Calculating the Minimum Height for a Steel "Boat" to Float

AI Thread Summary
To determine the minimum height for the sides of a steel "boat" to float, the buoyancy force must equal the weight of the boat. The weight of the boat's bottom is calculated as 2,385 kg, while the weight of the sides is expressed as 87.48h kg, where h is the height of the sides. The buoyancy force is derived from the water's density and the total volume of the boat, leading to the equation 2,385 kg + 87.48h kg = 54,000 kg(h + 0.05 m). Simplifying this equation allows for the calculation of h, which is essential for ensuring the boat floats in calm water. The final solution provides the necessary height for the sides of the boat.
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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
 
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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)
 


To calculate the minimum height for the sides of the steel "boat" to float, we need to determine the buoyancy force (F_B) and the weight of the boat (W_boat). We can use the formula F_B = W_boat, where W_boat is equal to the weight of the boat (W_B) plus the weight of the two sides (W_s1 and W_s2).

First, let's calculate the weight of the boat (W_B). This can be found by multiplying the density of steel (7900 kg/m^3) by the volume of the bottom of the boat (6.00 m x 9.00 m x 0.05 m). This gives us a weight of 2,385 kg.

Next, we need to calculate the weight of the two sides (W_s1 and W_s2). Since the sides are made of steel, we can use the same formula as above, but with a different volume. The volume of one side is 6.00 m x h m x 0.0046 m (where h is the height of the sides). So, the total weight of the two sides is 2 x (7900 kg/m^3) x (6.00 m x h m x 0.0046 m) = 87.48 h kg.

Now, we can plug these values into the formula F_B = W_boat. This gives us 2,385 kg + 87.48 h kg = F_B. We also know that the buoyancy force (F_B) is equal to the density of water (1000 kg/m^3) multiplied by the total volume of the boat (6.00 m x 9.00 m x (h + 0.05 m)). So, we can set these two values equal to each other and solve for h.

2,385 kg + 87.48 h kg = (1000 kg/m^3) x (6.00 m x 9.00 m x (h + 0.05 m))

Simplifying this equation gives us:

2,385 kg + 87.48 h kg = 54,000 kg x (h + 0.05 m)

Dividing both sides by 54,000 kg gives us:

0.04417 + 0.00162 h = h + 0.05

Subtracting h from
 
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