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Sliding rod on wall

  • Thread starter daniel_i_l
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  • #1
daniel_i_l
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Here's a question (not homework) that someone asked me a few days ago:
you have a rod with length L and mass M that's leaning against a frictionless wall and makes a 30deg angle with the floor. what does the static friction have to be inorder for the stick to stay inequilibrium?
What i did was to pretend that all the mass was at the center of the rod. then i split the force mg on that point into two parts - "along the rod" and perpendicular to the rod. the force along the rod pushes the bottom of the rod into the ground with and angle of 30 bellow the ground. the perp force induces a normal force which also has two parts - along the rod and perp to the rod going up. this perp force creates a torque which also pushes the bottom of the stick into the ground.
So basically i summed up all these forces and got the answer:
Ustatic >= 0.6sqrt(3)
But the real answer is
Ustatuc >= 0.5sqrt(3)
What it the correct way to deal with this problem?
Thanks.
 

Answers and Replies

  • #2
Doc Al
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Whether "homework" or not, all textbook/coursework-type questions should be posted in the appropriate "Homework Help" section. I'll move it. :smile:

I can't really follow what you are doing here. To solve this, just consider all the forces acting on the ladder:
(1) Gravity, which acts down through the center of mass (you started off with that)
(2) The force from the wall--since there's no friction, it acts horizontally
(3) The force from the floor--break it into two components: friction force (horizontal) and normal force (vertical)

Now just apply the conditions for translational and rotational equilibrium.
 
  • #3
daniel_i_l
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thanks, i got it now.
 

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