Question

A rubber ball filled with air has a diameter of 24.2 cm and a mass of 0.459 kg. What force is required to hold the ball in equilibrium immediately below the surface of water in a swimming pool? (Assume that the volume of the ball does not change. Indicate the direction with the sign of your answer.)

Answer 1

To solve this problem, it is necessary to apply the concepts related to Newton's second Law as well as to the expression of mass as a function of Volume and Density.

From Newton's second law we know that

F= ma

Where,

m = mass

a = acceleration

At the same time we know that the density is given by,

$\rho = \frac{m}{V} \rightarrow m = \rho V$

Our values are given as,

$g = 9.8m/s^2$

$m =0.459 kg$

D=0.242 m

Therefore the Force by Weight is

$F_w = mg$

$F_w = 0.459kg * 9.8m/s^2 = 4.498N$

Now the buoyant force acting on the ball is

$F_B=\rho V g$

The value of the Volume of a Sphere can be calculated as,

$V = \frac{4}{3} \pi r^3$

$V = \frac{4}{3} \pi (0.242/2)^3$

$V = 0.007420m^3$

$\rho_w = 1000kg/m^3 \rightarrow$ Normal conditions

Then,

$F_B=0.007420*(1000)*(9.8) = 72.716 N$

Therefore the Force net is,

$F_{net} = F_B -F_w$

$F_{net} = 72.716N - 4.498N =68.218 N$

Therefore the required Force is 68.218N