How Much Work is Done by Gravity When Lowering a Piano with Two Ropes?

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Homework Help Overview

The discussion revolves around calculating the work done by gravity when lowering a piano using two ropes. The problem involves understanding the forces acting on the piano, specifically the gravitational force and the tension in the ropes, as well as the distances involved in the lowering process.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between gravitational force and work done, with some questioning the role of tension in the ropes. There is an attempt to clarify the correct equations for calculating work done by the forces involved.

Discussion Status

The discussion is active, with participants providing insights into the correct approach for calculating work done by the tension in the ropes. Some have expressed confusion about the components of the forces and their directions, while others have clarified the necessary adjustments to the equations being used.

Contextual Notes

There is an emphasis on understanding the distinction between the gravitational force and the tension forces, as well as the need to consider the direction of displacement in the calculations. Participants are also navigating the complexities introduced by having multiple forces acting on the piano.

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Homework Statement


The two ropes are used to lower a 244.37 kg piano 9.47 m from a second-story window to the ground. How much work is done by gravity force? (T1=1805 N, q1=64°, T2=1112 N, q2=44°.)
C11P111.jpg


How much work is done by T1 force?
How much work is done by T2 force?

Homework Equations


Wt1= mgdTcos(theata)


The Attempt at a Solution



Wt1= mdTcos(theata)=RIDICULOUSLY HUGE and wrong lol. What is wrong with the way I am approaching this problem?
 
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The force on the piano due to gravity is just the weight. So the work done by gravity is (force)(distance moved in direction of force) = weight times the distance lowered.
 
mmhmmm. I understand that part. But the main problem is the tension parts
 
Oh sorry, I didn't see there were two more parts to the question.
 
I am sorry, I mean the work done by the first rope. I was thinking tension in my head. I thought the equation i came up with is right? What is wrong with it?
 
The vertical component of T1 is T1sin(θ), not T1cos(θ). You will also need to add a minus sign because the direction of displacement is opposite to this vertical component. So the work done by T1 will be -dT1sin(θ) (force times displacement in direction of force).
 
so the equation i should be using is Wt1=-d*sin(theata)*T

isn't T the force?
 
What is d? Isn't it the displacement?
 
Yep! I got it thanks for the help I got it right but more importantly undersand where the anwser came from.
 

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