Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Force parallel to a smooth wall?

  1. Nov 20, 2014 #1

    My question is quite a quick one- I was wondering whether it is ever possible to have a smooth wall exerting a force parallel to it (and not just perpendicular to it). For example, if you were to place a see-saw by a smooth wall so that the wall is holding one of the see-saw ends below the level of the other, is this only due to the moment because of the perpendicular, reaction force from the wall or is there also a force parallel to the wall exerted by it (in the line that friction would act). I seem to remember seeing a question like this before, although I cannot remember whether the force perpendicular to the smooth wall was only in the case of an object being attached (for example by a hinge) to the wall, or if it also applies to objects 'propped up', but not attached to, the wall as in the see-saw case.

    Also, when considering the see-saw and assuming that the pivot of the see saw is propped up on a knife edge to create the pivot, would the reaction force from this pivot act vertically upwards, or would it be perpendicular to the line of the see-saw?

    I apologise if I haven't explained this very clearly...

    Thank you in advance!

    EDIT: I think I figured out my second question about whether the force from the pivot is vertical or perpendicular to the see-saw plank. I think It must be perpendicular, because otherwise there the weight of the see saw, and the reaction force (and potentially the force at the wall which is parallel to the wall, whether it is friction or another force) would all be acting vertically and there would be no force to cancel the horizontal normal contact force at the wall... I think...
    Last edited: Nov 20, 2014
  2. jcsd
  3. Nov 20, 2014 #2


    User Avatar
    Science Advisor
    Homework Helper

    The expression "smooth wall" usually means there is no friction - so the reaction force must be perpendicular.
  4. Nov 20, 2014 #3
    Thank you for your reply! But can there really be no other force? I'm pretty sure that dealing with hinges, there was a force parallel to the wall despite the fact that the hing was smooth. Could the same principle not apply?
  5. Nov 20, 2014 #4


    User Avatar
    Science Advisor
    Homework Helper

    A hinge can be frictionless in rotation but must provide friction in other directions (parallel and perpendicular to the wall) or the door would fall off.
  6. Nov 20, 2014 #5


    User Avatar
    Science Advisor
    Homework Helper

    Perhaps this diagram helps.
    smooth wall.png

    The triangular pivot is fixed in space. The purple reaction force need not be perpendicular to the seesaw. Nothing is accelerating so all forces and torques sum to zero.

    The "red" force must be perpendicular to the smooth/frictionless wall. The magnitude of the red force depends on the geometry and can be calculated because the torques must sum to zero (The 10kg/100N blue force creates an anticlockwise torque that must be countered by the red force).

    Perhaps consider what happens if you reposition the seesaw so that it is almost horizontal. The magnitude of the red force would have to be very large, as would the red component of the reaction force at the pivot. If the pivot was just a knife blade it's possible there wouldn't be enough friction at the pivot.
  7. Nov 20, 2014 #6
    Thank you both!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook