# Two pulley problem

1. Jun 14, 2014

### ietr

Tried the problem but answer is not matching with the given answer. Can you tell me what is the correct answer and how you got it? (problem is attached as image)

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2. Jun 14, 2014

### ietr

The question is to find the accelerations of the two blocks

3. Jun 14, 2014

### Orodruin

Staff Emeritus
Why don't you show us what you tried and what your result was and we can point you in the correct direction? It will be much more instructive.

4. Jun 14, 2014

### ietr

Okay. I got that question now, but this one is really troubling me. This is what I've tried sofar:

First I draw FBD for each of the blocks. Applying F = MA for each block:

500 g block
----------------
500 g = 1/2 kg
The string attached to it has a tension T.
let it have an acceleration a1 downward: 1/2g - T = 1/2 (a1)

100 g block
----------------
100 g = 1/10 kg
taking the components, we have gsin(30) along the string and gcos(30) canceled by normal reaction.
Let it have an acceleration of a2 downward on the incline then
F = MA
(1/10) g sin(30) - T = (1/10) a2

50 g block
---------------
50 g = 50/1000 = 1/20 kg
let it have an acceleration a3 downward then
(1/20) g - T = (1/20) a3

Obviously there is some error in the above three equations because there are four unknowns: T, a1, a2 and a3.
I'm having trouble understanding this:
* Will the tension remain T along all segments of the string, unlike what I have assumed above?
* Since the string length is constant, the accelerations a1, a2 and a3 must be equal but if we set them all equal to 'a', then we have only two unknowns T and 'a' and three equations, and further on solving them we do not get the correct answer.

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5. Jun 14, 2014

### Orodruin

Staff Emeritus
If instead of being inclined by 30 degrees, the thread would be hanging vertically with a 100 g weight in the middle, would the tension be the same above and below the weight?

Yes, the accelerations have to be of equal magnitude - but be very mindful of their direction, if one of the ends accelerates up, the other end accelerates down.

6. Jun 14, 2014

### ietr

if a1 = a2 = a3, and we assume that their directions are:
a1 : of 500 g downwards
a2: of 100 g upwards along the incline
a3: of 50 g upwards

then the equations become (assuming that the tension is T along the length of the string)

(1/2)g - T = (1/2)a
T - (1/10)g sin(30) = (1/10)a
T - (1/20)g = (1/20)a

Solving these equations for 'a', we get a = (3/4)g from the first two equations and a = (9/11)g from the first and third equations. How can there be two values of 'a'?

7. Jun 14, 2014

### haruspex

It's not clear from your diagram how the 100g block is attached. Looks to like there are (logically, at least) two separate strings - one runs from the 500g block, over the top pulley, to the 100g block; the second runs from the 100g block, over the lower pulley, to the 50g block. There is no reason why the tensions should be the same in the two sections of string.

8. Jun 14, 2014

### Orodruin

Staff Emeritus
It does not matter much how it is attached. What matters is that it is attached (and parallel to the inlined plane).

ietr: You should also try to think a bit more about what happens for the weight on the inclined plane. What forces from the threads are acting upon it?

Also, think a bit what happens if you are in a tug of war contest. If the guy on your team who is in front of you happens to be so strong that he easily defeats the other team by himself, what is the rope tension you need to provide to the rope between you to win? What is the rope tension between him and the opposing team?

9. Jun 14, 2014

### haruspex

No, it does matter. Within the range of possibilities in the diagram, the 100g block would tip over, even if there is no friction.
But as I posted, ietr's error is in taking the tension as the same throughout. There are effectively two separate strings.