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Homework Help: Maximum force on a rope

  1. Jan 11, 2009 #1
    1. The problem statement, all variables and given/known data

    A mass m is attached by a rope to car. The car accelerates to speed V before the rope becomes taut. The rope is length R and will stretch and additional length Rs. Find the maximum force on the rope. Ignore friction and assume that the rope runs parallel to the ground.

    2. Relevant equations
    I tried this using both impulse and work, so


    t=Rs/V (where t is the total time of the stretching process; the stretching essentially occurs as fast as the car is moving)


    3. The attempt at a solution

    I reached a solution using both impulse and work, but the maximum force I get using impulse is twice what I get when using work. I haven't been able to figure out where I went wrong.

    Using impulse,

    J=mV=Ft however, solving for F would just give the average force. The force should increase linearly with respect to time, because as more time passes the more the rope has stretched, so the maximum force should be greater than the average. Since impulse is the integral of force with respect to time, and because the area of the force-time graph is a triangle, J=(Fmax*t)/2
    and Fmax=2J/t

    substituting then gives Fmax=2mV/t


    Using work,

    W=mV2/2=F*Rs similarly, solving for F would just give the average force. As the rope is stretched farther, the force on the rope should increase linearly, so the graph of force with respect to distance is a line. The integral of this graph should be equal to work, and once again the area under of the graph is a triangle so W=(Fmax*Rs)/2

    and substituting gives Fmax=mV2/Rs

    If you compare my result when using impulse when using work, it's pretty clear that impulse is twice as big. I've gone over my work a ton and can't find what's wrong. I suspect it's something with how I switched from finding the average force to finding the max force, but when looking at my impulse equations and graphs and my work equations and graphs separately they both seem to make sense. Anyone have any ideas?
  2. jcsd
  3. Jan 11, 2009 #2
    I believe the problem with your analysis is using the work approach. Think of the rope behaving like a spring. The the work done in stetching a spring is

    W = (1/2)kx2 where x = Rs.

    But, you do not know k for the rope. The force exists during the impulse time, and the force varies in time in a complex way so it cannot be assumed the force varies linearly with time. Since you were not given F as a function of time, only the average force can be calculated.
  4. Jan 11, 2009 #3


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    Homework Helper

    That's quite a nice pair of solutions, Shadow! Most impressive. I started out like chrisk suggested, with F = kx for the spring, and got into all kinds of trouble that you have neatly avoided with both of your approaches.

    The second equality looks suspicious. You get it from V = d/t = Rs/t but that uniform motion formula does not apply to this accelerated motion. You may have one of my difficulties in correcting this - we have no handy formulas for motion with increasing acceleration. Perhaps your assumption of steadily increasing acceleration will yield a solution or you could just go with your other one, which looks good to me.
  5. Jan 11, 2009 #4
    I should have made this clear, we do know that the rope will act like a spring (and the force is linear). I was using F=k*Rs/R (basically the spring equation, except that the force depends on the relative stretch of the rope--how much it stretches divided by how long it is; I've checked on this and am very sure it is correct). I was just trying to keep this a little simpler and didn't include it.

    Delphi, shouldn't uniform motion equations still work if I am using the average velocity and acceleration to find impulse? Even though the velocity is increasing, the total change in momentum should still be mV, and since the force increases linearly, mV should still equal Fmax*t/2

    I could understand that the equations wouldn't work at all if my first answer was completely different from my second, but it just seems that since it's exactly twice as much there must just be some relatively simple error that I'm missing somewhere. Thanks for the help so far.
    Last edited: Jan 11, 2009
  6. Jan 11, 2009 #5


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    Homework Helper

    There are lots of things we don't know. If the rope stretches to its limit before the mass is up to speed there will be an enormous force. So there is little choice but to assume the rope stretches gradually out to Rs and that is the point where the mass reaches speed V.
  7. Jan 11, 2009 #6
    Your expression for force

    F = k*Rs/R

    is not correct. The length units cancel out and it's no longer of the form F = kx. Since F is linear as is the case here then F = kx, the k value is the slope and this value has not be given.
  8. Jan 11, 2009 #7
    so do you think that the impulse equation is right?
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