For the brake shown in the figure, which one of the following is TRUE?

Option 1 : Self energizing for clockwise rotation of the drum

__Concept:__

When the frictional force helps the applied force in applying the brake, such type of brakes are said to be self-energizing brakes.

Self-energizing brake is the one in which torque due to Fr supports torque due to F.

**Calculation:**

**Given:**

The torque due to F is clockwise. The torque due to Fr = μR_{N} is clockwise. So, it is a self-energizing brake.

Let consider clockwise rotation

∑M_{pivot} = 0

F × ℓ - R_{N} × a + μ R_{N} × b = 0

When the wheel rotates in a clockwise direction we can see that **friction is helping the applied force** in applying the brake. Hence it is called a **self-energizing brake.**

Option 1 : Zero

**Concept:**

**When the frictional force is great enough to apply the brake with no external force, then the brake is said to be a self-locking brake.**

The figure shows the brake drum.

From the equilibrium of forces,

F × l + μR_{N }× b = R_{N }× a

F × l = R_{N }× a - μR_{N }× b

R_{N }(a - μb) = F × l

\(R_N=\frac{Fl}{(a-μ b)}\)...........................(1)

Braking torque on Drum is given as, T_{B} = F_{f }× R ⇒ T_{B}=μR_{N} × R

\(\therefore T_B=μ\times \frac{Fl}{(a-μ b}\times R\)

**For self-locking brake:**

From equation (1), the expression for the force "F" required to apply the brake is obtained as, \(F=\frac{R_N(a-μ b)}{l}\)

And if a < μ b, then** F will be negative or zero.** This means that no external force is required to apply the brake and hence the brake is self-locking.

**Self-energizing brakes**

When the frictional force helps the applied force in applying the brake, such types of brakes is said to be self-energizing brakes.

The brake should be self-energizing and not the self-locking.

A drum brake is shown in the figure. The drum is rotating in an anticlockwise direction. The coefficient of friction between drum and shoe is 0.2. The dimensions shown in the figure are in mm. The braking torque (in N-m) for the brake shoe is __________

**Concept:**

Referring to the FBD of the lever,

Taking moment about pin joint,

⇒ ΣM_{O} = 0 ...(1)

The braking torque for the brake shoe,

⇒ T_{b} = μNR ...(2)

**Calculation:**

__Given:__

P = 1000 N, μ = 0.2, R = 0.2 m

Using equation (1),

⇒ 1000× 0.8 = N × 0.48 + 0.2 × N × 0.1

⇒ N = 1600 N

Using equation (2),

⇒ T_{b} = 0.2 × 1600 × 0.2

A block brake with 400 mm diameter is used to brake a torque of 100 Nm as shown in the figure. If the coefficient friction is 0.25 at the brake surface what is the value of force F to be applied at the end of the lever.

Option 4 : 1000 N

**Concept:**

The FBD of above system is given below.

The wheel is rotating anti clockwise, so the friction force F_{t} acts on brake is in same direction.

Therefore, The moment about hinge,

∑M_{P} = 0

⇒ F(200 + 225) = R_{N} × 200 + F_{t} × 50

where, \( {F_t} = \frac{T}{r} \) and F_{t} = μR_{N}

R_{N} = normal reaction, μ = coefficient friction

**Calculation:**

**Given:**

D = 400 mm = 0.4 m, T = 100 Nm, μ = 0.25

\(\therefore {F_t} = \frac{T}{R} = \frac{{100}}{{0.2}}=500~N\)

And \({R_N} = \frac{{500}}{\mu } = \frac{{500}}{{0.25}}=2000 ~N\)

The moment about hinge,

⇒ F × 425 = 2000 × 200 + 500 × 50

**⇒ F = 1000 N**

A block-brake shown below has a face width of 300 mm and a mean coefficient of friction of 0.25. For an activating force of 400 N, the braking torque in Nm is

Option 3 : 45

Free body diagram of lever and brake dram are

For lever moment about O, ∑M_{o} = 0

⇒ (400 N) × (400 + 200) mm = N × 200 mm

\(\Rightarrow Normal\;force,\;N = \frac{{400 \times 600}}{{200}} = 1200\;N\)

Tangential force, F_{t} = μN = 0.25 × 1200 = 300 N

Braking torque = T_{b} = F_{t} × r = 300 × 0.15 = 45 N.m

Option 3 : Number of disk

**Explanation:**

The disc brake is an arrangement of disc and pad in which disc and pad both are having surfaces with high friction. From the figure, it can be seen that when the number of **pads gets increases** then the **contact surface will increase,** and hence the **braking torque will increase**. But if the number of **disks is increasing** then we have t**o place the disc co-axially **and the **effective contact area between the disc and pad will remain the same**. Therefore even after increasing the number of discs, the **braking torque will not increase.**

By increasing the number of disc the axial load bearing capacity will increase.

A short shoe external drum brake is shown in the figure. The diameter of the brake drum is 500 mm. The dimensions a = 1000 mm, b = 500 mm and c = 200 mm. The coefficient of friction between the drum and the shoe is 0.35. The force applied on the lever F = 100 N as shown in the figure. The drum is rotating anti-clockwise. The braking torque on the drum is ______ N.m (round off to two decimal places).

