Question

Two teams of nine members each engage in a tug-of-war, pulling in opposite directions on a horizontal rope. Each of the first team’s members has an average mass of 68 kg and exerts an average force of 1350 N horizontally on the ground as they pull on the rope. Each of the second team’s members has an average mass of 73 kg and exerts an average force of 1365 N horizontally on the ground as they pull on the rope in the opposite direction. (a) What is the magnitude of the acceleration of the two teams, and which team wins? (b) What is the tension in the section of rope between the teams?

a) Second team wins, 0.83 m/s²; 15 N
b) First team wins, 0.83 m/s²; 15 N
c) Second team wins, 1.83 m/s²; 15 N
d) First team wins, 1.83 m/s²; 15 N

Answer 1

The magnitude of acceleration for the two teams is approximately 0.106 m/s² in the direction of the second team, making the second team the winner of the tug-of-war. The tension in the section of rope between the teams is 1350 N.

Step-by-step explanation:

a) Magnitude of acceleration:

To find the magnitude of acceleration, we can use Newton's second law: F = ma. The total force acting on the system can be found by summing the individual forces from each team:
Ftotal = (mass1 x force1) - (mass2 x force2)
Substituting the given values and rearranging for acceleration, we get: a = Ftotal / (mass1 x total mass)

Using the given values:

mass1 = 68 kg, force1 = 1350 N
mass2 = 73 kg, force2 = 1365 N

a = (1350 N - 1365 N) / (68 kg + 73 kg)
a = -15 N / 141 kg
a ≈ -0.106 m/s²

So, the magnitude of acceleration is approximately 0.106 m/s² in the direction of the second team. Therefore, the second team wins the tug-of-war.

b) Tension in the rope:

To find the tension in the section of rope between the teams, we can use the Newton's third law: F = ma. The tension in the rope is equal to the force exerted by one team:

Tension = force1 = force2 = 1350 N

So, the tension in the section of rope between the teams is 1350 N.