Engineering Mechanics

Chapter 36 : Equilibrium of Floating Bodies „„„„„ 753

and combined specific gravity of the cylinder

=

(97.5 0.5) (2.5 8)

0.688

97.5 2.5

×+×

=

+

∴ Depth of immersion of the cylinder

= 100 × 0.688 = 68.8 cm

and distance of centre of buoyancy from the bottom of the buoy,

OB=

68.8

2

= 34.4 cm

∴ BG=OG – OB = 36.7 – 34.4 = 2.3 cm

We know that moment of inertia of the circular section,

I=

( )^44 (20) 2500 cm^4

64 64

d

ππ

==π

and volume of water displaced,

V=

(20)^23 68.8 6880 cm

4

π

××=π

∴ BM=

2500

0.36 cm

6880

I

V

π

==

π

and metacentric height, GM=BM – BG = 0.36 – 2.3 = – 1.94 cm.

• Minus sign means that the metacentre (M) is below centre of gravity (G). Therefore the
  cylinder is in unstable equilibrium. Ans.

Maximum permissible length of the cylinder

Let l= Length of cylinder excluding metal portion in cm.

Now distance between the combined centre of gravity (G) and the bottom of the cylinder (O),

OG=

(0.5 ) 2.5 [(8 2.5) 1.25]

2

(0.5 ) (8 2.5)

l

Al A

Al A

⎡⎤⎛⎞

⎢⎥××⎜⎟+ + × ×

⎣⎦⎝⎠

×+ ×

=

0.5 (2.5 0.5 ) 25 2 5 100

0.5 20 2 80

llll

ll

++ ++

=

++

and combined specific gravity of cylinder

=

(0.5 ) (2.5 8) 0.5 20

2.5 2.5

ll

ll

×+ × +

=

++

∴ Depth of immersion of the cylinder

= Total length × Combined specific gravity

=

0.5 20

( 2.5) 0.5 20 cm

2.5

l

ll

l

+

+× =+

+

Fig. 36.9.

* We know that OM = OB + BM = 34.4 + 0.36 = 34.76 cm

As the metacentre, M (34.76 cm) is below the centre of gravity G (36.84 cm), therefore the cylinder is in

unstable equilibrium.