Question

A belt is wrapped around a pulley as shown in the figure. The coefficient of friction between the belt and pulley is \(0.3\). If there is no slippage between the belt and pulley, the angle of wrap (\(\theta\)), in degrees, is . (Rounded off to two decimal places)

Hint:

Use the belt friction (capstan) equation relating the tight side and slack side tensions to the coefficient of friction and the angle of wrap.

Answer 1

Correct Answer: 132.38

Step 1: Identify which formula governs belt friction.
When a flat belt wraps around a pulley without slipping, the tension is different on the two sides because friction between the belt and the pulley surface holds part of the load. The relationship between the tight side tension \(T_1\) and the slack side tension \(T_2\) at the point of slipping is given by the belt friction (capstan) equation:
\[ \frac{T_1}{T_2} = e^{\mu\theta} \]
where \(\mu\) is the coefficient of friction and \(\theta\) is the angle of wrap in radians.

Step 2: Identify the tight and slack side tensions.
From the figure, the belt pulls with \(5.0\) kN on one side and \(2.5\) kN on the other side. The larger pull is the tight side, so \(T_1 = 5.0\) kN and \(T_2 = 2.5\) kN.

Step 3: Solve the capstan equation for the angle of wrap.
Taking the natural log of both sides of the equation,
\[ \ln\left(\frac{T_1}{T_2}\right) = \mu\theta \]
\[ \theta = \frac{1}{\mu}\ln\left(\frac{T_1}{T_2}\right) \]

Step 4: Substitute the given values.
\[ \frac{T_1}{T_2} = \frac{5.0}{2.5} = 2 \]
\[ \theta = \frac{\ln 2}{0.3} = \frac{0.693147}{0.3} = 2.31049 \text{ rad} \]
Converting radians to degrees by multiplying by \(180/\pi\):
\[ \theta = 2.31049 \times \frac{180}{\pi} = 132.38^{\circ} \]

Final Answer:
\[ \boxed{\theta = 132.38^{\circ}} \]