Question

The bottom of a steel "boat" is a 3 m ? 22 m ? 2 cm piece of steel (?_steel = 7900 kg/m³). The sides are made of 2.00 cm thick steel. What minimum height must the sides have for this boat to float in perfectly calm water?

Answer 1

volume of the steel boat, V_boat = 3m*22m*0.02m

density of the steel, rho_steel = 7900 kg/m^3

side thickness, t= 2 cm

let, height of the boat be h,

volume of the water displaced,V_displaced = (3*22*(h+0.02) m^3

mass of the dispaced water, m= rho_water*V_displaced =1000*3*22*(h+0.02)-----(1)

and volume of the steel in boat is,

v=V_boat+V_object = (3*22*0.02) + (2*(3*h*t)+ 2*(22*h*t))

V= 3*22*0.02+ 2*(3*h*0.02)+2*(22*h*0.02)

and

mass of the steel , m_steel=rho_steel*V

m_steel= 7900*(3*22*0.02+ 2*(3*h*0.02)+2*(22*h*0.02)) ---(2)

from Buoyancy principle,

mass of the dispaced water=mass of the steel

1000*3*22*(h+0.02)=7900*(3*22*0.02+ 2*(3*h*0.02)+2*(22*h*0.02))

=> h = 0.157 m

therefore , height, h = 0.157 m