# Window Washer Force Problem

1. Feb 10, 2008

### rwx1606

1. The problem statement, all variables and given/known data
A window washer pulls herself up using a pulley-bucket apparatus. How hard must she pull downward to raise herself slowly at constant speed? The mass of the person plus the bucket is 75 kg.

2. Relevant equations
F=ma

3. The attempt at a solution
I don't know what I'm doing wrong. Is the force she has to pull downward not equal to the weight? I get (9.8)(75)=735N. It says use two sig figs and I've tried both 7.4x10^2 and 7.3x10^2 N. I hate mastering physics!

2. Feb 10, 2008

### Dick

If she's using pulleys, then the force is NOT necessarily equal to the weight. That's the whole point to using pulleys. You'd better describe the pulley system.

3. Feb 10, 2008

### rwx1606

looks like this http://answerboard.cramster.com/physics-topic-5-139868-0.aspx [Broken]

Last edited by a moderator: May 3, 2017
4. Feb 10, 2008

### Staff: Mentor

Obviously the ropes must exert a total force on "washer + bucket" that equals its weight. Hint: How many times does the rope pull on "washer + bucket"?

5. Feb 10, 2008

### rwx1606

hmm so the ropes each carry half the weight. So 735N/2?

6. Feb 10, 2008

### Staff: Mentor

Yep. (Fix that typo.)

She only pulls on one rope, so she just needs to exert a force equal to the needed rope tension. Which is half the weight.

7. Feb 19, 2008

### AlexWyler

What if the problem is changed a bit?

Suppose there are two massless pulleys, and the massless rope is attached to the ceiling.
_______
| /O\ The 75kg window washer stands on the weightless platform and pulls on the
| | | string at point "x." How hard does she have to pull to slowly raise herself at
| | | constant velocity?
| x |
| _ |
\ O/

8. Feb 19, 2008

### Dick

If I'm reading your cartoon correctly, then you have three massless ropes connected only by massless pulleys. In that case the tension in all three ropes is the same. So they all exert the same tension T on the platform and washer. So 3*T=mg. Conclusion?