Calculating Work and Force for a Hanging Crate: Varying Force Homework Problem

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Homework Help Overview

The problem involves a 230kg crate hanging from a rope and being pushed horizontally with a varying force while moving 4.00m to the side. Participants are discussing the calculations of work done by different forces acting on the crate, including the weight of the crate and the tension in the rope.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are exploring the relationship between the varying force and the work done during the displacement. There are questions regarding the integration of work with respect to the angle and the assumptions about the function for force.

Discussion Status

Some participants have provided guidance on integrating work with respect to the angle of displacement, while others are questioning the assumptions made about the tension and its constancy. There is an ongoing exploration of the forces acting on the crate and their contributions to the total work done.

Contextual Notes

Participants note that the crate remains motionless before and after the displacement, and there are discussions about the implications of this on the work-energy theorem. There is uncertainty regarding the calculation of angles and the nature of the forces involved.

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Homework Statement


A 230kg crate hangs from the end of a 12.0 m rope. You push horizontally on the crate with a varying force F to move it 4.00m to the side.

What is the magnitude of F when the crate in the final position?[/B]

During the displacement, what are the work done on it, the work done by the weight of the crate, and the work done by the pull on the crate from the rope?

Knowing that the crate is motionless before and after displacement, use the answers to find the work your force does on the crate.

Why is the work of your force not equal to the product of the horizontal displacement and the initial magnitude of F?

Homework Equations


f= ma
w = fdcos(0)

The Attempt at a Solution


Magnitude of F is Fx = Tsin(Θ) = Fp
Work done by tension on the crate changes with (theta)
Work by weight = 0 because F is parallel to displacement
Work = ∫ from 0 to 4 (Tsin(Θ)) dΘ How do i do this? is this correct
The work of your force, is not equal to the product of horizontal displacement and the initial magnitude of F because F is varying. Also It is the sum of all the F*d from 0 to 4 m
 
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When you are integrating for work, your saying that it's from 0 to 4 which is the displacement but you have a ##d\theta## you can't integrate that, you should either get ##\theta## in terms of the displacements or easier thing to do is to find the angle of displacement for 4 m then integrate from 0 to that angle.
 
Suraj M said:
When you are integrating for work, your saying that it's from 0 to 4 which is the displacement but you have a ##d\theta## you can't integrate that, you should either get ##\theta## in terms of the displacements or easier thing to do is to find the angle of displacement for 4 m then integrate from 0 to that angle.

Thank you, that makes sense. So ∫ from o to 18.4 degrees
One extra question, is it right for me to assume the function for force is t(sinθ) ?
 
brycenrg said:
force is t(sinθ) ?
Im not a 100% sure. It should work, but what's your T(tension)? I mean, value. is that constant? I doubt it.
 
brycenrg said:
Thank you, that makes sense. So ∫ from o to 18.4 degrees
One extra question, is it right for me to assume the function for force is t(sinθ) ?
I don't think you calculated the angle correctly. When the crate is displaced 4 m horizontally, the cord length is still 12 m.
There is more than one force acting on the crate.
Which of those forces do work?
What is the total net work done by all forces? (HINT: use work-energy theorem.)
You don't have to use calculus.
 

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