The drawing shows a circus clown who weighs 813 N. The coefficientof static fric

Question

The drawing shows a circus clown who weighs 813 N. The coefficientof static friction between the clown's feet and the ground is0.569. He pulls vertically downward on a rope that passes aroundthree pulleys and is tied around his feet. What is the minimumpulling force that the clown must exert to yank his feet out fromunder himself?
note thatthis is problem 60, chapter 4 from Physics 8th by cutnell &johnson but with different variables.
note thatthis is problem 60, chapter 4 from Physics 8th by cutnell &johnson but with different variables.

Explanation / Answer

Given : .              Weight of the circus clown (W) = 813 N .               Coefficientof static friction ( s ) = 0.569 .              Using the newtons second law of motion : .                       FN + FP - W = 0   ; Where FP =   Pullingforce .                    and  ( fs )max - FP =0 .                     ==>     FP =  (fs )max   --------------------(1) .              But the maximum firctional force is given by : .                         (fs )max = s FN   .              The normal force is also given by : .                               FN   =   W - FP . .              Hence equation (1) becomes ; .                              FP   =    ( fs)max      .                                     =   s   * FN   = s  * (W - FP ) .                   or        FP   =   s W / (1 +s) .                                     = ( 0.569 * 813 ) / ( 1 + 0.569)    =   --------- N . Solve it . Hope this helps u!          

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