Clown physics problem
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Clown physics problem

[From: ] [author: ] [Date: 12-06-16] [Hit: ]
So in order to pull his feet sideways,The minimum force is 0.582 * 655N / 1.582 = 241N.......
http://edugen.wileyplus.com/edugen/courses/crs2216/art/qb/qu/c04/ch04p_54.gif

The drawing shows a circus clown who weighs 655 N. The coefficient of static friction between the clown's feet and the ground is 0.584. He pulls vertically downward on a rope that passes around three pulleys and is tied around his feet. What is the minimum pulling force that the clown must exert to yank his feet out from under himself?

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The rope must pull on his feet with enough force to overcome friction.

Assuming a massless rope and frictionless pulleys, the horizontal force from the rope equals the force with which the clown pulls.
It should also be noted that due to Newton's third law, the upper end of the rope pulls the clown up with the same force.
Call this force F.

The vertical forces, which are balanced, are: F (up), gravity (down), and normal force from the ground (up).
N = mg - F
Friction is µN = µ(mg - F)

So in order to pull his feet sideways,
F ≥ µ(mg - F)
F ≥ µmg / (1 + µ)

The minimum force is 0.582 * 655N / 1.582 = 241N.
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