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Equilibrium and friction problem

  • Thread starter mit_hacker
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  • #1
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Homework Statement



(Q) A uniform 100N ladder of length 5m rests on a rough floor for which the co-efficient of static friction is 0.8 and leans against a smooth wall. Determine the force P that must be applied at the center of such a rod in order to make it move.

Homework Equations



All the equations of equilibrium including moments.

The Attempt at a Solution



The reaction at the bottom has to be 100N as well since there is no other verticle force. Thus, the frictional force will be 80N. Taking moments about the upper tip of the ladder yields:

P(2) - 100(1.5) -80(4) +100(3) =0.

Therefore, P = 85. However, the answer at the back is 75. Please help me with this. I don't know where I am going wrong.
 

Answers and Replies

  • #2
Doc Al
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The reaction at the bottom has to be 100N as well since there is no other verticle force. Thus, the frictional force will be 80N.
[itex]\mu N[/itex] is the maximum value for static friction. The actual value depends on the angle of the ladder. (Are you given the angle?)
 
  • #3
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Nope!! The angle is not given.
 

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