Page 170 - WCS - Electrical

WORKSHOP CALCULATION & SCIENCE - CITS

Solution:

Normal reaction A=4500~kg
Coefficient of friction µ =0.30
Force of Friction = µ .R
=0.30 x 4500
=1350~kg
1
2
4 A glass block of 400 grams has been placed on the table. The glass is connected by a string to a 40 grams
scale pan. The string passes over a pulley. When a weight of 60 grams is placed on the scale pan, the block
F starts sliding. Find out the co-efficient of Inction between wood and glass
R
Solution:
Scale pan means a small container Meight of 60 grams placed on the scale pan. Weight of scale pan is given as
40 grams. Force required to move the glass block =60 grams + 40 grams =100 grams
F
R
F =Force required to start siding =100 grams
R=Normal Reaction weight of glass block = 400 grams
Let µ co-efficient of friction
Limiting Friction Force
Co − efficient of friction = Normal Reaction
We know:
F= µ .R
μ = F = 100 = 0.25
R 400
Co-efficient of friction=0.25 Ans.
Angle of Friction
The resultant force of limiting friction force and normal reaction makes an angle with normal reaction which is
called Angle of Friction. It is represented by µ (mew).

As is shown in Fig. (b). AB is Normal Reaction and AD is Limiting Friction Force in parallelogram, ABCD, the
diagonal AC would be the Resultant Force which makes an angle BAC with Normal Reaction. This angle is called
Angle of Friction.
Suppose Angle of Friction =0(theta)
Resultant force= S
If the Resultant force of AB and AD are divided into vertical and horizontal components, then
F=S sin Ø
R=S cos \ Ø
F/R=S sin Ø/S cos Ø=sin Ø/cos Ø
=tan Ø
∴ µ=tan Ø

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CITS : WCS - Electrical - Exercise 15 CITS : WCS - Electrical - Exercise 15