(^146) A Textbook of Engineering Mechanics
Fig. 8.23.
Solution. Given: Height of right circular cone (H) = 12 cm; Height of right circular cylinder
(h) = 3 cm; Radius of the base (r) = 2 cm and coefficient of friction (μ) = 0.5.
We have already found out in example 6.6 that the centre of
gravity of the body is at a height of 4.07 cm from the base. A little
consideration will show, that when the body is at the point of
toppling, the weight of the body will pass through the extreme point
B as shown in Fig. 8.23. Thus in the limiting case, the angle (θ 1 ) at
which the body is inclined with the horizontal is given by:
1
2
tan 0.49
4.07
BD
DG
θ= = =
Thus we see that the angle (θ 1 ) is less than the angle (θ). It is so, as
the value of tan θ 1 (equal to 0.49) is less than tan θ (equal to 0.5).
Or in other words, the angle of inclination is less than the angle of
friction. Thus the body will topple over before it begins to slide.
Ans.
Now in the second case, let us first find out the relation be-
tween h and H from the given condition that the centre of the body is at the centre of the common
base (G) as shown in Fig. 8.24.
(i) Cylinder
v 1 = π r^2 h
and 1
2
h
y =
(ii) Cone
2
2
1
3
vrH=π
and 2
4
H
y =
We know that the distance between centre of gravity of the body
from the base of the cylinder ()y
11 2 2
12
vy v y
h
vv
- =
22
22
1
23 4
1
3
hH
rh rH h
rh rH
⎡⎤ ⎛⎞⎡⎤
⎢⎥π×+π⎢⎥⎜⎟+
=⎣⎦ ⎝⎠⎣⎦
π+π
222222
22
32312
rHh rh rHh rH
rh
ππππ
π+ = + +
22 2 2
212
ππrh rH
= or
2
2
6
H
h = or
6
H
h=
In this case, when the body is at the point of toppling, the angle (θ 2 ) at which it is inclined
with the horizontal is given by :
2
6
tan
/6
rrr
hHH
θ= = =
Now if the body is to slide before toppling, the angle of friction should be less than the angle
of friction. Or in other words, tan θ should be less than tan θ 2 , Mathematically,
r 6
H
μ< or μ<Hr 6 Ans.
Fig. 8.24.