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Buoyant Force, Projectile Range

A block of wood is held in water, totally immersed, by a string attached to a pont in the base of the tank.
Calculation of range of jet stream flowing out of a tank.
Please see the attachment.


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Q1: (a) Let the volume of the wood be V m^3
Weight of wood = V*669 Kg
= V*669*9.8 N
Voume of water displaed = V m^3
Weight of water dispaled = V*1000*9.8 N
= Buoyant force
In equilibrium:

Buoyant force - weight of wood = Tension
V*1000*9.8 - V*669*9.8 = 0.92
V(1000 - 669)*9.8 = 0.92
V = 0.92/(331*9.8
= 0.0002836 m^3
= 2.836*10^-4 m^3

(b) If the string breaks, block will rise to the surface and starts floating, let us say with voume V' outside water surface.

V - V' = voume of block under water
While floating :
Upthrust = ...

Solution Summary

The solution shows the connection of buoyant force and weight of a body that is held in equilibrium by a string attached to a point in the base of the water container and the tension in the string. The Archimedes' principle and the law of flotation has been used to reach the required answer.
In the second question in the attachment the range of the water jet has been found by application of Bernoulli's principle.