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Homework Help: A force-related problem

  1. Nov 6, 2006 #1
    A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a force of 80 N at an angle of 25 degree above the horizontal. The box has a mass of 25 kg, and the coefficient of kinetic friction between box and floor is 0.300. a) Find the acceleration of the box b) The student now starts moving the box up a 10 degree incline, keeping her 80 N force directed at 25 degree above the line of the incline. If the coefficient of friction is unchanged, what is the new acceleration of the box?

    Thanks for the patience for reading the problem, my work so far is:
    a) For the y component,
    Fy = ma, since a is 0...
    Ty+Ny-Wy = 0
    80sin25+Ny-25*9.8 = 0
    Ny = 211N

    fx(friction) = uNy = 0.300*211N = 63.4

    For the x component,
    Fx = ma
    Tx-fx = ma
    80cos25-fx = 25a
    72.5-63.4 = 25a
    a = 0.364m/s2. This is my answer, but I'm not sure whether it's right or not, but the problem begins in part b.

    b) For the y component
    Fy = ma
    Ny + Ty + Wy = 0
    Ny + 80sin25-245cos10 = 0
    Ny = 207.5

    f = uN = 0.300 * 207.5 = 62.2N

    For the x component.
    F = ma
    Tx-Wx-fx = ma
    80cos25 - 245sin10-62.2 = 25a
    a = -1.29m/s2

    I do not believe the acceleration to be negative in the incline. If so, the student cannot pull the box up and the problem cannot have taken place. I checked my work for like 10 times. Can someone check what I did wrong? Thanks
  2. jcsd
  3. Nov 6, 2006 #2
    a) looks good, but I didn't check your calculations.

    b) is almost there. You just forgot to include the tilt of the incline when you computed the force of the student on the box.

  4. Nov 6, 2006 #3
    I didn't check your actual numbers (as I had broken the problem down slightly differently.) However, I agree (other than rounding) with your frictional force in part B. I too found that the parallel downhill component of the weight, along with the frictional force add up to more than the parallel uphill force from the string. i.e. it's not going up hill. (and therefore, friction would not actually be 62 Newtons)
  5. Nov 6, 2006 #4
    You sure about that? I may be mistaken, but I drew a free body diagram, and other than checking his final acceleration, I agreed with each of his values. :) Maybe I need help too!

    Besides, just estimating - the parallel component of the acceleration of gravity to the surface of the incline would be 9.81 m/s^2 sin 10 degrees ~1.7m/s^2. Significantly more than his answer in part A. From that, it seems apparent that it's not going to go uphill.
  6. Nov 6, 2006 #5
    "keeping her 80 N force directed at 25 degree above the line of the incline."

    The 25 degrees is above the incline, not above the horizontal... or am I having a complete brain fart here?
  7. Nov 6, 2006 #6
    Hi Sorry...

    I needed more coffee. Yes, I agree, the box won't be moving uphill under these conditions. My mistake, and sorry about that.

  8. Nov 6, 2006 #7
    Alright, thanks guys. I thought I made a mistake up there but I couldn't figure out why. So the box will never be moved under these conditions
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