Question

A father (weight W = 849 N) and his daughter (weight W = 394 N) are spending the day at the lake. They are each sitting on a beach ball that is just submerged beneath the water (see the figure). Ignoring the weight of the air in each ball, and the volumes of their legs that are under the water, find (a) the radius of father's ball and (b) the radius of daughter's ball.

Answer 1

the balls are in equilbrium, the net force acting on each of them must be zero.

Taking upward to be the positive direction,

F = F_B - W = 0                                 { eq. 1 }

where, F = net force

F_B = buoyant force

W = weight of the person

the buoyant force acting on the father's beach ball is greater, it must balances his greater weight.

According to archimede's principle,

F_B = _float V_displaced g                                        { eq. 2 }

where, g = acceleration due to gravity = 9.8 m/s²

_float = density of water = 1000 kg/m³

V_displaced = volume of the ball = 4/3 r³

inserting the values in eq.2,

F_B = (1000 kg/m³) (4/3 r³) (9.8 m/s²)

F_B = 41029.3 r³    N

the magnitude of buoyant force equals to the weight of the person.

(a) the radius of father's ball is given as :

using eq.1,      F_B = W_f

where, W_f = weight of the father = 849 N

(41029.3 r³) N = 849 N

r³ = (849 N) / (41029.3 N)

r = ³0.0206   m

r = 0.274 m

(b) the radius of daughter's ball is given as ::

F_B = W_d

where, W_d = weight of the daughter = 394 N

(41029.3 r³) N = 394 N

r³ = (394 N) / (41029.3)

r = ³0.0096 m

r = 0.212 m

the fathers ball has larger volume and larger radius than the daughter's ball.