A10 kg box resting on a horizontal, frictionless surface is attached to a 6 kg
weight by a thin, light wire that passes over a pulley . the pulley has the shape of a uniform solid disc of mass 2.5 kg and diameter 0.650 m. after the system is released, calculate
(i) the tension in the wire on both sides of the pulley.
(ii) the angular acceleration of the pulley.

T₁ = 34.1 N

T₂ = 38.4 N

α = 10.5 rad/s²

Explanation:

Sum of the forces on the box in the x direction:

∑F = ma

T₁ = m₁ a

Sum of the torques on the pulley:

∑τ = Iα

T₂ r − T₁ r = (½ mr²) α

T₂ r − T₁ r = (½ mr²) (a / r)

T₂ − T₁ = ½ ma

Sum of the forces on the weight in the y direction:

∑F = ma

T₂ − m₂ g = m₂ (-a)

T₂ = m₂ g − m₂ a

Substitute:

(m₂ g − m₂ a) − (m₁ a) = ½ ma

m₂ g = (m₁ + m₂ + ½ m) a

a = m₂ g / (m₁ + m₂ + ½ m)

Given m₁ = 10 kg, m₂ = 6 kg, and m = 2.5 kg:

a = (6) (9.81) / (10 + 6 + ½ (2.5))

a = 3.41 m/s²

Therefore:

T₁ = m₁ a

T₁ = (10) (3.41)

T₁ = 34.1 N

T₂ = m₂ g − m₂ a

T₂ = (6)(9.81) − (6)(3.41)

T₂ = 38.4 N

α = a / r

α = 3.41 / 0.325

α = 10.5 rad/s²

d.7.5cm is your answer

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