Two solid uniform spheres each of radius 5 cm are connected by a light and totally immersed in a tank of water. If the specific gravities of the pressure are 0.5 and 2, find the tension in the string and the contact force between the bottom of the tank and the heavier sphere.

Explanation

Let AC be the submerged part of the rod AB hinged at A as shown in figure. G is the centre of gravity of the rod and G' is the centre of buoyancy through which force of buoyancy `F_(B)` acts vertically upwards. Since the rod is uniform,

The weight of part AC will be ` (5)/(6)xx12=10 kg`
[Because AB=6 m and AC=5 m]
The boyance force on rod at G' is
`F_(B)=(10)/(0.5)=20kg` weight
(i) Let x be weight attached at the end B.
Balancing torques about A, we get
`W × AG + x × AB = F_(B) × AG'`
`12 + 3 + x × 6 = 20 × (5//2)`
[As AG'`=5//2`]
Solving we get `" "x=2.33` kg
(ii) Suppose R be the upward reaction on the hinge, then in equilibrium position, we have
`W+x=F_(B)+R`
or `" "R=W+x-F_(B)`
`=12+2.33-20=-5.67` kg. wt
Negative sign shows that the reaction at the hinge is acting in the downward direction. The magnitude of the reaction is `5.67` kg. wt.