(^750) A Textbook of Engineering Mechanics
Minus sign means that the metacentre (M) is below the centre of gravity (G). Therefore
cylinder is in an unstable equilibrium. Ans.
Example 36.7. A solid cylinder 500 mm long, 100 mm diameter has its base 10 mm thick
of specific gravity 7. The remaining part of the cylinder is of specific gravity 0·5. Determine, if it
can float vertically in water.
Solution. Given: Length of cylinder (l) = 500 mm; Diameter of the cylinder (d) = 100 mm
Base thickness = 10 mm; sp.gr. of base = 7 and sp.gr. of remaining portion = 0·5.
We know that distance between combined centre of gravity (G) and the bottom of cylinder (O),
OG=
490 10
0·5 490 10 7 10
22
(0·5 490) (7 10)
AA
AA
⎡⎤⎛⎞⎡ ⎤
⎢⎥×++××⎜⎟⎢⎥
⎣⎦⎝⎠⎣ ⎦
×+×
mm
...(where A is the area of cylinder)
62 825
199·4 mm
315
and combined specific gravity, =
(0·5 490) (7 10)
0·63
490 10
×+×
∴ Depth of immersion of the cylinder
= 0·63 × 500 = 315 mm
and distance of centre of buoyancy from the bottom of the cylinder
OB=
315
2
= 157·5 mm
∴ BG=OG – OB = 199·4 – 157·5 = 41·9 mm
We also know that moment of inertia of the circular section about its
centre of gravity,
I =
( )^44 (100) 1 562 500 mm^4
64 64
d
ππ
==π
and volume of water displaced,
V= (100)^23315 787 500 m
4
π
×= π
∴ BM=
1562500
2mm
787 500
I
V
π
π
and metacentric height, GM=BM – BG = 2 – 41·9 = – 39·9 mm.
- Minus sign means that the metacentre (M) is below the centre of gravity (G). Therefore
the cylinder is in an unstable equilibrium. Ans. - We know that OM = OB + BM = 1·2 + 0·235 = 1·435 m. As the metacentre M (1·435 m) is below the centre
of gravity G (1·5 m). Therefore the cylinder is in an unstable equilibrium.
Fig. 36.7. - We know that OM = OB + BM = 157·5 + 2 = 159·5 mm,
As the metacentre M (159·5 mm) is below the centre of gravity G (199·4 mm) therefore the cylinder is in an
unstable equilibrium.