All questions of Brakes & Clutches for Mechanical Engineering Exam

Proportional Relationship between Normal Pressure and Vertical Distance:
The intensity of normal pressure between the friction lining and the brake drum at any point is proportional to the square of the vertical distance from the pivot. This means that the relationship between normal pressure and vertical distance is not independent or inversely proportional, but rather linearly proportional.

Explanation:
- When the intensity of normal pressure is proportional to the square of the vertical distance, it means that as the vertical distance increases, the normal pressure also increases linearly.
- This relationship can be expressed as P ∝ h^2, where P is the normal pressure and h is the vertical distance from the pivot.
- This proportional relationship indicates that if the vertical distance is doubled, the normal pressure will increase by a factor of four (2^2 = 4).
- This linearly proportional relationship is important in understanding the distribution of normal pressure along the friction lining and brake drum, as it helps in determining the braking efficiency and performance of the system.
Therefore, the correct answer is option 'B': Proportional to vertical distance linearly.

CORRECT OPTION IS (B).
E=mk2ω2/2.

It can be either kinetic or potential.

Assumption in Block Brake with Short Shoe

Friction Force and Normal Reaction at Midpoint
In a block brake with a short shoe, it is assumed that the friction force and normal reaction are concentrated at the midpoint of the shoe. This assumption simplifies the analysis of the brake system and allows for easier calculations of the braking performance.

Reason for Concentration at Midpoint
The assumption that the friction force and normal reaction are concentrated at the midpoint of the shoe is based on the idea that the pressure distribution across the shoe is uniform. This simplifying assumption allows engineers to model the brake system more easily and accurately predict its performance.

Significance of Concentration at Midpoint
By assuming that the friction force and normal reaction are concentrated at the midpoint of the shoe, engineers can calculate the braking torque more accurately and design the brake system to effectively stop the moving object. This assumption also helps in determining the wear and heat generation within the brake system.

Conclusion
In conclusion, the assumption that the friction force and normal reaction are concentrated at the midpoint of the shoe in a block brake with a short shoe is valid and commonly used in the analysis and design of braking systems.

To calculate the angle through which the disk rotated during the braking period, we can use the concept of rotational kinetic energy.

1. Calculate the initial angular velocity:
The angular velocity (ω) is given in revolutions per minute (rpm). We need to convert it to radians per second (rad/s).
ω = (2π * n) / 60, where n is the number of revolutions per minute.
In this case, n = 350.
ω = (2π * 350) / 60 = 36.67 rad/s

2. Calculate the initial kinetic energy:
The formula for rotational kinetic energy (KE) is:
KE = (1/2) * I * ω^2, where I is the moment of inertia and ω is the angular velocity.
Given that the square of the radius of gyration (k^2) is 0.2, we can calculate the moment of inertia (I) using the formula:
I = m * k^2, where m is the mass of the disk.
Given that the mass (m) is 1000 kg, the moment of inertia is:
I = 1000 * 0.2^2 = 40 kg·m^2

Substituting the values, we can calculate the initial kinetic energy:
KE = (1/2) * 40 * (36.67)^2 = 26694.33 J

3. Calculate the final kinetic energy:
The final kinetic energy is zero because the disk comes to a stop.

4. Calculate the work done by the brake:
The work done by the brake is equal to the change in kinetic energy. Since the final kinetic energy is zero, the work done by the brake is equal to the initial kinetic energy:
Work = 26694.33 J

5. Calculate the angle through which the disk rotated:
The work done by the brake is given by the equation:
Work = τ * θ, where τ is the torque exerted by the brake and θ is the angle through which the disk rotated.
Since the torque (τ) is constant, we can rearrange the equation to solve for θ:
θ = Work / τ
To find τ, we need to determine the braking time (t) and the moment of inertia (I):
τ = I * α, where α is the angular acceleration.
The angular acceleration can be calculated using the formula:
α = ω / t
Given that the braking time (t) is 1.6 s, we can calculate the angular acceleration:
α = 36.67 / 1.6 = 22.92 rad/s^2
Substituting the values, we can calculate the torque:
τ = 40 * 22.92 = 916.8 N·m

Now we can calculate the angle:
θ = 26694.33 / 916.8 = 29.18 rad

Therefore, the angle through which the disk rotated during the braking period is approximately 29.3 rad (option C).

To calculate the distance of the pivot from the axis of the drum in a pivoted double block brake, we can use the geometry of the brake system.

