Using static friction to find tension

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The discussion focuses on calculating the minimum pulling force a clown must exert to overcome static friction and yank his feet out from under himself. The clown weighs 860 N, and the coefficient of static friction is 0.41, leading to a frictional force of 353.6 N. The participant uses the equation F=ma to analyze the forces, concluding that the net force when pulling down must equal the weight minus the frictional force. Questions arise regarding the roles of normal force and tension in the rope, as well as the direction of forces involved. Understanding these concepts is crucial for solving the problem accurately.
jenador
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I am having a lot of problems with the tension/force questions on my homework. Here is the last question I don't understand in my homework set:

The drawing shows a circus clown who weighs 860 N. The coefficient of static friction between the clown's feet and the ground is 0.41. 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?

http://www.webassign.net/CJ/04_58.gif

I know that the clown is in equilibrium when he is not moving and if he yanks his feet off the floor. Using the F=ma equation, I figured he would move in the y-direction. therefore F=ma=-(his weight)+(friction)=-860N+(860*0.41)=507.4N.

Is that correct? Is it right to say that friction is felt in the y direction in this case because the clown is just standing on the floor? Does F(normal force) play a role in this problem?
 
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If the clown pulls on the rope with a force F, what is the tension in the rope? What is the vertical force exerted by the rope on the clown? What is the normal force on the ground due to the clown? What is the horizontal force exerted by the rope on his feet?
 
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