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Q52PE

Expert-verifiedFound in: Page 397

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Referring to Figure $11.21$** **, prove that the buoyant force on the cylinder is equal to the weight of the fluid displaced (Archimedes’ principle). You may assume that the buoyant force is ${F}_{2}-{F}_{1}$** ** and that the ends of the cylinder have equal areas A. Note that the volume of the cylinder (and that of the fluid it displaces) equals ** $({h}_{2}-{h}_{1})A$** .**

The buoyant force acting on the body is equal to the weight of the object which is submerged into the liquid.

**Fluid statics, often known as hydrostatics, is a branch of fluid mechanics that investigates the state of balance of a floating and submerged body, as well as the pressure in a fluid, or imposed by a fluid, on an immersed body.**

The ratio of the object's density to the density of the fluid in which it is immersed determines the fraction submerged in the fluid.

Sketch of the figure

Here the ${F}_{by}$_{ }= buoyant force acting on the body

The two surfaces are named as A and B there the force is acted.

${F}_{by}={F}_{b}-{F}_{T}$

${F}_{by}={F}_{b}-{F}_{T}\phantom{\rule{0ex}{0ex}}={P}_{b}{A}_{b}-{P}_{T}{A}_{T}\phantom{\rule{0ex}{0ex}}=\left({h}_{2}{\rho}_{w}g\right)A-\left({h}_{1}{\rho}_{w}g\right)A\phantom{\rule{0ex}{0ex}}={\rho}_{w}\times g\times A\left[{h}_{2}-{h}_{1}\right]$

Further simplifying, we get

${F}_{by}={\rho}_{w}\times g\times {V}_{wd}\phantom{\rule{0ex}{0ex}}={m}_{wd}g\phantom{\rule{0ex}{0ex}}={W}_{wd}$

Hence, the buoyant force acting on the body is equal to the weight of the object which is submerged into the liquid.

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