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?

a) a = - 0.106 m/s^2 (←)

b) T = 12215.1064 N

Explanation:

If

F₁ = 9*1350 N = 12150 N (→)

F₂ = 9*1365 N = 12285 N (←)

∑Fx = M*a = (M₁ + M₂) * a (→)

F₁ - F₂ = (M₁ + M₂) * a

→ a = (F₁ - F₂) / (M₁ + M₂) = (12150-12285) N / (9*68+9*73) Kg

→ a = - 0.106 m/s^2 (←)

(b) What is the tension in the section of rope between the teams?

If we apply ∑Fx = M*a for the team 1

F₁ - T = - M₁*a ⇒ T = F₁ + M₁*a

⇒ T = 12150 N + (9 * 68 Kg) * (0.106 m/s^2)

⇒ T = 12215.1064 N

If we choose the team 2 we get

- F₂ + T = - M₂*a ⇒ T = F₂ - M₂*a

⇒ T = 12285 N - (9 * 73 Kg) * (0.106 m/s^2)

⇒ T = 12215.1064 N