Question #40068

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Question #40068

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Answer 1 · 2016-08-18T20:47:21

Bouyant force will decrease as it makes the volume of the object

Explanation:

As Archimedes' principle explains:

The apparent weight of an object, $W_{apparent}$ $W_"apparent"$ , in a fluid equals to its weight, $W$ $W$ , minus the weight of the fluid which occupes its volume, $B$ $B$

$W_{apparent} = W - B$ $W_"apparent" = W - B$

where $B$ $B$ is buoyant force.

If you put a body on a fluid, it will occupe some volume of fluid. If this fluid has a density $ρ$ $rho$ , and the volume of the object is $V$ $V$ , the mass of fluid which will be occupied is:

$m_{fluid} = ρ V$ $m_"fluid" = rho V$

And the weight of this mass is right the buoyant force:

$B = m_{fluid} \cdot g = ρ V g$ $B = m_"fluid" cdot g = rho V g$

As you see, buoyant force does not depend on the mass of the submerged body, but its volume. If you decrease its size, $B$ $B$ will reduce.

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Question #40068

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