What is the tension in the center of the rope in a tug-of-war?

[SnackMan78]

Textbook question: In a tug-of-war, each man on a 5-man team pulls with an avg force of 500N. What is the tension in the center of the rope?

Is the answer 2500N because each mans force is added together, or since the question uses avg, is the answer 500N? OR am I wrong with both conclusions?

 

[berkeman]

Seems like 2500N, unless I'm missing something.

 

[blimkie]

Yes its 2500 the average force they each pull with is 500. you have to combine them all to find the net force because it asks for tension.

 

[daniel_i_l]

Is there only one team, or two pulling against each other? If the latter is true then the answer should be double.

 

[topsquark]

Is there only one team, or two pulling against each other? If the latter is true then the answer should be double.

I agree. EACH team member exerts 500 N, so one team exerts 2500 N. Thus if there are two teams we have 2*2500 N = 5000 N on the rope.

-Dan

 

[NeRdHeRd]

Wouldn't the net force actually be zero. Each team is applying an average force of 2500N but in opposite directions thus the net force in the center of the rope would be 2500N-2500N = 0.

 

[rootX]

what would be the tension in the rope that has M mass hanging on its both ends, and goes through a pulley?

I got Mg

T-Mg=0
Mg-T=0
subtract them.

 

[PhanthomJay]

If team A pulls on one end with a force, and team B pulls on the other end with an equal but opposite force, there is no net external force on the rope; thus, the rope is in equilibrium (F_net = 0). However, that does not imply that the internal force in the rope is 0. That rope is being strained tremendously, and if it could feel pain, it'd know that the force on it would be greater than 0. So is it 500, 2500, or 5000? Draw a free body diagram around the left side of the rope that cuts thru its center to find the result, noting that F_net=0.