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

Click For Summary
SUMMARY

The minimum height for the sides of a steel "boat" to float in calm water is determined by balancing the buoyancy force (F_B) with the weight of the boat (W_boat). The weight of the boat, calculated from its bottom dimensions (6.00 m x 9.00 m x 0.05 m) and the density of steel (7900 kg/m³), is 2,385 kg. The weight of the sides, dependent on their height (h) and thickness (0.460 cm), contributes an additional 87.48h kg. By equating the total weight to the buoyancy force derived from the water density (1000 kg/m³) and the total submerged volume, the equation simplifies to find the minimum height required for the sides.

PREREQUISITES
  • Understanding of buoyancy principles and Archimedes' principle
  • Familiarity with density calculations and unit conversions
  • Basic algebra for solving equations
  • Knowledge of material properties, specifically steel density (7900 kg/m³)
NEXT STEPS
  • Study the principles of buoyancy and Archimedes' principle in detail
  • Learn about density and volume calculations for various materials
  • Explore the effects of shape and size on buoyancy in fluid mechanics
  • Investigate real-world applications of buoyancy in marine engineering
USEFUL FOR

Engineers, naval architects, and students studying fluid mechanics or marine design will benefit from this discussion, particularly those interested in the practical applications of buoyancy in boat design.

quick
Messages
42
Reaction score
0
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
 
Physics news on Phys.org
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
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
2
Views
7K
  • · Replies 13 ·
Replies
13
Views
4K
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
9K
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K