The schematic of an external drum rotating clockwise engaging with a short shoe is shown in the figure. The shoe is mounted at point Y on a rigid lever XYZ hinged at point X. A force 𝐹 = 100 𝑁 is applied at the free end of the lever as shown. Given that the coefficient of friction between the shoe and the drum is 0.3, the braking torque (in Nm ) applied on the drum is _______ (correct to two decimal places).

The correct answer is 8.18 (or 8.18 Nm).

1. Identify the given values from the schematic diagram:
- Applied force at the free end Z of the lever, $F=100\text{N}$
- Coefficient of friction between the shoe and the drum, $\mu =0.3$
- Radius of the drum, $R=100\text{mm}=0.1\text{m}$
- Horizontal distance from hinge point X to the applied force line of action at Z = $300\text{mm}$
- Horizontal distance from hinge point X to the center of the shoe at Y = $200\text{mm}$
- Vertical distance from hinge point X to the shoe contact point Y = $300\text{mm}$

2. Determine the direction of the friction force:
The drum rotates in a clockwise direction. Therefore, the tangent velocity of the drum's top surface at the contact point Y is directed from left to right (towards the right). By Newton's third law, the friction force exerted by the rotating drum on the brake shoe (lever XYZ) acts in the direction of drum rotation, which is to the right.

3. Set up the moment equilibrium about hinge X:
Taking the sum of moments about the hinge point X to find the normal reaction force $R_N$:
$\sum M_X=0$

The moments acting on the lever about point X are:
- The moment due to the downward vertical force $F=100\text{N}$ acts counter-clockwise at a distance of $300\text{mm}$.
- The normal reaction force $R_N$ acts vertically upwards on the shoe, producing a clockwise moment at a horizontal distance of $200\text{mm}$.
- The friction force $F_f=\mu R_N$ acts horizontally to the right at Y, producing a counter-clockwise moment at a vertical distance of $300\text{mm}$ below X.

Equating the counter-clockwise moments to the clockwise moments:
$(F\times 300)+(\mu R_N\times 300)=R_N\times 200$

Substitute the given values:
$(100\times 300)+(0.3\times R_N\times 300)=R_N\times 200$
$30000+90R_N=200R_N$
$30000=110R_N$
$R_N=\frac{30000}{110}\approx 272.73\text{N}$

4. Calculate the braking torque applied on the drum:
The braking torque $T_b$ is given by the product of the friction force and the drum radius:
$T_b=\mu R_{N}R$
$T_b=0.3\times 272.73\text{N}\times 0.1\text{m}\approx 8.18\text{Nm}$