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Homework Help: Force exerted on a swimmer

  1. Apr 19, 2013 #1
    1. The problem statement, all variables and given/known data
    So the question i have involves finding the force exerted on a swimmer using only arm strokes through the water.

    A snapshot of the swimmer is taken by a camera underwater when the swimmer's arm is vertical and pointing at the camera.
    Here is a diagram of the hand:
    nWj5Gzg.jpg
    (forgot to label the top stream as A and bottom stream as B)

    The question asks to calculate the Force component that drives the swimmer forward (in this case the x direction), at the time the image was taken.

    Given values are u1=u1A=u1B= 3.5m/s
    d1=d1A=d1B= 20*10^-3 m
    Angle [itex]\alpha[/itex] = 25 degrees
    d2A= 23*10^-3 m
    d2B = 15*10^-3 m

    Given Assumptions

    The effective area of the hand is a rectangle, 0.15m *0.25m.
    The density of water is 1000kg/m^3
    The flow around the time the image was taken is approximately steady.
    The pressure upstream where u1 was measured is the same on both sides of the hand, i.e. P1=P1A=P1B
    The flow pattern is effectively two-dimensional, i.e, you would see the same pattern near the fingertips as near the wrist.



    2. Relevant equations

    A1V1=A2V2

    Bernoulli's Equation form of

    1/2(V2^2-V1^2)=-1/ρ(P2-P1)



    3. The attempt at a solution

    To start off with i found the velocities of each of the flow streams near the end (point 2)

    To do this i used the ratios of the d's to get

    V1A1=V2AA2 (height will be the same so)
    V1d1=V2Ad2A
    V2A= d1/d2A * V1

    and i did the same for V2B

    So the values for the velocities i got were
    V2A=3.04 m/s
    V2B= 4.67 m/s

    Using these values i put them into Bernoulli's Equation

    (V2A^2-V1^2)=-1/ρ(P2A-P1)
    (V2B^2-V1^2)=-1/ρ(P2B-P1)
    sub values and get:

    (P2A-P1)= -1504.2 pascal
    (P2B-P1)= 4779.45 pascal

    I then subbed for P1 and got

    (P2A-P2B)=-6283.65 pascal So this value is the Pressure difference between the flow streams.

    I then used
    (P2A-P2B)*Area(hand) to get the force
    =-235.63 N
    then for the force in the x direction (since the force acts normal to the object)
    -235.63*Cos(115degrees) = 99.58 N

    The value sort of sounds reasonable but I'm not sure if I've made any errors going about this.

    In particular is there anything in the y direction i need to include?

    Any feedback on the working would be appreciated
     
    Last edited by a moderator: Apr 19, 2013
  2. jcsd
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