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General

Module 1. Tractor Mechanics

Module 2. Traction

Module 3. Introduction to Transmission System

Module 4. Clutch System

Module 5. Gear Box

Module 6. Differential and Final drive

Module 7. Brakes

Module 8. Steering system

Module 9. Hydraulics

Module 10. Power Transmission

Module 11. Human Factors

## Lesson 21. Calculation of braking effort.

**Weight Transfer:**

The figure 2 shows a body moving towards the left with an inertial force. Brake is applied, which produces a retarding force at the road surface. This retarding force at the road surface causes an overturning couple on the body due to which weight transfer takes place from the rear to the front wheels.

In figure 3, the body is coming down a gradient of angle f when the brakes are applied.

W = Weight of the vehicle

R_{f}= Normal reaction on the front wheel

R_{r}= Normal reaction on the rear wheel

b = Wheelbase

h = Height of CG from the surface of the road

∞ = Retarding force due to breaking

μ = Coefficient of friction between the wheels and the road surface

μR_{f}= Braking force on the front wheels

μR_{r}= Braking force on the rear wheels

When brakes are applied on all the wheels

R_{f} + R_{r} = W Cos \[\phi \] -1

μR_{f} + μR_{r} = W Sin \[\phi \] + W .∞ -2

g

In case of equilibrium ∑ M=0 at point B

W . ∞.h + W Sin \[\phi \]. h + W Cos \[\phi \]. x - R_{f}. b

g

From equation 1 and 2

Μ W Cos \[\phi \] = W Sin \[\phi \] + W. ∞

g

∞ = μ.Cos \[\phi \] - Sin \[\phi \] -4

g

From equation 3 and 4

R_{f} = W(x +μh). Cos \[\phi \]

b

R_{r}=W(b- x- μh) . Cos \[\phi \]

b

Case of Indian Agricultural Tractors

While most of the automobiles have an all wheel braking but most of the Indian Agricultural Tractors, being rear wheel drive, have braking only in the rear wheels.

The weight transfer in case of tractors with braking in the rear wheels only is analyzed as follows:

R_{f} + R_{r} = W Cos \[\phi \] -5

μR_{r} = W Sin \[\phi \] + W . ∞ -6

g

Taking moments above point A

R_{r} .b + W.h.Sin \[\phi \] + W . ∞. h – W(b-x)Cos \[\phi \] = 0 -7

g

From equation 6 and 7

∞ =μ(b-x) Cos \[\phi \] - Sin \[\phi \]

g (b + μh)

R_{f}=W.(b-x). Cos \[\phi \]

(b + μh)

R_{r} = W(x + μh). Cos \[\phi \]

(b + μh)

In case the body (or the tractor) is moving up the slope, the angle of the slope may be taken as negative and the derived expression can be used.