The drawing shows a circus clown who weighs 780 N. The coefficient of static fri
ID: 1321779 • Letter: T
Question
The drawing shows a circus clown who weighs 780 N. The coefficient of static friction between the clowns feet and the ground is 0.600. 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?
The drawing shows a circus clown who weighs 780 N. The coefficient of static friction between the clown 1/2 s feet and the ground is 0.600. 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?Explanation / Answer
Note that
Ff = usN
Along y, the summation of forces is
Fnety = P - W + N = 0
where P = the force applied, N = the normal force, W = the weight of the clown.
Solving for N,
N = W - P
Thus,
Ff = us(W - P)
Alongx, thus, the summation of forces is
Fnetx = 0 = P - Ff
Using our expression for Ff earlier,
0 = P - us(W - P)
Plugging in us = 0.600, W = 780 N,
0 = P - 0.600(780 - P)
0 = P - 468 + 0.600P
0 = 1.600P - 468
468 = 1.600P
Thus
P = 292.5 N [ANSWER]
DONE! It was nice working with you!
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