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Buoyant force on a steel boat

  1. Feb 16, 2008 #1
    [SOLVED] Buoyant force on a steel boat

    The bottom of a steel "boat" is a 7.00 m \times 10.0 m \times 4.00 cm piece of steel ( \rho _{{\rm steel}}=7900\;{\rm kg}/{\rm m}^{3}). The sides are made of 0.550 cm-thick steel. What minimum height must the sides have for this boat to float in perfectly calm water?

    I think I have to find the point where the weight of the boat is equal to the buoyant force of the water below. So, rho_b*V_b*g = rho_w*V_w*g. I'm confused on what to plug in for the volume of the boat, though. Wouldn't it be the volume of the base (7*10*0.04) plus the volume of two sides (2*7*0.0055*h) + (2*10*0.0055*h), where h is the height of the sides? But I don't know what to plug in for V_w, the volume of the water displaced. Is it equal to the volume of the boat? I'm really not sure if my equation or variables are correct, and I would really appreciate it if someone could help me out. Thanks.
     
  2. jcsd
  3. Feb 16, 2008 #2

    Kurdt

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    The water displaced just has to hold the weight of the metal. Then you can solve for the volume of water which will be the volume of the boat.
     
  4. Feb 16, 2008 #3
    I don't quite understand. The water displaced will not hold the weight of the metal base (7*10*0.4), so how could adding weight onto the boat make it float if it doesn't increase the surface area? I'm picturing a rectangular base with four sides that extend upwards, forming an open-topped box.

    I found that rho*V_steel*g = 216776 (downward force) and if the steel base is completely submerged, the buoyant force rho_water*V*g = 27440. There is a difference of 189336, meaning the buoyant force needs to increase by that much. The density of water and value of g are constant, so I divided 189336 by (1000*9.8) and got 19.32, which means I need to add 19.32 onto the volume of the boat in order to make it float. But any volume I add on multiplies the weight downward by a factor of 7900, while the force upward is only multiplied by 1000. How can they ever match up?
     
  5. Feb 16, 2008 #4

    Kurdt

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    What you want though is the water displaced to equal the weight of the boat. Then the volume of that water displaced will be the volume of the boat not just the steel.
     
  6. Feb 16, 2008 #5
    The weight of the boat is equal to the weight of the base (rho*v*g) plus the weight of the sides (rho*v*g). The volume of the sides is equal to 0.187x, where x is the height of the sides. So the weight of the boat is 216776 + 14477.5x, and this is the weight of the water displaced. The volume of the water displaced is then (219776 + 14477.5x)/(1000*9.8), which equals 22.4261+1.4773x, which is equal to the volume of the boat (Ah). So 22.4261+1.4773x = 70*(x+0.04). Solving for x gives me 32.7 cm. Neither 32.7cm or 36.7 cm are correct. Am I close?
     
  7. Feb 16, 2008 #6
    Got it! I made an error in the above calculation when typing it in. I fixed it, and got the correct answer. Thank you for your help!
     
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