Engineering Mechanics

(Joyce) #1

(^98)  A Textbook of Engineering Mechanics
Now we find that the answer obtained is more than the right hand side of equation (i). Thus the
value of (h) is less than 39.5 mm. Now let us substitute the h = 39.4 mm in the left hand side of
equation, (i).
(39.4)^4 – 78.67 (39.4)^3 = – 2 401 900
We find that is answer is very close to the right hand side of the equation and there is no need
of further calculations. Thus the value of h = 39.4 mm Ans.
EXERCISE 6.3

1. A circular hole of 50 mm diameter is cut out from a circular disc of 100 mm diameter as
shown in Fig. 6.31. Find the centre of gravity of the section from A.[Ans. 41.7 mm]

``Fig. 6.31. Fig. 6.32.``

1. Find the centre of gravity of a semicircular section having outer and inner diameters of 200
mm and 160 mm respectively as shown in Fig. 6.32. [Ans. 57.5 mm from the base]

2. A circular sector of angle 45° is cut from the circle of radius 220 mm Determine the centre
of gravity of the remainder from the centre of the sector. [Ans. 200 mm]

3. A hemisphere of radius 80 mm is cut out from a right circular cylinder of diameter 80
mm and height 160 mm as shown in Fig. 6.33. Find the centre of gravity of the body
from the base AB.[Ans. 77.2 mm]

``Fig. 6.33. Fig. 6.34.``

1. A right circular cone of 30 mm diameter and 60 mm height is cut from a cylinder of 50
mm diameter at 120 mm height as shown in Fig. 6.34. Find the position of the centre of
gravity of the body from its base. [Ans. 60.7 mm]

QUESTIONS

1. Define the terms ‘centre of gravity’.

2. Distinguish between centre of gravity and centroid.
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