Page 169 - WCS - Electrical

WORKSHOP CALCULATION & SCIENCE - CITS

Laws of Limiting Friction Force

This force acts in the opposite direction to be taken by the moving object.
This force is proportional to normal reaction or there is a definite ratio between the limiting friction force and
normal reaction.

The limiting friction force depends on the nature and state of the surfaces but is free from their shape and area.
(See Fig. above).
Normal Reaction R=W kg (which is the weight of the object).
Co-efficient of Friction

The ratio between the limiting friction force and the normal reaction is called the co-efficient of friction.
Suppose, by applying a force P kg, the object is just fit to move, then limiting friction force will be produced in
between the two surfaces. The limiting friction force will be equal to external force applied and will work in the
1
opposite direction.
F= P kg 2
According to the second law of limiting friction force, the tiction force will be proportional to normal reaction.
F
F ∝ R (∝(sign is proportional to )
1
R
2
F = R x constant
Or 1 F
F 2 R
= Constant
R
This constant between objects is called Co- efficient of Solution:. This is represented by µ.
F
µ= or F = µ.R
F R Limiting Friction Force

friction =

efficient
of
Co −
R
Co-efficient of Friction = Normal Reaction
Co efficient of friction is always constant for any two objects and it has no unit.
F

R
Examples μ = F = 100 = 0.25
400
R Limiting Friction Force
Co − efficient of friction =
1 Weight of a machine vice placed on the table of machine tool is 20 kgf. What is the minimum horizontal bice
Normal

Reaction
needed which will start it sliding, if the co-efficient Siding friction is 0.19
Solution: Limiting Friction Force
F
100
Co − efficient = of friction = Normal Reaction
0.25
=
μ =
400
R
Co-efficient of friction u=0.19

Normal reaction R=20~kgf.
F
100
μ = = = 0.25
R
400
Force of friction = µ .R
0.19 x 20 =3.8~kgf
Minimum horizontal force required = 3.8 kgf
2 A solid weighing 50 kg is placed on a solid surface. How Minimum horizontal force required to move the block
when coefficient of friction is 0.25 between the block and the surface.
Solution:
Normal reaction R=50kg
Coefficient of friction µ =0.25
Force required to move the block = µ.R
=0.25 x 50 =12.5 kg
3 A bus with full =0.25\times50=12.5~kg 4500 kg is running on the road. If the coefficient of friction between the
types and the road is 0.30, then calculate the force of friction.
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CITS : WCS - Electrical - Exercise 15