A rod is in equilibrium as shown in figure. What can be the weight of this
0.5 m
w
19.25 m
0.75 m
F = 100 N
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Answer:
Length of rod inside the water =1secθ=secθ
Upthrust, F=(
2
2
)(secθ)(
500
1
)(1000)(10)
⇒F=20secθ
Weight of rod, w=20×10=20N
For rotational equilibrium of rod, net torque about O should be zero.
∴F(
2
secθ
)(sinθ)=w=(1sinθ)
⇒
2
20
sec
2
θ=20
⇒θ=45
o
⇒F=20sec45
o
F=20
2
N
For vertical equilibrium of rod, force exerted by the hinge on the rod will be (20
2
−20)N downwards i.e. 8.3N downwards.
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