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Homework Help: Position of normal force

  1. Apr 28, 2007 #1
    1. The problem statement, all variables and given/known data
    A 2D rectangular block is being pulled at a constant speed with a height of 2.2m and width of 51cm and weighs 403N. The pulling force of 225N is being applied at an 18 degree angle at a height of .666cm from the base of the box. What distance from the lower left corner does the normal force act?

    2. Relevant equations
    Normal force = 403 - 225sin18 = 333.471N
  2. jcsd
  3. Apr 28, 2007 #2


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    The applied force will have a torque about the center of mass of the block. to balance this torque the position of normal force will be shifted accordingly. consider the torque due to friction if any.
  4. Apr 29, 2007 #3
    So the torque of friction should equal the torque of the applied force in order to maintain equilibrium. So I need to find the torque of the x and y components of the applied force to find the applied force's torque and then divide that by the force of friction in order to get the distance.
  5. Apr 29, 2007 #4


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    Friction is a shear force. The net torque (sum of the moments) is zero, otherwise the block would rotate. Is the block being pulled left or right?

    If the applied force is at an angle upward, then there is an upward component that lessens the normal force (e.g. weight downward).
  6. Apr 29, 2007 #5
    I found the torque of the applied force which is 1.12268(225sin18)=78.05 but I just do not know where to go from here.
  7. May 1, 2007 #6


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    Resolve the applied force in horizontal (Fx) and vertical (Fy) direction.

    Friction force is equal to the horizontal component but in opposite direction

    the other forces are the normal force and the weight of the block.

    consider the distance of the normal force equal to x from lower left corner.

    Now calculate net torque due to all forces about any point (consider appropriate sign) and it should be zero. This will give the equation in x solve this to get x.
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