Question Bank • QB_FM_079 NAT • 2 Marks

Section 2: Farm Machinery

Machine Design

Subtopic: Design and selection of machine elements – gears, pulleys, chains and sprockets and belts

Question Statement:

An open flat belt drive connects two pulleys of diameters $300\text{ mm}$ and $600\text{ mm}$ with a center distance of $2.5\text{ m}$. Calculate the angle of lap on the smaller pulley in $\text{radians}$.

Answer Key:

3.02

Step-by-Step Solution & Explanation

For an open belt drive: $\sin \alpha = \frac{D - d}{2 C} = \frac{0.60 - 0.30}{2 \times 2.5} = \frac{0.30}{5.0} = 0.06$ $\alpha = \arcsin(0.06) \approx 0.06004\text{ rad} = 3.44^\circ$ Angle of lap on smaller pulley: $\theta = \pi - 2\alpha = 3.14159 - 2(0.06004) = 3.14159 - 0.12008 = 3.0215\text{ rad}$
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