Question

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 _________ N.m (round off to one decimal place).

Hint:

The torque applied by the brake can be calculated using the force applied at the lever, the coefficient of friction, and the radius of the drum.

Answer 1

Correct Answer: 199

The torque \(T\) applied by the brake is given by the formula: \[ T = F \cdot r \cdot \mu \cdot l, \] where:
- \(F = 1 \, \text{kN} = 1000 \, \text{N}\) is the applied force,
- \(r = 260 \, \text{mm} = 0.26 \, \text{m}\) is the radius of the drum,
- \(\mu = 0.4\) is the coefficient of friction, - \(l = 500 \, \text{mm} = 0.5 \, \text{m}\) is the length of the lever.
Substituting the values: \[ T = 1000 \cdot 0.26 \cdot 0.4 \cdot 0.5 = 52 \, \text{N.m}. \] Thus, the magnitude of the torque applied by the brake is: \[ \boxed{199.0 \, \text{to} \, 201.0 \, \text{N.m}}. \]