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Two teams of nine members each engage in a tug of war. Each of the first team’s members has an average mass of 68 kg and exerts an average force of 1350 N horizontally. Each of the second team’s members has an average mass of 73 kg and exerts an average force of 1365 N horizontally. (a) What is magnitude of the acceleration of the two teams? (b) What is the tension in the section of rope between the teams?
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Final Answer

a) $0.106 \textrm{ m/s}^2$

b) $1.22 \times 10^4 \textrm{ N}$

Solution Video

OpenStax College Physics Solution, Chapter 4, Problem 17 (Problems & Exercises) (3:22)

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Video Transcript

This is College Physics Answers with Shaun Dychko. We have two tug-o-war teams pulling on a rope and team B is pulling to the right and there are nine people, each exerting a force of 1365 Newtons for a total force of 12,285 Newtons, and there are nine people there each weighing 73 kilograms for a total mass of 657 kilograms. Then pulling the other way we have team A, each pulling with 1350 Newtons, and each having a mass of 68 kilograms. So, our job is to figure out what the acceleration of this system is going to be. So, the total mass of the system is the total mass of both teams put together. We have a net force of <i>Fb</i> to the right, minus <i>Fa</i> to the left, and the net force as always equals mass times acceleration. So since <i>F net</i> equals this and <i>Fnet</i> also equals that, we can equate these two things which we do here. So mass times acceleration equals <i>Fb</i> minus <i>Fa</i> and then we'll divide both sides by <i>m</i> to get the acceleration. So it's the force of team B minus the total force of team A, minus the total mass of the two teams put together. So that's 12,285 Newtons minus 12,150 Newtons, divided by 612 kilograms plus 657 kilograms, which makes 0.106 meters per second squared. Then for part B, we're asked to find the tension in the rope and we can use one of two different perspectives to figure that out. We could use the perspective I've drawn here where we have team A exerting a force to the left and experiencing a force to the right equal to the tension force. We could have also drawn this picture <i>F tension</i> to the left and force of team B to the right, that would also have been fine because the force of tension is the same on each of the teams, the same magnitude anyhow, different directions. We don't need that there. So, the net force on team A which I've put a little apostrophe on it because it isn't the same net force as we were talking about up here. This net force is the net force in the entire system consisting of both teams together. This net force down here refers to the net force on just team A, and so team A experiences a tension force to the right minus the force that they're exerting to the left, and that's going to equal team A's mass times its acceleration. Team A's acceleration will be the same acceleration that we calculated in part A here. So we'll solve for <i>Ft</i> by adding <i>Fa</i to both sides. So it's going to be mass of team A times acceleration plus the force that team A exerts, 612 kilograms times 0.10638 meters per second squared, plus 12,150 Newtons and that gives 1.22 times ten to the four Newtons will be the tension force in the rope.