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Math Help - Buoyant Force

  1. #1
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    Buoyant Force

    1. The problem statement, all variables and given/known data
    A buoyant measuring device of m = 0.5 kg is projected straight down into a lake at initial velocity v0 = 20 m/s. Two forces act on the device: (1) buoyancy pulls up with a force equal to the weight of the displaced water, in this case 10 newtons; (2) water resistance resists the motion of the device in the water with force 1/2 |v|, where v is the velocity. (For this question, we'll ignore the force of gravity on the object.)
    (a) Find (analytically) the time at which the device reaches its maximum depth.
    (b) Find the maximum depth that is reached.
    (c) Find expressions for the velocity and position of the device as it returns to the lake's surface.
    (d) Estimate the total time for which the device will be under water


    2. Attempt to solve:
    a) m(dv/dt) = Fb+1/2v
    dv/dt = 10+0.5v/m
    (1/20+v)dv=dt
    ln |20+v| = t +C
    v = Ce^t-20 [where C=40 after subbing in initial condition]

    so then I sub in v=0 to find the time of max depth and get t=0.7s

    b) integral (v) = integral (40e^t-20)
    x(t) = 40e^t-20t
    and sub in the time from (a) to get max depth of 66.5m

    c) m(dv/dt) = Fb-1/2v
    (1/20-v)dv = dt
    v = -Ce^t+20 [sub in v=0 since that would be the velocity when it's back on surface?]
    to get: v=-20e^t+20
    Now take the integral of it to find distance: x(t) = -20e^t+20t

    d) This is where I got stuck. I think something in my previous solutions is incorrect or I just don't think about it right... I just can't get an answer...

    Thanks!
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  2. #2
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    It looks like you might not have been careful enough in defining your coordinate system. Are you defining the velocity positive downward or positive upward?

    Also, in part (a), you didn't divide the 10 N by the mass, though you divided everything else by it. I'm guessing you were just sloppy with your parentheses in your second step. It should have been

    dv/dt = (10+0.5v)/m. It looks like you fixed that problem in the next line.

    You didn't separate out the variables correctly. The next step after the one I just mentioned should have been

    dv = (20+v) dt.

    I see that, again, in your next step you corrected the mistake. You really have to be a lot better with your parentheses! Don't write so that people can understand you. Write so that no one can misunderstand you!

    Your solution, v = 40 e^t - 20 cannot possibly be correct. An exponentially increasing velocity? Surely common sense would argue that that is an impossibility when the only two known forces are opposing the direction of motion. So, I would go back to your original DE, and carefully check out the signs. Your general idea is correct, but I think you have a sign error, or maybe errors, in there somewhere. Do the same for the rest of the problem.

    Make sense?
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  3. #3
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    Yea.. the first equation should be v=-40e^t-20. when I sub 0 for velocity to find the time of max depth I end up needing to take the ln of a negative number.
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  4. #4
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    the first equation should be v=-40e^t-20
    I disagree. I think your original DE is wrong. Check that out, and then solve that correctly.
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  5. #5
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    Well, if we decide that the direction of the velocity [downwards] would be positive. then when the object hits the water, a buoyant force will affect it upwards, and the water resistance will act against the motion, aka, upwards.
    so m(dv/dt)=-Fb-(1/2)v but that gives me the v=-40e^t-20 when I solve it.
    Can you please explain what's your view on the problem?
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  6. #6
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    I like your DE for the descent now. But I don't think you've solved it correctly. Can you show your work more carefully, please?
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  7. #7
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    Yes if course:
    m(dv/dt)=-Fb-(1/2)v
    dv/dt = [-10-(1/2)v]/0.5
    dv/dt = (-20-v)
    1/(-20-v)dv = dt
    -ln (20+v) = t + C
    -(20+v) = e^(t+C)
    v = -Ce^t -20
    Sub in v=0 to find C, which is equal to -40.
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  8. #8
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    This step, from

    -ln (20+v) = t + C to

    -(20+v) = e^(t+C)

    is incorrect. You need to multiply by -1 before exponentiating.
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  9. #9
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    so should it be -ln|v+20| = t+C
    ln|v+20| = -t -C
    v+20 = e^-t-C
    v = Ce^(-t) -20 ?
    which becomes: v = 40e^(-t) -20.. with IC.
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  10. #10
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    so should it be -ln|v+20| = t+C
    ln|v+20| = -t -C
    v+20 = e^-t-C
    v = Ce^(-t) -20 ?
    which becomes: v = 40e^(-t) -20.. with IC.
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  11. #11
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    Yeah, I agree with that solution. If you plot it, it makes sense now. So now you should be able to solve parts (a) and (b). What is your DE for parts (c) and (d)?
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  12. #12
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    I am thinking of using m(dv/dt)=-Fb+(1/2)v, since the object now goes upwards the water resistance works in the positive direction [downwards]. But I am not sure how my m(dv/dt) term relates, and moreover, which values I'd use as initial condition when solving for the constants once the object is coming back to surface.
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  13. #13
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    I don't think your LHS will change at all, since you're not changing your coordinate system. The -Fb is good still: it doesn't change just because the velocity changes direction. So, in looking at the v/2 term, let me think: the velocity will be negative, because the object is going upwards. Therefore, the term +v/2 will be negative. But the velocity and the resistive force must have opposite signs, correct? The resistive force is going to be in the positive direction, since the velocity is negative. So I don't think your DE changes, believe it or not.
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  14. #14
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    I see what you mean. However, what values do I use to find my constant for the equation of velocity and position they ask for in (c)?
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  15. #15
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    Well, you've found the time when it changes direction, correct? And you've found the position at which that happens, correct? Those are the answers to part (a) and (b), respectively. What is the velocity of the object when it changes direction?
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