In a three pulley system a man who weighs 875.0N is standing upright holding on

Question

In a three pulley system a man who weighs 875.0N is standing upright holding on one end of the rope while the other end is tied to his feet. The coefficient of static friction between the man's feet and the ground is 0.41. He pulls vertically downward on a rope that passses around three pulleys and is tied around his feet. The pulleys and rope are assumed to be massles and frictionless.

A.) List five forces acting ono the clown and specify their directions.

B.)What is the minimum pulling force that the clown must exert to yank his feet out from under himself?

C.)Estimate the normal force from the ground.

D.)Estimate the static friction from the ground just before he starts moving.

Explanation / Answer

A)1) Frictional force in negative x direction

2) Tension in string in positive x direction

3) weight of clown acting downwards

4) Normal force acting upwards

5) Tension in string on hands acting upwards

B)u*N = F

N = mg-F

so 0.41*(mg-F) = F

1.41F = 0.41*875

so F = 254.43 N

C) Normal foce = mg-F = 875-F

if F = 254.43N then normal force = 875-F = 650.57 N

D) Frictional force = u(Mg-F) = 0.41*(875-F) = F

If F = 254.43 N

then Frictional force = 254.43 N

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