Question

A plastic container in semispheric form have 80kg. Determined the minimum radio that container must have to avoid fully sunk in the water, when we add a mass of 5.00 * 10^3 kg inside the container.

Answer 1

Solution:

Volume of a sphere is (4/3)πr³. Hence the volume of a semispheric container is half the volume of a sphere;

V = (4/3)πr³/2

V = (4/6)πr³

The mass of container and the load together,

m = 80 kg + 5.00*10³ kg

m = 5080 kg

The buoyant force is given by,

f_b = m_f*g                   where m_f = mass of the fluid displaced by the floating object

But when the object of weight F_g = m*g floats in the fluid then the buoyant force equals to the weight of the object;

F_g = F_b

Hence,

m*g = m_f*g

It means a floating object displaces the fluid whose weight is equal to the weight of the object.

m = m_f

we know density = mass/volume, hence for water, m_f = σ_f*V_f

m = σ_f*V_f

where σ_f = density of water = 1000 kg/m³ and V_f* = volume of water that is being displaced by the object.

V_f = m/σ_f

V_f = 5080 kg/(1000 kg/m³)

V_f = 5.080 m³

Thus volume of water that is displaced by the floating object of mass m is 5.080 m³.

Now the plastic container (object) is just fully submerged, thus its volume V must the same as that of the displaced volume of water V_f

Hence

V = 5.080 m³.

(4/6)πr³ = 5.080 m³

r³ = 6*(5.080 m³)/(4π)

r = ³√[6*(5.080 m³)/(4*3.1416)]

r = 1.3436 m

Thus the minimum radius of the container should be 1.3436 m. If the radius greater than this then object will float.

Hence the answer is 1.3436 m