Allied & ICAR Exam Bank • ALLIED_2798NAT • 2 Marks
Section 2: Farm Machinery
Machine Design
Subtopic: Overload safety devices used in farm machinery
Question Statement:
In a single-plate friction slip clutch with mean friction radius $R_m = 100\text{ mm}$, coefficient of friction $\mu = 0.35$, and total axial spring clamping force $W = 4000\text{ N}$ with two friction surfaces ($n = 2$), the slip torque capacity in $\text{N}\cdot\text{m}$ is:
Answer Key:
280
Step-by-Step Solution & Explanation
Torque $T = n \mu W R_m = 2 \times 0.35 \times 4000\text{ N} \times 0.1\text{ m} = 280\text{ N}\cdot\text{m}$.