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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. What is the tension in the section of rope between the teams?

Respuesta :

Answer:

65.01 N

Explanation:

N = 9

m1 = 68 kg

m2 = 73 kg

F1 = 1350 N towards left

F2 = 1365 N towards right

Fleft = N x F1 = 9 x 1350 = 12150 N

FRight = N x F2 = 9 x 1365 = 12285 N

Fnet  = Fright - Fleft = 12285 - 12150 = 135 N right

Let a be the acceleration

a = Fnet / N(m1 + m2) = 135 / 9(68 + 73) = 0.1064 m/s^2

Let T be the tension in the rope

By the Newton's second law

Fnet - T = 9 m2 x a

135 - T = 9 x 73 x 0.1064

T = 65.01 N

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