# Buoyancy force

1. Dec 7, 2004

### quick

The bottom of a steel "boat" is a 6.00 m x 9.00 m x 5.00 cm piece of steel(density of steel = 7900 kg/m^3) . The sides are made of 0.460 cm-thick steel.

what minimum height must the sides have for this boat to float in perfectly calm water? in cm

i have that F_B (buoyancy force) is equal to W_boat (weight of boat) is equal to W_B + 2*W_s1 + 2*W_s2. where w is the weights and it equal rho*g*V. F_B = density of water*g*total volume and total volume is equal to 6*9*(h+ .05)

any suggestions would really help

2. Dec 7, 2004

### Tide

Give this a try:

$$Weight_{bottom} + 2\rho_{steel} g \ h (L + W) = \rho_{water} g (L-2\times0.046)(W-2\times0.046)(h-0.05)$$