An Equilibrium Question

  • Thread starter e(ho0n3
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  • #1
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There is a table besides a wall (so that one edge of the table is touching the wall). Both wall and table have frictionless surfaces. Say I place a block on the table so that it's touching the wall.

Drawing a free-body diagram depicting the situation gives me three forces: the normal to the table and the weight of the block which cancel out, and the normal to the wall. If the block is in equilibrium then the normal force to the wall is zero. Somehow this is bothering me. Now if the table had a rough surface, then I could say that the normal to the wall and static friction cancel each other out. But if there is no static friction, what happens? Can anybody clarify on this please?
 

Answers and Replies

  • #2
well, if the wall is straight up and down and everything is at right angles, i see no reason for it to exert any type of a normal force on the block. thus you only have to consider the vertical forces, which cancel.

saw your message on your other thread, no prob.
 
  • #3
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DarkEternal said:
well, if the wall is straight up and down and everything is at right angles, i see no reason for it to exert any type of a normal force on the block. thus you only have to consider the vertical forces, which cancel.
So the wall is not exerting a force and the block and the block is not exerting a force on the wall? So the wall and block are in contact and in perfect harmony!?
 
  • #4
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Why would there be any force between the wall and table? I see no need for the wall to support the table. In fact, if the wall magically disappeared, what would happen? If nothing, then how can the wall even be a factor in the problem?
 
  • #5
well the normal force stems from the compression of atoms on the surface by the body, because a force on the body transmits this compression to the surface. if there is no horizontal force to compress the wall surface, how can there be a reaction force from the wall?
 
  • #6
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So two bodies can be in contact with each other without exerting forces on each other? OK, I'll buy that.
 

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