The diagram isn\'t that great, but the outline of thesquare is the rope, the cir

Question

The diagram isn't that great, but the outline of thesquare is the rope, the circles are the pulleys, and thebottom right corner of the square is where the clown is standingand pulling down on the rope. The drawing shows a circus clown who weighs 801 N. Thecoefficient of static friction between the clown's feet and theground is 0.493. He pulls vertically downward on a rope that passesaround three pulleys and is tied around his feet. What is theminimum pulling force that the clown must exert to yank his feetout from under himself?

Explanation / Answer

the friction force of the clown F=0.493*N N is reflextive force from the ground acting on theclown N=mg-T; T is tension force of string, which is the same at allpoits along the string. to yank his feet: F=T => 0.493(mg-T)=T => T=264.5 N, which isequal to the force caused by the clown

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