1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Buoyancy - block of concrete is being raised from a lake

  1. Mar 23, 2014 #1
    Buoyancy -- block of concrete is being raised from a lake


    A 50kg block of concrete is being raised from a lake. What fraction of its weight in air is required to lift it while submerged?

    My attempt,

    I'm pretty sure this question pertains to Archimedes principle

    I calculated the weight of the block (force due to gravity) and noted that the buoyancy force is equal to the weight of fluid displaced

    Buoyancy = ρgV = ρgAy

    And this is where it hit the buffers I'm not so sure how to proceed?

    I think that it would be something like, calculate the buoyancy force and the force which would be needed to lift the block (in air) and then subtract the two as the buoyancy force would aid the elevation and then take some kind of ratio of that mass required and the given mass?

    Any help would be great I think I have given enough information:)

    Thanks for any help in advanced:)

  2. jcsd
  3. Mar 23, 2014 #2


    User Avatar
    Science Advisor
    Homework Helper
    2017 Award

    Buoyancy = ρgV = ρgAy is fine. Stick to V, because a "block" doesn't necessarily mean a rectangular block.
    Something similar can be written down for the mass in terms of V and for the weight in air (check if you can ignore the buoyancy in air).

    Since they want a fraction as a numerical answer, you will need the values of ##\rho## for concrete, water (and perhaps also for air).

    Oh, and "submerged" can be taken to mean "fully submerged" I suppose.
  4. Mar 23, 2014 #3
    Ok I will give it a go:)

    Thanks a bunch I really appreciate it
  5. Mar 24, 2014 #4
    So when I write mass in terms of ρV my volume is constant so I can cancel it and then go from there?
  6. Mar 24, 2014 #5
    Do I start with the initial condition

    mg = ρgV where density is that of concrete? Just so I know I'm on the correct path?
  7. Mar 24, 2014 #6
    This is not initial condition. It is true all the time for the given block.
    But it is one of the equations you need, yes.
  8. Mar 24, 2014 #7
    So I would I then add the force applied to lifting the block
  9. Mar 24, 2014 #8
    OK. What will be this force when the block is in air?
    What will it be when it is submerged?
  10. Mar 24, 2014 #9
    mg would be still acting downwards and force of pulling would be

    F= ma?
  11. Mar 24, 2014 #10
    When submerged the buoyancy aids the pulling force, and the only force acting down would be mg? Disregarding viscous forces and drag?
  12. Mar 24, 2014 #11
    What is "a" here?
    In Newton's law, F must be the net force. The sum of all the forces acting on the body.
    Here you can assume uniform lifting, with no acceleration.
  13. Mar 24, 2014 #12
    Ahh ok that helps a lot! I usually don't assume such things unless the problem states it explicitly, I'm the typical student who finds it hard to think on their own:(

    So I am omitting acceleration as it is zero in this case?

    ρVg = m(0)+ ρgV

    Where the first rho is concrete density and the second is water?
  14. Mar 24, 2014 #13
    What is m(0)?
  15. Mar 24, 2014 #14
    That's what i don't get if the acceleration is zero the

    F= ma would be zero?
  16. Mar 24, 2014 #15
    Yes, the net force will be zero. This means that the forces acting on the body are balanced.
    For the case in air, you have two forces: the weight of the block and the force pulling up.
    You said that the weight is ρgV. What is the force pulling up, taking into account that the two have to be balanced?
  17. Mar 24, 2014 #16
    ρgV- mg =0

    ρgV= mg
  18. Mar 24, 2014 #17
    No, ρVg and mg are two expressions of the same force.
    You have two forces, gravity, G and the pulling force, F.
    Newton's law applied to this situation is

    Now you can express G=mg=ρVg
    so going back to the first equation,
    F-ρVg=0 and

    This is the pulling force in the first case.

    Now do the same thing for the case you have buoyant force too. Now you have three forces.
    Again, find the pulling force, let say F'.
  19. Mar 24, 2014 #18
    Buoyancy force is ρgV also though so would I just add that force
  20. Mar 24, 2014 #19
    F= ρgV for the case in air

    So buoyancy (when submerged)

    mg= ρgV

    Where rho would be that of water in this case where as it would be concrete in the first case
  21. Mar 24, 2014 #20
    Just write the equilibrium of the forces first. Then you will know if (and where) you add it.
  22. Mar 24, 2014 #21
    So in water

    mg= ρgV + F

    Where F is the force discussed previously?
  23. Mar 24, 2014 #22
    Yes, if the ρ is density of water. You should use different symbols for different quantities.
    Using ρ for both concrete and water is bad practice.
    And the pulling force is different now so you should not use F which was the pulling force when in air.
    Call it F' for example.

    Now calculate this F' and compare with F found previously.
  24. Mar 25, 2014 #23
    mg =gV(ρ+φ)

    Where phi is the density of concrete
  25. Mar 25, 2014 #24
    Now, you don't listen.
    The "F" in here is not the same as for the air. It is not ρVg or φVg.
    It is a new force, F'. This is what you are supposed to find.
    Maybe a drawing showing the forces may help you understand.
  26. Mar 25, 2014 #25


    User Avatar

    Staff: Mentor

    What value are you going to use as the density of concrete?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted