Two questions are relevant to the problem posed:
(a) Determine the acceleration of the system. m/s2
(b) Determine the tension T in the rope
Solution:
1) Split the drag force of 89.0N into its vertical and horizontal components, Fy and Fx.
Fx = F × cos(22°) = 89.0 N × cos(22°) = 82.5 N
Fy = F × sin(22°) = 89.0 N × sin(22°) = 33.3 N
2) Calculate the normal forces.
i) m₁ = 14.0 kg ⇒ N₁ = 14.0 kg × 9.81 m/s² = 137.3 N
ii) m₂ = 22.0 kg ⇒ N₂ = 22.0 kg × 9.81 m/s² = 215.8 N
3) Calculate the friction forces
i) Ff₁ = μk × N₁ = 0.128 × 137.3 N = 17.6 N
ii) Ff₂ = μk × N₂ = 0.128 × 215.8 N = 27.6 N
3) Total mass of the system: m = m₁ + m₂ = 14.0 Kg + 22.0 Kg = 36.0 Kg
4) Horizontal net force:
Fr = Fx - Ff₁ - -Ff₂ = 82.5 N - 17.6 N - 27.6 N = 37.3
5) Acceleration:
Newton’s Law: F = m × a ⇒ a = F / m
a = 37.3 N / 36.0 kg = 1.04 m/s² ← acceleration
6) Tension force, T
Draw the forces around the object over which it is being applied the force of 89.0N, which is m₂.
T + Ff₂ = Fx ⇒ T = Fx - Ff₂
T = 82.5N - 27.6N = 54.9N ← tension in the rope