__Concept:__

Torque T = r × F

Force due to friction F = μN

Refer the figure shown.

Consider the pinned lever and Brake-shoe as one rigid member.

Since drum is rotating counter - clockwise so braking torque must act clockwise.

For braking torque clockwise friction force on drum must be towards (+ x)

⇒ From newtons third law, force on brake shoe will be in -x direction.

Now taking moment about 0 equal to zero.

⇒ N(b) – Fa – (μN) C = 0

⇒ N (0.5) – 100 (1) – (0.35) N (0.2) = 0

⇒ N (0.43) – 100 = 0

⇒ N = 232.558 N

⇒ Braking torque, T = (μN) r = (0.35) (232.558) (0.25) = 20.3488 Nm.

⇒ Braking torque = 20.35 Nm (Rounded to two decimal)

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).

**Concept:**

Braking torque

T = f × radius of drum

f = μ × RN

RN = Normal reaction on the drum, f = Force of friction on the lever due to rotation of the drum.

μ = coefficient of friction between the shoe and the drum

**Calculation:**

For the given situation, as the lever (xyz) is under equilibrium, so the total moment about x will be zero. Drawing the FBD of the (xyz) lever

F = 100 N; μ = 0.3

Now Σ M_{x} = 0 balancing moment about x

⇒ F × 300 + f × 300 – R_{N} × 200 = 0

⇒ 3F + 3f – 2R_{N} = 0

⇒ 3 × 100 + 3 × μR_{N} – 2R_{N} = 0

⇒ 300 + (3 × 0.3 - 2) R_{N} = 0

⇒ 300 + (0.9 - 2) R_{N} = 0

⇒ 300 = 1.1 R_{N}

⇒ R_{N} = 272.72 N

Now f = μR_{N} = 0.3 × 272.72 = 81.816 N

Braking torque

T = f × radius of drum = 81.816 × 100 × 10^{-3 }= 8.18 Nm

**Points to remember:** Take care of the sign while calculated the moment.

A single block brake with a short shoe and torque capacity of 250 N·m 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 _________.

__Concept:__

Principle of static equilibrium in 2-D,

ΣF_{x} = 0, ΣF_{Y} = 0, ΣM_{z} = 0 ...(1)

__Calculation:__

__Figure: free body diagram of brake system__

Torque capacity, M_{t} = 250 Nm

⇒ μN.a = 250 × 10^{3} Nmm …(1)

ΣM_{x} = 0;

\(P.\left( 2.5a \right)=Na+\mu N.\frac{a}{4}\)

\(2.5P=N+\frac{\mu N}{4}\)

P = 2000 N, μ = 0.25

\(\Rightarrow \left( 2.5 \right)\left( 2000 \right)=N\left[ 1+\frac{0.25}{4} \right]\)

⇒ N = 4705.88 N ...(2)

\(\Rightarrow a=\frac{\left( 250 \right)\left( 1000 \right)}{\left( 0.25 \right)\left( 4705.88 \right)}=212.50~mm\)

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).

__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

A band brake having band-width of 80 mm, drum diameter of 250 mm, coefficient of friction of 0.25 and angle of wrap of 270 degrees is required to exert a friction torque of 1000 N-m. The maximum tension (in kN) developed in the band is

Option 4 : 11.56

μ = 0.25, θ = 270^{o} = 270 × ( π / 180) = 4.7124 radians

Torque = 1000 = (T_{1} – T_{2}) r ............(1)

\(\frac{{{T_1}}}{{{T_2}}} = {e^{\mu \theta }} = 3.248\) ...............(2)

From equation (1) and (2), we get

T_{2} = 3.558 kN, and T_{1} = 11.558 kN

A force of 400 N is applied to the brake drum of 0.5 m diameter in a band brake system as shown in the figure, where the wrapping angle is 180°. If the coefficient of friction between the drum and the band is 0.25, the braking torque applied, in Nm is

Option 2 : 54.4

__Concept:__

For a brake:

**\(\frac{{{T_1}}}{{{T_2}}} = {e^{\mu \theta }}\)**

where T_{1} = tension on tight side, T_{2} = tension on slack side, μ = coefficient of friction, θ = wrapping angle

And **braking torque** can be calculated by:

**BT = (T _{1} – T_{2}) × R**

where R is the radius of the brake drum.

__Calculation:__

__Given:__

As the drum is rotating in the anti-clockwise direction, T1 will be tight side & T2 will be slack side.

T_{1} = 400 N, R = 0.25 m, μ = 0.25

θ = 180° = π radian

Now, we know that

\(\frac{{{T_1}}}{{{T_2}}} = {e^{\mu \theta }}\)

\(\frac{{400}}{{{T_2}}} = {e^{0.25 \times \pi }} = 2.193\)

\( \Rightarrow {T_2} = \frac{{400}}{{2.193}} = 182.375\;N\)

**Braking torque (BT)**

**BT = (T _{1} – T_{2}) × R**

BT = (400 – 182.375) × 0.25

BT =** 54.40 Nm**