# Two blocks and two pulleys

1. Nov 13, 2004

### tigerseye

Based on the fact that the string is massless, and the system and table are frictionless, how would I find the tension in the string acting on mass 2?

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2. Nov 13, 2004

### CartoonKid

Since the table is frictionless, the top pulley would be just free to move horizontally. The only contribution to the tension would be from m1. I think the tension would be just m1g, is it?

3. Nov 13, 2004

### Andrew Mason

The tension in the string is equal to 1/2 of the force accelerating the block (since the tension in the string both above and below the top pulley is acting on the block). So:

Ftension = mblockablock/2

Also, for the same reason, ablock=am1/2

The key here is to realize that while the downward force is m1g, it provides the acceleration for both masses.

So:

m1g = (m1am1+mblockablock)
= (m12ablock +mblockablock)
= ablock(2m1+mblock)

ablock = m1g/(2m1+mblock)

Resulting in:

Ftension = 1/2 x mblockm1g/(2m1+mblock)

(I think).

AM

4. Nov 14, 2004

### Heart

But I think they only want the tension/answer in terms of m2 and a2!?

5. Nov 14, 2004

### Andrew Mason

a2? We aren't given a2. We have to figure it out first.

AM

6. Nov 14, 2004

### Heart

I got it now. Thanks for all the replies. Thanks to tigerseye too for posting this question.

Last edited: Nov 14, 2004
7. Dec 1, 2004

### pmrazavi

It's just about free body diagrams...draw the free body diagram for both of the masses..then apply newton's second law: for the hanging mass you will get m1g-T=m1a
and for the block on the table you will get 2T=m2a..add up these 2 equations you will get T=(m1+m2)a-m1g..this is the tension in the string..since we have two tension forces on the block on the table, the exerting force on it is 2T.
I'm sure about my answer...Andrew is making mistake when saying that the mg in the first block provides the a for both of the masses..it is technically WRONG..becuase the cause of acceleration of the block on the table is tension force as you can see in the FBD.

8. Dec 1, 2004

### Andrew Mason

I disagree. You seem to be overlooking the fact that the acceleration of the block is one half that of m1. While T = m1(g-a) it is not true that 2T = m2a (where a is the acceleration of m1).

Of course the block is accelerated due to the tension of the string, but that tension is due entirely to the force of gravity on m1. Gravity causes m1's acceleration and T: m1g = T + m1a

So solving for a (= acceleration of m1):

$$T = \frac{1}{2}m_2a_2 = \frac{1}{4} m_2a_1$$

So substituting this into: T = m1(g-a) I get:

$$m_1(g-a) = \frac{1}{4}m_2a_1$$

$$a = m_1g/(m_2/4 + m_1)$$

AM

Last edited: Dec 1, 2004