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Friction and Forces

  1. Nov 24, 2007 #1
    1. The problem statement, all variables and given/known data

    A block of mass 10kg on a rough plane inclined at 30deg to the horizontal. A horizontal force P acts on the block. The coefficient of friction between the block and the plane is 0.4, and it is sufficiently small for the block to slide down the slope if P does not act. Find the range of possible values of P if the block remains stationary.

    2. Relevant equations



    3. The attempt at a solution

    Ive made an attempt but I think I may be wrong, before I bother post it, can I just check a few of my assumptions are correct?

    - The surface reaction is perpendicular to the "rough plane".
    - The friction force is directed directly up the "rough plane".

    Im not too confident on those 2 assumptions and my initial attempt goes on those 2 being true :S

    can someone please clarify?

    thanks
     
  2. jcsd
  3. Nov 24, 2007 #2

    rock.freak667

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    This should be in the physics section but oh well

    Draw the inclined plane with the mass anywhere on it...Consider the weight of the block(mg) and then consider the components of the weight...one acts parallel to the plane and the other acts perpendicularly to it. The normal reaction will be equal to the component perpendicular to the plane.

    P will act in the same direction as the component of the weight parallel to the surface...

    if [itex]F_1[/itex] is the component of the weight parallel to the plane and you know that that the frictional force([itex]F_R[/itex],which opposes motion down the plane) is equal to [itex]\mu*R[/itex] (R= the normal reaction) what condition should be fulfilled so that the block doesnt move?
     
  4. Nov 24, 2007 #3

    dlgoff

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    I think this wikipedia page might help.
     
  5. Nov 25, 2007 #4
    oo thanks :)

    so to make the block not move, would I be right in saying all the forces would have to be in equilibrium? im guessing by resolving in one direction (lets say right), and then in another perpendicular (lets say up)?
     
  6. Nov 25, 2007 #5

    rock.freak667

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    Well that is halfway correct...you see the resultant force perpendicular to the plane is Zero so that those forces are always in equilibrium....

    now if all the forces down the plane("forward motion") is equal to all the forces resisting motion, then the block won't move...i.e. [itex]F_1+P=F_R[/itex]. Now consider what happens if the forces resisting motion is greater than the forces down the plane..i.e. [itex]F_R\geq F_1+P[/itex]
     
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