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Finding min/max force (net force sliding friction problem)

  1. Sep 24, 2009 #1
    1. The problem statement, all variables and given/known data
    A block of mass m is pulled across a level surface by a rope that makes an angle θ with the horizontal. The coefficient of friction is μ.

    (a) Determine the amount of force F required to slide the block at a constant velocity.

    (b) Determine the optimum angle at which to pull on the block (so that the required force is minimized).

    (c) If the force of the rope is 15.0 N acting on a block of mass 2.00 kg where μ = 0.35, what is the maximum acceleration possible?


    2. Relevant equations

    Net force in x and y directions; Fk = [tex]\mu[/tex]Fn; vector components


    3. The attempt at a solution

    a) The solution to part a) is found by combining the information gained when writing the net force equations in the x and y directions. The final, simplified solution for the force required to slide the block at constant velocity is F = ([tex]\mu[/tex]mg)/(cos[tex]\theta[/tex] + [tex]\mu[/tex]sin[tex]\theta[/tex])

    b) Not sure...To find the optimum angle (presumed to correspond to min force) I think I would want to find the derivative of my solution (wrt theta) to the force equation found in part a) and set the derivative equal to zero and then solve for theta. I have been unsuccessful in doing this thus far because the derivative I found for the force equation is complex and not easy to solve for [tex]\theta[/tex] when set equal to zero. Help!

    c) I would assume you could find the maximum acceleration by finding the maximum force which would be accomplished by finding the derivative as in part b), but since I've had little success there, I'm stuck here. One might assume that the max acceleration would occur at an angle of zero, but this is not necessarily true I believe because any verticle force component would reduce the frictional force, thus increasing the net acceleration of the box. Regardless, I need help here!


    Finally, the answers to parts a, b, and c are listed below. Alas, while I have the answers, it is the solutions I seek. Please help :)

    a. F = ([tex]\mu[/tex]mg)/(cos[tex]\theta[/tex] + [tex]\mu[/tex]sin[tex]\theta[/tex])

    b. [tex]\theta[/tex] = arctan([tex]\mu[/tex])
    c. 4.5 m/s2, 0°
     
    Last edited: Sep 24, 2009
  2. jcsd
  3. Sep 24, 2009 #2

    Delphi51

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    Homework Helper

    Welcome to the forum, Ultimate! You know what you are doing. That part (a) is difficult.

    Your approach for (b) is certainly correct, but it is tricky to find the angle that makes the derivative zero. Notice that it has a constant umg (which you can ignore) divided by a squared quantity that is always positive (so you can ignore that, too). Concentrate on the remaining factor - the only way the derivative can be zero is for that factor (u*cos - sin) to be zero. Sorry, I don't know how to make a theta here.
    Setting it equal to zero and solving for theta should give you the answer you are looking for.

    In (c) you can just use the angle you found in (b) - no need to do the derivative again.
     
  4. Sep 24, 2009 #3
    For (b) you start with what you found in (a).
    F = ([tex]\mu[/tex]mg)/(cos[tex]\theta[/tex] + [tex]\mu[/tex]sin[tex]\theta[/tex])
    You can take the derivative of that, which really isn't that complex, then just set the numerator equal to zero (aka set the entire thing to zero then multiply by the denominator). Theta is easy to solve for there. Another way you could solve it is to realize that the top of F=... is constant. The only thing that changes is the bottom. F is smallest when the bottom is largest. So you can just take the derivative of the bottom and find when that is zero.
    For (c) just use what you found in (b) and plug that into (a).
     
  5. Sep 24, 2009 #4
    Thanks for the responses. Ah, I was so close in part b)! I had the derivative but focused on how difficult it was to solve rather than simply making observations about the possible ways it could equal zero.

    Ah...I think I misunderstood the question in part c). It says "find the maximum acceleration possible" when in reality it might as well say "find the acceleration."

    Ok I should be able to finish it off now. Thanks guys!

    Edit: Got it. Super duper. Thanks again.
     
    Last edited: Sep 24, 2009
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