake in a my previous math

19. Jul 15, 2017

### TSny

OK, good. That's what I get, too.