A tug of war is held between two teams of five men each. The mass of each man is

Question

A tug of war is held between two teams of five men each. The mass of each man is 80 kg and he can initially pull on therope with a force of 1000 newtons.  The teams start evenlymatched, but as the men tire, the force with which each manpulls decreases accordingly: where T is the mean tiring time.  The team on the lefthas a mean tiring time of TL = 10s while the teamon the ight has a mean tiring time of TR =20s.  Find the following: a.)  the velocity of the two teams b.) the position of the teams as function of time c.) evaluate both velocity and position after one minute'stime. A tug of war is held between two teams of five men each. The mass of each man is 80 kg and he can initially pull on therope with a force of 1000 newtons.  The teams start evenlymatched, but as the men tire, the force with which each manpulls decreases accordingly: where T is the mean tiring time.  The team on the lefthas a mean tiring time of TL = 10s while the teamon the ight has a mean tiring time of TR =20s.  Find the following: a.)  the velocity of the two teams b.) the position of the teams as function of time c.) evaluate both velocity and position after one minute'stime.

Explanation / Answer

FR = 1000 * e-(t /TR) FL= 1000 *e-(t / TL FR -FL = 1000 * e-t *(e1/TR - e1/TL ) = 1000* et *(e-.05 - e-.1) = 1000 *et * (.951 - .905) FR -FL = 1000 * e-t *(e1/TR - e1/TL ) = 1000* et *(e-.05 - e-.1) = 1000 *et * (.951 - .905) So the net force F = 46 * et Now you can use F = m * a v = a* t s = 1/2 * a * t^2 etc. Where dv = a * dt ds = v *dt The integrations are simple because det / d t = et Now you can use F = m * a v = a* t s = 1/2 * a * t^2 etc. Where dv = a * dt ds = v *dt The integrations are simple because det / d t = et

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