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Kinetic Friction ProblemNeed Help

  1. Nov 1, 2006 #1
    Kinetic Friction Problem..Need Help!!

    Okay...so I really think I'm doing this right..but the CAPA thing won't take my answer so there has to be something I'm doing wrong.
    The Question is ...
    A 32.0 kg box is being pulled along a horizontal surface by means of a rope that exerts a force of 265 N at an angle of 54.5o above the horizontal. The coefficient of kinetic friction between the box and the surface is 0.330. What is the acceleration of the box?

    So I drew my free body diagram and I had the Normal Force and the Weight and cosine of the angle in the y direction and I had the kinetic friction (normal force times kinetic friction coefficient) and weight and sine of the angle in the X direction.

    Then I worked it all out and I got
    Fy = may
    Fy = 0
    N - W(cos54.5) = 0
    N - mg(cos54.5) = 0
    N = 32.0kg (9.81 m/s^2)(cos54.5)
    N = 182.3

    Fx = max
    -fk + T - w(sin54.5) = max
    T - (N x uk) - mg(sin54.5) = max
    265 - (182.3 x 0.330) - (32kg)(9.81m/s^2)(sin54.5) = 32kg(ax)
    then I solved It all and it worked out to be -1.59 m/s^2
    but its saying its not right..so I'm confused beyond belief now.
    so Please help me .. lol:confused:
     
  2. jcsd
  3. Nov 1, 2006 #2
    Check your trig functions in calculating N.
     
  4. Nov 1, 2006 #3
    I've switched the trig functions around and tried that answer but still nothing. and I'm almost positive I have my trig done right anyways...but thanks.
     
  5. Nov 1, 2006 #4
    So you're sure [tex]N - w \cos{54.5} = 0[/tex]?
     
  6. Nov 1, 2006 #5
    well there is no acceleration in the y direction so I guess it would have to be that.
     
  7. Nov 1, 2006 #6
    Look at the x- and y-components of all the forces, and determine what you have.

    [tex]\Sigma F_{x} = T_{x} - F_{f}[/tex]

    [tex]\Sigma F_{y} = N + T_{y} - mg[/tex]
     
  8. Nov 1, 2006 #7
    Why the cosine? Like I said, rethink your trig.
     
  9. Nov 1, 2006 #8
    I didn't think you had to incorporate the tension into the Y direction...isn't it just in the X?
     
  10. Nov 1, 2006 #9
    Re-read the question, and redraw your free body diagram. The surface is horizontal, but the rope is not.

    There is certainly a y-component of the tension, and it alleviates some of the normal force.
     
  11. Nov 1, 2006 #10
    okay I'll go do that and hopefully figure this out..thanks
     
  12. Nov 1, 2006 #11
    Okay I just redid it and realized my stupid mistake. I thought that the box was on a slope..but then when I reread it I realized that the box was on a straight horizontal and it was the rope that was on the angle. Thanks a bunch! :D
     
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