A single block brake with a short shoe and torque capacity of 250 Nm is shown. The cylindrical brake drum rotates anticlockwise at 100 rpm and the coefficient of friction is 0.25. The value of a, in mm (round off to one decimal place), such that the maximum actuating force P is 2000 N, is __________.

The correct answer is 212.5.

To find the value of $a$ in mm, we analyze the forces acting on the lever and the drum.

1. Braking Torque and Normal Force:
The braking torque $T$ for a single block brake with a short shoe is given by:
$T=\mu \times N\times R$
where:
• $T=250\text{ Nm}$ is the torque capacity.
• $\mu =0.25$ is the coefficient of friction.
• $N$ is the normal reaction force between the brake shoe and the drum.
• $R=a$ (in meters) is the radius of the brake drum as shown in the diagram.

Substituting the given values into the torque equation:
$250=0.25\times N\times a$
$N=\frac{1000}{a}$

2. Moment Equilibrium of the Lever:
From the schematic diagram, we identify the forces acting on the lever:
• Actuating force $P=2000\text{ N}$ acting downwards at the left end at a distance of $1.5a+a=2.5a$ from the pivot.
• Normal reaction force $N$ acting upwards at the brake shoe at a horizontal distance of $a$ from the pivot.
• Friction force $F=\mu N$ acting at the contact surface. Since the drum rotates anticlockwise, the friction force on the drum is directed to the right to oppose rotation. By Newton's third law, the friction force on the brake shoe (lever) is directed to the left. The vertical distance from the pivot to the line of action of this friction force is $\frac{a}{4}$.

Taking moments about the pivot hinge:
$P\times 2.5a=N\times a+\mu N\times \frac{a}{4}$

Dividing both sides by $a$:
$2.5P=N\left(1+\frac{\mu }{4}\right)$

Substituting $P=2000\text{ N}$ and $\mu =0.25$:
$2.5\times 2000=N\left(1+\frac{0.25}{4}\right)$
$5000=1.0625N$
$N=\frac{5000}{1.0625}\approx 4705.88\text{ N}$

3. Calculating the value of a:
Substitute the value of $N$ back into the torque equation:
$4705.88=\frac{1000}{a}$
$a=\frac{1000}{4705.88}\approx 0.2125\text{ m}$

Converting to mm:
$a=0.2125\times 1000=212.5\text{ mm}$