(adsbygoogle = window.adsbygoogle || []).push({}); 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

J=mV=Ft

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

W=(mV^{2})/2=F*Rs

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

=2mV^{2}/Rs

Using work,

W=mV^{2}/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=mV^{2}/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?

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# Homework Help: Maximum force on a rope

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