Question Bank • QB_FP_062 NAT • 2 Marks

Section 3: Farm Power

Tractors and Power Tillers

Subtopic: Tractor clutches and brakes

Question Statement:

A differential band brake installed on a tractor final drive has a brake drum diameter of $D = 350\text{ mm}$ ($R = 0.175\text{ m}$). The angle of contact of the friction band is $\theta = 240^\circ$ ($4.189\text{ rad}$), and the friction coefficient between lining and drum is $\mu = 0.35$. If the tension on the slack side of the band is $T_2 = 800\text{ N}$, the braking torque exerted on the brake drum in $\text{N}\cdot\text{m}$ is ________ (round off to 1 decimal place).

Answer Key:

466.5

Step-by-Step Solution & Explanation

1. Tension ratio across band brake: $\frac{T_1}{T_2} = e^{\mu \theta} = e^{0.35 \times 4.189} = e^{1.46615} \approx 4.3325$ $T_1 = 800 \times 4.3325 \approx 3466.0\text{ N}$ 2. Braking torque ($T_b$): $T_b = (T_1 - T_2) \times R = (3466.0 - 800) \times 0.175 = 2666.0 \times 0.175 \approx 466.55\text{ N}\cdot\text{m} \approx 466.5\text{ N}\cdot\text{m}$
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