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    • Thread Starter

    This is one question I haven't exactly done before:
    A builder holds up a vertical piece of particle having weight 40 N by pressing the two sides, force P Newtons with his hands. If mu = 1/4 (0.25) what is the least value of P needed if the particle is not to slip.
    I know, Friction = 1/2 W
    So Friction = 20N
    What do I do from here?
    • PS Helper

    PS Helper
    • Thread Starter

    That one liner is the one of the best

    F = uR
    (u = mew!)

    = 0.25 x 40 = 10N

    So in one direction (downwards) you have 40N.
    Friction acts in the opposite direction, with a magnitude of 10N.
    You want an overall force of magnitude 0N.
    P will act in the same direction as the friction; it's stopping this 'vertical particle' (wtf XD) from slipping down.
    So what is P?

    Edit: I think I may be wrong.

    First of all, the frictional force, is it per hand or what?
    Secondly, I'm not sure that R is 40N. I think R = P, or something.

    40 = P + 0.25R
    R = P
    40 = 1.25P
    P = 32N

    And if the frictional force is per hand, then:

    40 = P + 9.5R
    R = P
    40 = 1.5P
    P = 26.67N (2dp)

    At least one of these has to be right.
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