Tug-of-war with competing decay of force equations

[calebs17]

Homework Statement

Two frats compete in a tug of war. Each team has 5 players of mass 80kg each. Each person is capable of pulling with 100N of force. The force of one team decays according to F(t) = F0(e^(-t/2)) and the other, F(t) =F0(e^(-t/1)). What is the position of the (very lightweight) rope as a function of time?

Homework Equations

Fnet = ma

The Attempt at a Solution

(m)(a) =F0(e^(-t/1)) - F0(e^(-t/2)) where F0 = 500 for both team. I am not sure if the mass is simply 5*80 or double that because of the two teams. Then I suppose you could integrate this and find x from the acceleration.

 

[TomHart]

I am not sure if the mass is simply 5*80 or double that because of the two teams.

If you're not sure if ma or 2ma, you could draw a free body diagram for each team, sum the forces for each team, and then solve the equations.

Welcome to Physics Forums.

Edit: I just noticed that this post is in "Advanced Physics Homework". I'm not a mentor on this forum, but based on what I've seen, it seems like it would be better if it was posted in "Introductory Physics Homework". You may get much quicker responses there too.

 

[kuruman]

The force of one team decays according to ...

I am not sure what is meant by this. The force of one team acting on what object? If, as the problem states, the mass of the rope is negligible, shouldn't the force exerted by team A on team B be equal and opposite to the force exerted by team B on team A at all times because the tension is the same at the two ends of the rope?

Tugs of war rely on friction to work. So, I suppose, the problem is telling us that the ground is getting more slippery as time increases but by differing amounts for each team. When you draw your FBDs as TomHart suggested, to draw them correctly, you need to take friction into account and assume that it is that friction that decays exponentially. It doesn't mater if you draw one FBD for the two teams together or separate FBDs. It should be clear what mass to use in each case.

 

[TomHart]

shouldn't the force exerted by team A on team B be equal and opposite

I agree with what you are saying. Rather than saying, "Each person is capable of pulling with 100N of force., I think the problem should have been worded: "Each person is capable of exerting a 100 N friction force on the floor as he pulls on the rope."

As a side note: 100 N = 22.5 lb. Those are some pretty wimpy frat boys. They start out weak and quickly go downhill from there.

 

[kuruman]

Those are some pretty wimpy frat boys.

Don't be so quick to condemn them as wimps. Remember, it is the force of static friction that counts. The maximum force of static friction that the ground can exert on one of these boys is the 100 N. This means that the coefficient of static friction is μ_s = F_max/mg = 100 N/800 N = 0.125. Pretty slippery. Probably, the boys are barefoot on linoleum and it is raining thus getting even more slippery.

 

[TomHart]

Probably, the boys are barefoot on linoleum and it is raining thus getting even more slippery.

I'm sorry, but with all due respect, I have to disagree. I think they are outside on a grass field wearing metal cleats. Wimps!