Question

What hanging mass (in kg) must be placed on the cord to keep the pulley from rotating (see the following figure)? The mass on the frictionless plane is 4.3 kg. The inner radius of the pulley is 20 cm, and the outer radius is 39 cm. (Assume the 4.3 kg mass and the hanging mass are only connected to the inner and outer radii, respectively.)

Answer 1

The hanging mass that must be placed on the cord to keep the pulley from rotating is 1.103 kg.

Torque in the cord

To keep the pulley from rotating, the torque in the cord due to the sliding mass must be equal to the torque in the cord due to the hanging mass.

The torque is calculated as follows;

τ₁ = τ₂

F₁r₁ = F₂r₂

m₁gsin(θ)r₁ = m₂gr₂

m₁sin(θ)r₁ = m₂r₂

m₂ = m₁sin(θ)r₁/r₂

where;

m₁ is the sliding mass

• r₁ is the inner radius of the pulley
• r₂ is the outer radius of the pulley
• θ is the angle of inclination of the plane

Substitute the given parameters and solve for m₂

$m_2 = \frac{m_1r_1sin(\theta)}{r_2} \\\\m_2 = \frac{4.3 \times 0.2 \times sin(30)}{0.39} \\\\m_2 = 1.103 \ kg$

Thus, the hanging mass that must be placed on the cord to keep the pulley from rotating is 1.103 kg.

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