- #1
Frozen Stair
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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?
Work = F*d
P1
Work = ∫F * dl.
P2
dl = Rdθ
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).
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 theta(0).