Increased acceleration in a pully system

AI Thread Summary
To raise herself slowly at a constant speed, the window washer must exert a force of 320 N, calculated using the equation 2Ft = mg. When the force is increased by 15%, the new force becomes 368 N. The correct method to find the acceleration involves using the total upward force and the weight, leading to an acceleration of 1.47 m/s². The initial attempt at calculating acceleration was incorrect due to the misapplication of equations. The final answer aligns with the book's solution of 1.5 m/s², confirming the calculations.
coneheadceo
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Homework Statement


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?

Homework Equations


2Ft - mg = ma

a = (Mb - Mp)/(Mb+Mp) ... I think

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.
 
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coneheadceo said:

Homework Statement


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?

Homework Equations


2Ft - mg = ma

a = (Mb - Mp)/(Mb+Mp) ... I think

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.

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
 
Your new upward force is 2(320+48)
 
So am I literally taking (9.8 m/s^2)(0.15)= 1.47 m/s^2 to get the acceleration?
 
coneheadceo said:
So am I literally taking (9.8 m/s^2)(0.15)= 1.47 m/s^2 to get the acceleration?

Yes.

ehild
 

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