- #1

grouchy

- 73

- 0

http://geocities.com/grouchy187/untitled.bmp

My attemp:

T = m1a T - m2g = m2a

m1a = m2g + m2a

a= m2g / m1-m2

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- Thread starter grouchy
- Start date

- #1

grouchy

- 73

- 0

http://geocities.com/grouchy187/untitled.bmp

My attemp:

T = m1a T - m2g = m2a

m1a = m2g + m2a

a= m2g / m1-m2

- #2

Doc Al

Mentor

- 45,433

- 1,886

What tension force acts on m2? (Consider the pulley as part of m2.)

What about that hint?

What about that hint?

- #3

grouchy

- 73

- 0

for m2, would the weight W, be equal to 2T?

also I think a1 = a2

also I think a1 = a2

- #4

Doc Al

Mentor

- 45,433

- 1,886

No. But thosefor m2, would the weight W, be equal to 2T?

No. Try to think this through. (Using a piece of string to work it out may help--I'm not kidding.) If m1 moves 1 m to the right, how far down does m2 move?also I think a1 = a2

- #5

grouchy

- 73

- 0

i dunno, I'm thinking about it and I just dont get this one

- #6

alphysicist

Homework Helper

- 2,238

- 2

If you're having trouble visualizing the amount of extra rope, just start with your orginal diagram. Now draw in the new position of m2 and the lower pulley 1 meter below where they were. Note that you have to extend the lines for the ropes on each side of the pulley. How much extra rope do these new lines represent?

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