# Tension and Pulleys

1. Oct 14, 2005

### amcavoy

I have attached a picture of what I am describing below (Edit: also a link if the attachment didn't go through):

http://img368.imageshack.us/img368/949/mech036fig012zi.jpg [Broken]

If a mass m1 sits on top of a frictionless table (assume the pulley is as well), attached to a massless string with a mass m2 on the other end hanging off the table, what is the accel. and tension in the string assuming that m1=2m2?

I figured that if it were attached to a wall (rather than m1), the tension would be |T|=m2g. However, since the mass m1 will move, this is what I did:

$$\vec{a}=\frac{F}{m}=\frac{m_2g}{m_2+2m_2}=\frac{g}{3}$$

$$T=m\vec{a}=\frac{m_2g}{3}$$

Am I correct?

Thanks for the help

Alex

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2. Oct 14, 2005

### Staff: Mentor

Good!
Careful: $$T = m_1 a$$

Just for the exercise, I recommend doing the problem a second way: Apply Newton's 2nd law to each mass seperately and combine the two equations. (You'll get the same answer, of course.)

3. Oct 14, 2005

### Fermat

The acceleration is correct, but T is the tension, providing acceleration, on m1, not m2.

4. Oct 14, 2005

### amcavoy

Hmm... Could you explain a bit more why it isn't m2? It seems like the hanging mass (m2) would determine the tension.

Thanks again,

Alex

5. Oct 14, 2005

### amcavoy

Alright I tried your suggestion Doc Al:

$$F_1=\frac{2m_2g}{3}$$

$$F_2=\frac{m_2g}{3}$$

$$F_1+F_2=m_2g$$

I'm not coming up with m1g like you suggested. Where did I go wrong?

Thank you.

Alex

6. Oct 14, 2005

### Staff: Mentor

I suggested starting over and looking at each mass separately:
(1) What forces act on mass 1? Write Newton's 2nd law for mass 1.
(1) What forces act on mass 2? Write Newton's 2nd law for mass 2.​
When you do that, you'll get two equations. They will allow you to solve for the acceleration of the masses and the tension in the cord. (And you'll also see why $T = m_1 g$.)

7. Oct 14, 2005

### amcavoy

Edit: Nevermind, I see what you're saying. The force on the mass m1 is equal to the tension, thus T=m1a. Thanks a lot for the help, it's appreciated.

Alex

Last edited: Oct 14, 2005
8. Oct 14, 2005

### Staff: Mentor

You only need to consider the forces parallel to the direction of the acceleration:
on $m_1$: The only horizontal force is the tension in the string
on $m_2$: There two vertical forces, the tension in the string and the weight