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Normal Force

  1. Oct 5, 2009 #1
    1. The problem statement, all variables and given/known data
    Particle P has a mass of 1 kg and is confined to move along the smooth vertical slot due to the rotation of the arm AB. Assume that at the instant as shown, the acceleration of particle P is 2.46 ms-2 upward. Determine the normal force Nrod on the particle by the rod and the normal force Nslot on the particle by the slot in the position shown. Friction forces are negligible.


    2. Relevant equations
    F = ma

    3. The attempt at a solution
    I think that the Nslot is directed out of the page and equals to W so Nslot = 1*9.8 = 9.8 N

    Then, Nrod is directed to the top left and is perpendicular to AB. So, Nrod can be projected to Nrod cos 30o directed vertically. Hence :
    Nrod cos 30o = m*a
    Nrod = 2.84 N

    But there is no such answer in the multiple choices....:grumpy:

  2. jcsd
  3. Oct 5, 2009 #2


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    Homework Helper

    Nslot is normal to the walls of the slot, so it is in-plane and horizontal. As for Nrod, you are right, Nrod is perpendicular to AB, but there is the weight of the body which is vertical and points downward.

  4. Oct 5, 2009 #3
    Hi ehild

    Ok I think that Nslot is the projection of the normal force in horizontal direction and yes I forgot to take the weight into account. I think I figure out the answer

    Thanks a lot !
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