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Boy pushing box along floor - friction

  1. Sep 20, 2005 #1
    A boy is pushing a 80 kg box along the floor. The force is exerted at the top of the box with an -30 degree angle with respect to the horizon. The kinetic friction is 0.40, and g is 9.8m/s^2.

    Find the force needed to keep the box moving with a constant velocity.

    80kg = 784 N

    Force needed to keep it moving: 784*0.40= 313.6 N

    Since we want constant velocity, the kinetic forces and the applied forces need to be in equilibrium.

    313.6N = mg*sin 30 + mg*cos30
    313.6N - (mg*sin 30 + m*cos 30) = 0

    Is this correct?
     
  2. jcsd
  3. Sep 20, 2005 #2

    HallsofIvy

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    The question asked you to find the force necessary to keep the box moving. You immediately calculated
    "80kg = 784 N

    Force needed to keep it moving: 784*0.40= 313.6 N"

    which doesn't use the angle at all. You then went ahead and wrote:
    "313.6N - (mg*sin 30 + m*cos 30) = 0".

    For what reason? Hadn't you already found the force? Oh, and what does "m" represent in that equation? Surely not the mass of the box- you know that is 80 kg.
    Doesn't that seem a bit peculiar?

    The force you calculated, 784 N, is the horizontal force necessary to keep the box moving. The problem is asking for the total force (at 30 degrees) whose horizontal component is 784 N.
     
  4. Sep 20, 2005 #3
    I'm lost here. Is it as simple as saying
    mg*sin30*0.40 + mg*cos30 = force needed to keep it moving at constant velocity?
     
  5. Sep 20, 2005 #4

    Doc Al

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    Approach it systematically:

    What forces act on the box? (I see four.)
    What are the horizontal and vertical components of these forces?
    Apply the conditions for equilibrium for each direction.
     
  6. Sep 20, 2005 #5

    HallsofIvy

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    Didn't I just point out that there is NO "m" in this problem? The only mass involved is the 80 kg you are given. Further, the problem asks you to find force but there is no unknown representing the force in that equation!

    Further, you NEVER calculate mg sin 30+ mg cos 30 for a force or acceleration- those are components of the vector, they are not added together.

    You calculated that the weight of the was 784 N and so the friction force resisting motion (assuming by "The kinetic friction is 0.40" you mean the coefficient of kinetic friction) is 0.4*784= 313.6 N.

    The horizontal component of force must be equal to that to move the box without acceleration. Call the magnitude of the force applied (that's what you are looking for) F. Since that force is at 30 degrees to the floor, what is the horizontal component of force (it will depend on F, of course) and set it equal to 313.6 N. Solve for F. It's that simple!
     
    Last edited: Sep 20, 2005
  7. Sep 20, 2005 #6

    mukundpa

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    HallsofIve

    I think you have forgotten the vertical component of the force applied, which is going to change the normal reaction and hence the friction force.

    regards
     
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