# Another pulley problem?

1. Sep 26, 2009

### kashiark

1. The problem statement, all variables and given/known data
A 47N weight is pulled on a frictionless surface by a frictionless pulley that is attached to a 25N weight by a cord.
A. What is the acceleration of the 47N weight?
B. What is the force on the cord?

2. Relevant equations
F=ma

3. The attempt at a solution
A.
25=(47/9.8)a
B. no clue whatsoever

2. Sep 26, 2009

### tiny-tim

Hi kashiark!

(is this a horizontal surface? is the weight hanging vertically off the end? )

Start again …

write out Newton's second law three times: once for the 47N weight on its own, once for the 25N weight on its own, and once for the pair of weights together.

3. Sep 26, 2009

### kashiark

I'm understanding the problem to mean that one weight is on the ground and the other end is in the air, but if the heavier end is on the ground, wouldn't the cord just go limp and pull the lighter weight up to the top of the pulley?

47=9.8m
m= 4.8

25=9.8m
m= 2.6

I'm not sure how I would compose the third equation.

4. Sep 26, 2009

### tiny-tim

No, that must be wrong …
the question clearly states that the weight stays on the surface …

I assume the rope starts horizontal, then goes over the pulley and hangs down vertically.

5. Sep 26, 2009

### kashiark

Your understanding definitely makes more sense though, but if that's the case, then the weight on the vertical cord would be pulling the weight on the surface horizontally, and you wouldn't have to account for its weight at all in A because it's pulling vertically right?

6. Sep 26, 2009

### kashiark

Ok, I just read that, and it's very ambiguous. Let's call WA the weight that's on the surface and WB the weight that's not. In part A, you wouldn't need to include WA's weight because it would be pulling it down, and WB's weight would be pulling it horizontally right?

7. Sep 26, 2009

### tiny-tim

Physics is equations.

Stop waffling … write out the equations for Newton's second law three times … what do you get?

8. Sep 26, 2009

### kashiark

47=9.8m
m= 4.8

25=9.8m
m= 2.6

Ok I'll stop for now, but I would greatly appreciate it if you could explain it to me. How's this:
22=a(2.6+4.8)
22=7.4a
a=3.0 m/s²