Question

A child playing in a swimming pool realizes that it is easy to push a small inflated ball under the surface of the water whereas a large ball requires a lot of force. The child happens to have a styrofoam ball (the shape of the ball will not distort when it is forced under the surface), which he forces under the surface of the water. If the child needs to supply 554 N to totally 1.000 x 103 kg/m2, the density of submerge the ball, calculate the diameter d of the ball. The density of water is Pw 9.81 m/s2 95.0 kg/m2, and the acceleration due to gravity is g styrofoam is pfoam d = m.

Answer 1

Points to remember:

Buoyant Force = Density of fluid*Volume of displaced fluid *g

Mass = Density*Volume

Since the ball is fully submerged, the volume of displaced fluid is equal to the volume of the ball.

The volume of sphere, $V=\frac{4}{3}\pi r^{3}$

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Consider the styrofoam ball

Weight of the ball and the applied force are acting downwards which is balanced by the buoyant force which is acting upwards.

$W+F_{applied}=F_{buoyant}$

$mg+F_{applied}=\rho _{fluid}*V_{displaced}*g$

$mg+F_{applied}=\rho _{water}*V_{ball}*g$

$\rho _{foam}*V_{ball}*g+F_{applied}=\rho _{water}*V_{ball}*g$

$95kg/m^{3}*V_{ball}*9.81m/s^{2}+554N=1000kg/m^{3}*V_{ball}*9.81m/s^{2}$

$95*V_{ball}*9.81+554=1000*V_{ball}*9.81$

$931.95*V_{ball}+554=9810*V_{ball}$

$554=9810*V_{ball}-931.95*V_{ball}$

$554=(9810-931.95)*V_{ball}$

$554=8878.05*V_{ball}$

$\frac{554}{8878.05}=V_{ball}$

${\color{Red} V_{ball}=0.0624m^{3}}$

$\frac{4}{3}\pi r^{3}=0.0624m^{3}$

$r^{3}=\frac{3*0.0624}{4\pi }$

$r=\sqrt[3]{\frac{3*0.0624}{4\pi }}$

${\color{Red} r=0.2461m}$

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Diameter, $d=2*r=2*0.2461m$

ANSWER: ${\color{Red} d=0.4922m}$

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