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).

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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

Correct option is

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 Oacting 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 RH = 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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Where  is the density at sea-level, A is a constant and h is the height. Calculate the atmospheric pressure at sea-level. Assume g to be constant. 

Q10

 

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[RD of ice = 0.9]