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Nice NSL problem w/ calculus

  1. Aug 7, 2004 #1
    A block rests on a horizontal surface with a rope attached. Assume the block is a pointlike particle in order that any moments on the box can be neglected. If kinetic friction opposes the motion of the block as it is dragged by a tensile force induced on the block by the rope, then what angle above the horizontal should the rope be pulled if the acceleration of the box is to be maximized for any given mass and tension? Hint: the solution is quite a simple answer
     
  2. jcsd
  3. Aug 7, 2004 #2
    Hoo boy. I thought I did well in physics but I question for the first time how I pulled off doing well ever, oops. No matter what the mass is, the tension must be greater than the original static friction, then we also know that the tension in the rope must be greater than the kinetic friction. Now my memory is getting fuzzy. Ok so here's my final answer then I guess in my stupor having just woken up from a nap: If k > 1, the angle should be 90 degrees, if k < 1, it should be 0 degrees. If k = 1, it doesn't really matter perpendicular or parallel. Any other angle I can only see it maximize the friction + gravity components. Oh and k is the coefficient of kinetic friction. I remember seeing in a book there was some substance with a coefficient of like 1.2 but feel free to correct me, I really am drawing a blank and my grasp of physics on fundamental levels like this is laughable.
     
    Last edited by a moderator: Aug 8, 2004
  4. Aug 8, 2004 #3

    Galileo

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    The horizontal component of the net force is:
    [tex]F\cos\theta - F_r[/tex]
    Where [itex]F_r[/itex] is the frictional force.

    [tex]F_r=\mu N[/tex]
    Where N is the (magnitude of the) normal force.
    [tex]N=mg-F\sin\theta[/tex]

    Therefore:

    [tex]F\cos\theta-\mu(mg-F\sin\theta)=ma[/tex]

    Which gives for the acceleration:

    [tex]a=\frac{F}{m}(\cos\theta+\mu\sin\theta)-\mu g[/tex]

    You can work out the angle [itex]\theta_m[/itex] for which this is a maximum and you'll find:

    [tex]\theta_m=\arctan(\mu)[/tex]

    It seems the answer is dependent on the coefficient of friction alone.
    If there is alot of friction, the angle higher. If there is no friction you should pull horizontally.

    So if you want to drag along a heavy object by a rope, you should first
    pull more or less vertically (high angle) to overcome the static friction, then
    when it moves you should pull more horizontally.
    I don't know what a typical coefficient of friction is (for stone for example),
    but it may yield some practical information for when you're moving out of the house :smile: .
     
  5. Aug 8, 2004 #4
    typically coefficients of friction lie between 0 and 1, a very very rough surface would have a coefficient of friction of 1, if the coefficient was one, then in order to push a 100kg block on the ground you would need to exert a force of 9800 N
     
  6. Aug 8, 2004 #5
    Galileo is correct with his answer (isn't he always correct...i mean...about moons and Jupiter...acceleration on Earth independent of mass.....except maybe for his non-relativistic work on motion for frame of references). Anyways, the answer to the problem is that the angle = arctan u. If the coeficient were 0, then the angle would be 0, which makes sense. If the coefficient were 1, then the angle would be 45 degress, which also makes sense.
     
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