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

  • Thread starter brycenrg
  • Start date
  • Tags
    Force
In summary: The net work done by all forces is equal to the product of the tension force and the displacement. The tension force is the force you are pushing, and the displacement is the horizontal distance the crate moved.
  • #1
brycenrg
95
2

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
 
Physics news on Phys.org
  • #2
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.
 
  • #3
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) ?
 
  • #4
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.
 
  • #5
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.
 

1. How do you calculate work?

Work is calculated by multiplying the force applied to an object by the distance the object moves in the direction of the force. The formula for work is W = F * d, where W is work, F is force, and d is distance.

2. What is the formula for calculating force?

The formula for calculating force is F = m * a, where F is force, m is mass, and a is acceleration. In this problem, the force applied to the crate is the weight of the crate, which is calculated by multiplying the mass of the crate by the acceleration due to gravity, which is 9.8 m/s^2.

3. How do you determine the direction of the force?

The direction of the force is determined by the direction in which the object is moving. In this problem, the crate is hanging from a rope, so the force is acting downwards towards the ground.

4. What is the significance of varying force in this problem?

Varying force in this problem means that the force applied to the crate changes over time. This could be due to different amounts of weight being added or removed from the crate, or the force being applied at different angles or in different directions.

5. How does the angle of the force affect the calculations?

The angle of the force affects the calculations by changing the direction in which the force is applied. This can change the amount of work that is done on the crate, as the distance the crate moves may be different depending on the angle of the force. To account for this, the force must be broken down into its x and y components, and the work and force must be calculated separately for each component.

Similar threads

  • Introductory Physics Homework Help
Replies
14
Views
5K
  • Introductory Physics Homework Help
Replies
10
Views
323
  • Introductory Physics Homework Help
Replies
5
Views
10K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
2K
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
2K
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
6K
Back
Top