Constant, Horizontal Force on Swing

[Frozen Stair]

1. Homework Statement
This problem is connected to an example in the textbook.

Here's some of the key info from the example, first of all:
You are appointed to push your cousin Throckymorton in a swing. His weight is w, the length of the chains is R, and you push Throcky until the chains make an angle θ(0) with the vertical. To do this, you exert a varying horizontal force F that starts at zero and gradually increases just enough so that Throcky and the swing move very slowly and remain very nearly in equilibrium.
In this example, the textbook found that the work "I" do by exerting the force F is equal to wR(1-cosθ(0)).

Here is the problem:

In the example, instead of applying a varying horizontal force F that maintains Throcky very nearly in equilibrium, you apply a constant a constant, horizontal force on Throcky with magnitude F=2w, where w is Throcky's weight. Consider Throcky to be a particle and neglect the small weight of the chains and seat. You push Throcky until the chains make an angle θ(0) with the vertical.
(a) Calculate the work done on Throcky by the force F that you apply.
(b) How does the work done by F in this exercises compare to that in the example?

Homework Equations

Work = F*d

P1
Work = ∫F * dl.
P2

dl = Rdθ

The Attempt at a Solution

Part (a)
θ(0) θ(0) θ(0)
W = ∫F(tan)dl = ∫(Fcosθ)(Rdθ) = 2wR∫cosθdθ = 2wRsinθ(0)
0 0 0

Part (b)
This is the part I have the most trouble with. When I graph y = 2sinθ(0) and y = 1-cosθ(0) because these graphs resemble, by a constant factor, the work equations for the two different situations, it's clear that the work done for the situation in the exercise is greater than the work done for the situation in the example. I've tried to find an explanation for this, but I can't think of one.Edit: Sorry, I don't know how to type in the limits of integration! But the lower limit is always 0 and the upper is theta(0).

 

[SammyS]

1. Homework Statement
This problem is connected to an example in the textbook.

Here's some of the key info from the example, first of all:
You are appointed to push your cousin Throckymorton in a swing. His weight is w, the length of the chains is R, and you push Throcky until the chains make an angle θ(0) with the verticle. To do this, you exert a varying horizontal force F that starts at zero and gradually increases just enough so that Throcky and the swing move very slowly and remain very nearly in equilibrium.
In this example, the textbook found that the work "I" do by exerting the force F is equal to wR(1-cosθ(0)).

Here is the problem:

In the example, instead of applying a varying horizontal force F that maintains Throcky very nearly in equilibrium, you apply a constant a constant, horizontal force on Throcky with magnitude F=2w, where w is Throcky's weight. Consider Throcky to be a particle and neglect the small weight of the chains and seat. You push Throcky until the chains make an angle θ(0) with the verticle.
(a) Calculate the work done on Throcky by the force F that you apply.
(b) How does the work done by F in this exercises compare to that in the example?

Homework Equations

Work = F*d

P1
Work = ∫F * dl.
P2

dl = Rdθ

The Attempt at a Solution

Part (a)

     θ(0)        θ(0)              θ(0)
W = ∫F(tan)dl = ∫(Fcosθ)(Rdθ) = 2wR∫cosθdθ = 2wRsinθ(0)
      0          0                0

Part (b)
This is the part I have the most trouble with. When I graph y = 2sinθ(0) and y = 1-cosθ(0) because these graphs resemble, by a constant factor, the work equations for the two different situations, it's clear that the work done for the situation in the exercise is greater than the work done for the situation in the example. I've tried to find an explanation for this, but I can't think of one.

Edit: Sorry, I don't know how to type in the limits of integration! But the lower limit is always 0 and the upper is θ(0).

Hello Frozen Stair. Welcome to PF !

When θ is small, how does the force in the example compare with that in the exercise ?

 

[Frozen Stair]

Hi. :)

When θ is small...in the example, the force would be very small (F=wtanθ). In the exercise, it would be 2w. I was thinking about this, but I can't figure out how exactly this connects to the work that is done.

 

[SammyS]

Hi. :)

When θ is small...in the example, the force would be very small (F=wtanθ). In the exercise, it would be 2w. I was thinking about this, but I can't figure out how exactly this connects to the work that is done.

Both forces are in the horizontal direction.

Both move along the same path.

One force is constant and relatively large.

The other starts small. It's w/2 @ ≈ 26.6°. It's w at 45°. It's equal to 2w at what angle? At that point there is not much displacement remaining in the horizontal direction.

Also, what happens to all of that extra work that's done?

 

[Frozen Stair]

I see what you mean. F=2w at θ=1.11, which is already quite far.

But what do you mean by the "extra work"?

 

[SammyS]

I see what you mean. F=2w at θ=1.11, which is already quite far.

But what do you mean by the "extra work"?

In the exercise, the amount of work to reach any particular θ₀ is quite a bit larger than the amount of work done in the example, to reach the same θ₀. What is the result of expending that extra amount of work in the exercise as compared to in the example?

 

[Frozen Stair]

Oh, I think I see now.

Instead of Throcky being pushed so that he is at a constant speed, he has an acceleration. And, by the work-energy theorem, W = ΔK, so this would suggest that the exercise has such a greater value for work than the example because the speed is increasing.

 

[SammyS]

Oh, I think I see now.

Instead of Throcky being pushed so that he is at a constant speed, he has an acceleration. And, by the work-energy theorem, W = ΔK, so this would suggest that the exercise has such a greater value for work than the example because the speed is increasing.

Yes. That's what I was getting at.

Should I ask any more questions?

 

[Frozen Stair]

No, thank you so much! :)