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
[tex]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[/tex]
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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