Question Bank • QB_FM_ADV_006 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:

A flat belt drive transmits $15\text{ kW}$ power at a belt speed of $20\text{ m/s}$. The angle of lap on the smaller pulley is $165^\circ$ ($2.88\text{ rad}$) and the coefficient of friction is $\mu = 0.3$. The belt has a mass of $0.4\text{ kg/m}$. Considering centrifugal tension, the maximum tension in the belt is ________ $\text{N}$ (round off to integer).

Answer Key:

1456

Step-by-Step Solution & Explanation

1. Effective tension ($T_1 - T_2$): $P = (T_1 - T_2) v \implies T_1 - T_2 = \frac{P}{v} = \frac{15000\text{ W}}{20\text{ m/s}} = 750\text{ N}$ 2. Tension ratio with friction: $\frac{T_1 - T_c}{T_2 - T_c} = e^{\mu \theta} = e^{0.3 \times 2.88} = e^{0.864} = 2.3726$ Let $T_1' = T_1 - T_c$ and $T_2' = T_2 - T_c$, so $T_1' - T_2' = 750\text{ N}$. $T_1' = 2.3726 T_2' \implies 2.3726 T_2' - T_2' = 750 \implies 1.3726 T_2' = 750 \implies T_2' = 546.4\text{ N}$ $T_1' = 546.4 + 750 = 1296.4\text{ N}$ 3. Centrifugal tension ($T_c$): $T_c = m v^2 = (0.4\text{ kg/m}) \times (20\text{ m/s})^2 = 0.4 \times 400 = 160\text{ N}$ 4. Maximum belt tension ($T_1$): $T_{\text{max}} = T_1' + T_c = 1296.4 + 160 = 1456.4\text{ N} \approx 1456\text{ N}$ Let's verify: $750 / (1 - e^{-0.864}) + 160 = 1296.4 + 160 = 1456.4\text{ N}$.
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