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Coefficient of friction for the block to be in equlibrium

  1. Aug 18, 2014 #1
    1. The problem statement, all variables and given/known data

    a square box held with a person's fingers pushing downward one quarter of the way along the top of the box and with their thumb pushing side wards on the bottom of the side face of the box. Friction between the person's fingers and the box is obviously very important if the box is not to fall.Determine the required coefficient of friction to keep the box from falling.

    θ=tan-1 (1/2) and α= 45°

    2. The attempt at a solution

    I have drawn he following forces to show you

    the equations which i have got is:

    μN2= N2 + Mg

    and by balancing the torque I have got :

    N2 x root 5a x sinθ = μN2 x 2 root 2 x cosα[
     

    Attached Files:

  2. jcsd
  3. Aug 18, 2014 #2
    can any one help me??
     
  4. Aug 18, 2014 #3

    andrevdh

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    The forces also cancel in the x-direction.
    The torque is the product of the force and its lever arm,
    which is the perpendicular distance to its line of action.
     
  5. Aug 18, 2014 #4

    ehild

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    Gold Member

    It is μN2=N1+Mg
    The torque equation is not correct. Multiply each force by the distance of the line of the force from the CM.


    ehild
     
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