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If a person who weighs 612N sits in a hammock 3m long and sags 1m below the point of support, what force is exerted on the ropes suporting the hammock. Please help me by showing how you got that answer

2007-12-02 07:19:52 · 2 answers · asked by Anonymous in Science & Mathematics Physics

2 answers

I'm assuming the hammock makes a perfect "V"-shape when the person's sitting in it, so there's a straight stretch of hammock 1.5 meters long extending diagonally up from the person on each side. Right?

Each half of the hammock forms a certain angle θ with the horizontal. And there's a certain tension ("T") in the hammock.

Vertical forces on person:
1. Weight (mg), pulling down
2. T(sinθ) pulling up, due to left half of hammock;
3. T(sinθ) pulling up, due to right half of hammock;

Since the person is not accelerating, the vertical forces cancel out. (The horizontal forces also cancel out; but that's irrelevant to this problem.)

That is:
downward forces = upward forces
mg = T(sinθ) + T(sinθ)
mg = 2T(sinθ)

The "force exerted on the ropes" is just "T". Solve for that:

T = mg/(2sinθ)

Now, if you draw a diagram, you should be able to see that:

sinθ = (1m / 1.5m)

That gives you all you need to know.

2007-12-02 07:37:27 · answer #1 · answered by RickB 7 · 0 0

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2016-10-02 05:50:22 · answer #2 · answered by Anonymous · 0 0

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