A rod of length `6m` has a mass of `12 kg`. It is hinged at one end of the rod at a distance of `3m` below the water surface.
a.What must be the weight of a block that is attached to the other end of the rod so that `5 m` of the rod's length is under water?
b. Find the magnitude and direction of the force exerted by the hinge on the rod. The specific gravity of the material of rod is `0.5.`

Explanation

The rod makes angle with the vertical shape cose`theta=(3)/(5)` and sin `theta=(4)/(3)` Let .a. be the area of cross section of the rod.
`therefore` weight of the rod mg (6) 500 g N, Force of buoyancy F = (5a) 1000 g N
IF W N be the weight at the end (required), then taking moments about the hinge and equating it to zero
for rotatory equilibrium),(5000ag)2.5 sin `theta`)-(3000 ag)3 sin `theta`-W (6 sin `theta`)=0
Since sin `thetanetheta` ,so cancelling ag sin `theta` throughtout.
12500-9000-`(6W)/(ag)=3500implies(w)/(g)=(3500)/(6)a`

Now, the mass of the rod=12 kg `implies(6a)xx500=12impliesa=(1)/(250)m^(2)`
`therefore (W)/(g)=(3500)/(6xx250)=2.33 kg` wt. or W=22.837 N
ii)Now, upthrust F=`5000 xx(1)/(250)` g=20 gN
IFR be the verical reaction (up) at the hinge, then for vertical equlibrium of the rod, R+F-mg - W=0
`impliesR+20g-12g-2.33g=0implies`R=-5.67kg 2t.(or) -55.57N
The -Ve sign implies that ,the reaction is downwards of magnitude 55.57N