A short shoe drum (radius 260 mm) brake is shown in the figure. A force of 1 kN is applied to the lever. The coefficient of friction is 0.4.

The magnitude of the torque applied by the brake is _______ Nm (round off to one decimal place).

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GATE ME 2021 Official Paper: Shift 1

CT 1: Ratio and Proportion

2846

10 Questions
16 Marks
30 Mins

__Concept:__

Principle of static equilibrium in 2-D

ΣFx = 0, ΣFY = 0, ΣMz = 0

Draw the FBD diagram and find the moment to calculate the unknown.

Torque applied = Friction force × radius = μN × R

[**NOTE**: The friction force 'μN' always acts in the direction of rotation of the drum for FBD force resolving in lever.]____

__Calculation__:

__Given__:

R = 260 mm ⇒ 0.26 m, P = 1 kN ⇒ 1000 N, μ = 0.4

Taking moment about point 'Z'

(F × 500) + [μF × (310 - 260)] - (P × 1000) = 0

F × [500 + (0.4 × 50)] = 106

∴ F = 1923.076 N

Torque applied (Tf) = μN × R

⇒ 0.4 × 1923.076 × 0.26

∴ Tf = 200 Nm

Taking moment about point 'Z'

(F × 500) + [μF × (310 - 260)] - (P × 1000) = 0

F × [500 + (0.4 × 50)] = 10^{6}

∴ F = 1923.076 N

Torque applied (T_{f}) = μN × R

⇒ 0.4 × 1923.076 × 0.26

∴ T_{f} = 200 Nm