Question

What is the braking torque at leading shoe if resultant frictional force acts at a distance of 250 mm from the brake drum center,  coefficient of friction between the shoe and the drum as 0.5, the free ends of the two shoes are pushed apart with a force of 300 N which is acting at a distance of 320 mm from anchor, and two shoes are anchored together 170 mm away from the brake drum center?

(a) 276.6 Nm

(b) 256.6 Nm

(c) 266.6 Nm

(d) 246.6 Nm

I got this question during an internship interview.

This intriguing question comes from Braking System topic in section Braking System of Automobile Engineering

Answer 1

The correct option is (c) 266.6 Nm

Explanation: TL =\(\frac{W * L * \mu * R}{M-(\mu * R)} = \frac{320 * 300 * 0.5 * 250}{170-(0.5 * 250)}\) = 266666.66 Nmm = 266.6 Nm where L is the distance at which the force acts from the anchor, W is the force from the anchor, μ is the coefficient of friction, R is the distance of resulting friction force from brake drum center, M is the distance between the two anchors.