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.)

Respuesta :

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,

[tex]\rho = \frac{m}{V} \rightarrow m = \rho V[/tex]

Our values are given as,

[tex]g = 9.8m/s^2[/tex]

[tex]m =0.459 kg[/tex]

D=0.242 m

Therefore the Force by Weight is

[tex]F_w = mg[/tex]

[tex]F_w = 0.459kg * 9.8m/s^2 = 4.498N[/tex]

Now the buoyant force acting on the ball is

[tex]F_B=\rho V g[/tex]

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

[tex]V = \frac{4}{3} \pi r^3[/tex]

[tex]V =  \frac{4}{3} \pi (0.242/2)^3[/tex]

[tex]V = 0.007420m^3[/tex]

[tex]\rho_w = 1000kg/m^3 \rightarrow[/tex] Normal conditions

Then,

[tex]F_B=0.007420*(1000)*(9.8) = 72.716 N[/tex]

Therefore the Force net is,

[tex]F_{net} = F_B -F_w[/tex]

[tex]F_{net} = 72.716N - 4.498N =68.218 N[/tex]

Therefore the required Force is 68.218N