Question

How do you compute the net force acting on the rope in a game of tug of war?

Answer 1

If you pull the leftmost piece with 20N and rightmost with 30N, you would get 10N on a 0 mass object, meaning infinite acceleration. Therefore you have to assume there are two bodies (actually only one would be enough). So, one person pulling on each side of the rope. Assuming the rope is massless, and is consisted of lots of tiny pieces, we can see that each tiny piece has two forces on it. FlFl from the left and FrFr from the right. Since acceleration of each tiny piece is a=Frâ’Flma=Frâ’Flm and m->0 then FrFr = FlFl, otherwise aa would be infinite. Also, this means that any piece of the rope, including the ones that connect to the bodies, pull adjacent pieces with the same force. That is, body A and body B are both pulled by the rope with the same force TT, which is the tension. There is FA=20NFA=20N on the left body and FB=30NFB=30N on the right, plus TT and â’Tâ’T respectively. Since bodies A, B and the rope have the same acceleration (if they didn't, they'd move apart from each other), we get: a=Tâ’FAmA=FBâ’TmBa=Tâ’FAmA=FBâ’TmB , meaning T=mAFB+mBFAmA+mBT=mAFB+mBFAmA+mB. If mBmB->0 then FBâ’T=0FBâ’T=0, and T=FB=30NT=FB=30N. Likewise if mAmA->0 then T=FA=20NT=FA=20N. You can not assume that mB=mA=0mB=mA=0 without FA=FBFA=FB, therefore you need at least one non 0 mass object attached to the rope

Answer 2

A force is simply a "push" or a "pull" measured in Newton Force, let us assume that there are 10 kids pulling on both opposite end of a rope, and each kid has an average mass of 40 Kg, this means each kid can pull on a rope of equivalent of 400 N. If we have 10 kids on each end of the rope, therefore the total tension on the rope could be around 4000 N. This would clearly explain that there are two 4000 N in each rope, creating a huge tension horizontally on both ends of the rope.