Given:
Drum radius = 280 mm
Angle subtended by the shoes = 100°

Let's break down the solution into steps:

Step 1: Find the length of the arc subtended by the shoes.
The length of the arc can be calculated using the formula:
Arc length = (angle / 360°) × 2πr
Where r is the radius of the drum.

Arc length = (100° / 360°) × 2π × 280 mm
Arc length = (5/18) × 2π × 280 mm
Arc length ≈ 245.94 mm

Step 2: Find the distance between the pivot and the center of the drum.
In a pivoted double block brake, the pivot is located at the midpoint of the line joining the centers of the two shoes. Since the shoes subtend an angle of 100°, the center of the drum lies at the midpoint of the arc subtended by the shoes.

Therefore, the distance between the pivot and the center of the drum is half the length of the arc subtended by the shoes.

Distance = 245.94 mm / 2
Distance ≈ 122.97 mm

Step 3: Find the distance of the pivot from the axis of the drum.
The distance of the pivot from the axis of the drum is equal to the radius of the drum minus the distance between the pivot and the center of the drum.

Distance of pivot from axis = 280 mm - 122.97 mm
Distance of pivot from axis ≈ 157.03 mm

Therefore, the correct answer is option (B) 314.3 mm.

Introduction:
Internally expanding brakes are a type of braking system commonly used in automobiles, motorcycles, and other vehicles. They rely on the principle of friction to slow down or stop the vehicle's motion. However, one of the drawbacks of internally expanding brakes is their limited heat dissipating capacity.

Explanation:
Internally expanding brakes consist of a pair of brake shoes that are pressed against the inner surface of a drum or rotor to generate the necessary friction and slow down the vehicle. When the brakes are applied, the shoes create frictional forces that convert the kinetic energy of the moving vehicle into heat energy. This heat energy needs to be dissipated efficiently to prevent the brakes from overheating and losing their effectiveness.

Limited Heat Dissipating Capacity:
The limited heat dissipating capacity of internally expanding brakes is primarily due to their design and construction. Unlike externally ventilated brakes, which have cooling fins or holes to promote airflow and dissipate heat, internally expanding brakes do not have such features. This means that the heat generated during braking is trapped within the brake assembly, leading to a buildup of heat.

Consequences of Heat Buildup:
The buildup of heat within the brake assembly can have several negative consequences:

1. Reduced Braking Performance: As the temperature of the brake shoes and drum/rotor increases, the coefficient of friction decreases. This can result in reduced braking performance, longer stopping distances, and increased wear on the brake components.

2. Brake Fade: Brake fade is a phenomenon that occurs when the brakes become less effective due to excessive heat. This can happen when the temperature of the brake components exceeds their thermal capacity. Brake fade can lead to a loss of control over the vehicle and increased risk of accidents.

3. Brake Component Damage: Excessive heat can also damage the brake components, such as the brake shoes, drum/rotor, and other associated parts. The high temperatures can cause thermal stress, warping, and even cracking of the brake components, leading to the need for costly repairs or replacements.

Conclusion:
In conclusion, internally expanding brakes have a limited heat dissipating capacity, which can lead to reduced braking performance, brake fade, and damage to the brake components. It is important to consider these limitations when designing and using this type of braking system, and implement appropriate measures to manage the heat generated during braking.

Given data:
Mass of the disk, m = 1000 kg
Rotational speed, ω = 350 rpm
Diameter of the disk, D = 1 m
Time taken to stop the disk by brake, t = 1.6 s
Square of radius of gyration, k^2 = 0.2

To find:
Torque capacity of the brake

Solution:
1. Convert the rotational speed from rpm to rad/s:

ω = 2πN/60 = 2π(350)/60 = 36.6 rad/s

2. Calculate the moment of inertia of the disk:

I = (1/4)md^2 = (1/4)(1000)(1)^2 = 250 kg·m^2

3. Calculate the torque required to stop the disk:

τ = Iα

where α is the angular acceleration of the disk.

Using the formula for angular acceleration:

α = (ωf - ωi)/t

where ωi is the initial angular velocity (ω) and ωf is the final angular velocity (0).

Substituting the given values:

α = (0 - 36.6)/1.6 = -22.9 rad/s^2

Therefore,

τ = Iα = (250)(-22.9) = -5725 N·m

Note that the negative sign indicates that the torque is in the opposite direction of the rotation of the disk.

4. Calculate the torque capacity of the brake:

The torque capacity of the brake is equal to the torque required to stop the disk, which is:

|τ| = 5725 N·m

Therefore, the correct answer is option B: 4583.6 N·m.