Using static friction to find tension

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SUMMARY

The discussion focuses on calculating the minimum pulling force required for a clown weighing 860 N to yank his feet out from under himself using static friction. The coefficient of static friction between the clown's feet and the ground is 0.41. The participant correctly identifies that the clown is in equilibrium and applies the equation F=ma to derive the pulling force, resulting in a calculated force of 507.4 N. The discussion also raises questions about the roles of normal force and tension in the rope during this scenario.

PREREQUISITES
  • Understanding of static friction and its coefficient
  • Knowledge of Newton's laws of motion, particularly F=ma
  • Familiarity with concepts of equilibrium in physics
  • Basic understanding of tension in ropes and pulleys
NEXT STEPS
  • Study the principles of static friction and its calculations
  • Learn about forces in equilibrium and how to analyze them
  • Explore tension in systems involving pulleys and ropes
  • Investigate the role of normal force in various physical scenarios
USEFUL FOR

Students studying physics, particularly those tackling problems involving forces, tension, and static friction. This discussion is beneficial for anyone looking to deepen their understanding of equilibrium and force analysis in practical applications.

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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