The drawing shows a circus clown who weighs 890 N. The coefficientof static fric
ID: 1737192 • Letter: T
Question
The drawing shows a circus clown who weighs 890 N. The coefficientof static friction between the clown's feet and the ground is 0.53.He pulls vertically downward on a rope that passes around threepulleys and is tied around his feet. What is the minimum pullingforce that the clown must exert to yank his feet out from underhimself?
The drawing shows a circus clown who weighs 890 N. The coefficientof static friction between the clown's feet and the ground is 0.53.He pulls vertically downward on a rope that passes around threepulleys and is tied around his feet. What is the minimum pullingforce that the clown must exert to yank his feet out from underhimself?
The drawing shows a circus clown who weighs 890 N. The coefficientof static friction between the clown's feet and the ground is 0.53.He pulls vertically downward on a rope that passes around threepulleys and is tied around his feet. What is the minimum pullingforce that the clown must exert to yank his feet out from underhimself?
Explanation / Answer
the force due to friction is equal to the applied force up to themaximum friction force, which is Ffmax=mgcos() where m=mass g=gravity = the angleoff the horizontal (zero in this case) and = thecoeffcient of friction (.53 in this case) m*g=890 N (in this case) so, Ffmax=(890N)*cos(0)*(.53)=471.7N
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