Increased acceleration in a pully system

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Homework Help Overview

The problem involves a window washer using a bucket and single pulley system, with a focus on the forces required to lift themselves at a constant speed and the resulting acceleration when the force is increased by 15%. The subject area includes mechanics and dynamics.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the forces acting on the system, including tension and gravitational force. There is an exploration of the equations governing the motion, with some questioning the appropriateness of certain equations for the scenario. Attempts to calculate the necessary pulling force and resulting acceleration are presented.

Discussion Status

The discussion includes various attempts to solve the problem, with some participants providing guidance on the correct application of equations. There is recognition of potential confusion regarding the equations used, and multiple interpretations of the problem are being explored without a clear consensus on the correct approach.

Contextual Notes

Participants note that the original equations may not fully apply to the scenario, and there is a mention of the book's answer differing from the calculations presented. The problem context includes constraints related to the forces involved and the specific setup of the pulley system.

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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