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Homework Help: Box moving up a slope by a force

  1. Nov 17, 2012 #1
    A 40.8 kg box is being pulled up a 13.5 deg slope by a force of 294 N which is parallel to the slope. The coeffcient of kinetic friction between the box and the slope is 0.21. what is the acceleration of the box?
    Equations: F=MA, Fk=μk(MA)
    Attemped at the solution

    deg= 13.5 M= 40.8 kg F=294 μκ= 0.21

    force x-direct---------------------------------------force y-direct

    294(cos13.5)(0.21)=(40.8 kg)A------------------294(sin13.5)(.21)=(40.8)A
    A=1.47 m/s^2-------------------------------------A= 0.35 m/s^2

    acceleration= √(1.47^2+0.35^2)
    acceleration= 1.51 m/s^2

    The answer is 2.91 m/s^2, which I don't understand.
    Last edited: Nov 17, 2012
  2. jcsd
  3. Nov 17, 2012 #2
    I don't really understand your calculations here.

    Have you drawn a free-body diagram with all 4 forces? Put the x-axis along the slope of the incline, and then show us what your expressions for the x and y components are.
  4. Nov 17, 2012 #3
    I have my free body diagram attached.

    Attached Files:

  5. Nov 17, 2012 #4
    For the y-direction ( perpendicular to slope ),
    normal force = weight * cos(13.5) = 389N

    For the x-direction ( up the slope ),
    applied force - friction - weight_x = m*a

    What do you get when you work this out?
    ( remember that friction = μ * normal force )
  6. Nov 17, 2012 #5
    normal force=40.8cos13.5(9.81)=389.19N

    friction=μκ(Nf)=81.73 N

    x-direction= applied force - friction - weight= MA

    212.27 N - 40.8 kg= 40.8 kg A

    am I doing everything right so far?
  7. Nov 17, 2012 #6
    This is all good apart from the weight part - you need the x-component of the weight, ie mgsinθ.
    Once you put that in, it will work out correctly.
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