# Increased acceleration in a pully system

1. Aug 8, 2012

1. The problem statement, all variables and given/known data
A window washer pulls themself upward using a bucket/single pulley system. Their weight with the bucket is 65 Kg.
a) How hard must she pull doward to raise herself slowly at a constant speed?
b) If they increase the force by 15% what will her accelertion be?

2. Relevant equations
2Ft - mg = ma

a = (Mb - Mp)/(Mb+Mp) .... I think
3. The attempt at a solution
a) I've got no problem...

to get constant speed accel = 0 so....

2Ft - mg = 0
2Ft = mg
Ft = mg/2 -> (65)(9.8)/2 = 320 N

b) This is where I think I have the wrong equation to attempt this as
(320)(0.15)= 48, so ..... [(368-320)/(368+320)](9.8)= 0.68 m/s^2

answer in the book is 1.5 m/s^2 so need something else but don't know what to use.

2. Aug 8, 2012

### ehild

The equation in red is irrelevant here. The person exerts F force, and the tension in the chord equals to Ft=F. The person in the bucket is pulled by 2Ft upward, and gravity pulls downward. So the equation you wrote 2Ft - mg = ma is correct.

The other equation refers to the case of a pulley with two masses.

ehild

3. Aug 8, 2012

### azizlwl

Your new upward force is 2(320+48)

4. Aug 9, 2012