Two blocks connected by a rope of negligible mass are being dragged by a force at a 22° angle above horizontal (see figure below). suppose f = 89.0 n, m1 = 14.0 kg, m2 = 22.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.128.

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

explanation:

wheres the pic