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Homework Help: Newton Application Problem

  1. Oct 16, 2014 #1
    Unless otherwise stated, all objects are located near the Earth's surface, where g = 9.80 {\rm m/s^2}.
    A person has a choice while trying to push a crate across a horizontal pad of concrete: push it at a downward angle of 29∘ , or pull it at an upward angle 29∘.

    If the crate has a mass of 47.0kg and the coefficient of kinetic friction between it and the concrete is 0.740, calculate the required force to move it across the concrete at a steady speed for both situations.

    2. Relevant equations
    I'm stuck on this problem, I don't know if I have my equations mixed up or I'm calculating the wrong thing.
    Fk (is Kinetic Friction)
    mg is mass * gravity (weight of the object)
    muk (coefficient of kinetic friction - given as 0.74).
    N is the Normal force (same as Fn as explained below).

    Fn= mg*sin(theta)
    Fk= muk*N
    plug Fn into N since the coefficient of friction is given.

    The frictional force when pulling is µW1:
    460.6 - sin(29)Fp (29 degrees is given - plugged into place of theta).
    The frictional force when pushing is µW2:


    Ff(force of friction) Pull:
    muk (w - sin(theta)
    = (0.74) (47*9.8-sin(29) = 340.8 - 0.359Fp
    Ff Push:
    (0.74)(47*9.8+sin(29) = 340.8+0.359Fp

    The horizontal component is Fp·cos(29°)
    Fp*cos(29) = 0.875Fp
    The net force when pulling is:
    0.875Fp - (340.8 - 0.3588) = 0 -> 0.875Fp+0.3588 Fp -340.8 -> = 1.23 Fp - 340.8
    The net force when pushing is:
    0.875Fp - (340.8+0.3588) -> 0.875Fp - 0.3588 Fp -340.8 -> = 0.5162Fp - 340.8

    Since it's moving at constant speed, does that mean the horizontal component Fp equal to the frictional force in each case?

    This is basically where I'm lost at. I try setting Fp = Ff but I'm not getting the correct answers.
    Last edited: Oct 17, 2014
  2. jcsd
  3. Oct 16, 2014 #2


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    Homework Helper

    Welcome to PF!

    Life would be easier if you explained what your notations mean.
    You are right, constant velocity means that both the horizontal and vertical components of the net force are zero.

  4. Oct 16, 2014 #3
    Sorry I tried writing out what the notations meant before working the equation out. I'll edit it, and then bold it for you.
  5. Oct 17, 2014 #4
    Nevermind I think I got it.
    For Pull: it would be cos(29) + (0.74)*sin(29) = 1.23
    340.8 N / 1.23 = 276 N
    For Push: cos(29) - (0.74)*sin(29) = 0.516
    340.8 / 0.515 = 661 N
  6. Oct 17, 2014 #5


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    Homework Helper

    It is correct now.
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