## Question

A rod of length 6 m has a mass 12 kg. It is hinged at one end at a distance of 3 m below water surface. (a) What weight must be attached to the other end of the rod so that 5 m of the rod are submerged? (b) Find the magnitude and direction of the force exerted by the hinge on the rod.

(Specific gravity of rod is 0.5).

### Solution

8 g, 5.67 kg

As shown in figure, the forces acting on the rod are :

(1) The weight of rod 12 g N acting downwards through the CG of the rod, i.e., at a distance of 3 m from the hinge.

(2) Force of buoyancy through the CG of displacement liquid vertically upwards. As

Force of buoyancy

and acts at a distance 2.5 m from the hinge

(3) Extra weight *w* at the other end of the rod at a distance 6 m from *O*acting vertically downwards.

(4) Reaction *R* at the hinge at *O* will be vertical (as here all the forces are vertical, so for horizontal equilibrium of the rod *R _{H}* = 0)

So for translator equilibrium of rod,

And for rotational equilibrium of rod (taking moments about *O*)

Substituting the value of *w* from eq. (ii) in (i) and solving for *R*, we get

Negative sign implies that *R* is directed vertically downwards.

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