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

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The discussion revolves around calculating the force and work done on a 230kg crate hanging from a rope while being displaced horizontally. The magnitude of the force F is derived from the tension in the rope and the angle of displacement, which changes as the crate moves. The work done by the weight of the crate is zero since it is parallel to the displacement, while the work done by the applied force is not simply the product of force and distance due to its varying nature. Participants suggest integrating the force over the angle of displacement to find the total work done, while also questioning the calculation of the angle and the assumption of tension being constant. The conversation emphasizes the importance of considering all forces acting on the crate and applying the work-energy theorem for a comprehensive analysis.
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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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