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Fluid Mechanics: Cart and tank of water

  1. Feb 6, 2016 #1
    1. The problem statement, all variables and given/known data
    A cart has a tank of water on it which has a nozzle. Water jets out from the nozzle onto a turn vane oriented at theta = 30° which deflects the flow upward. The cart is held stationary by a rope as shown in the image below. Find the force in the rope if the velocity of the jet is 68.5 [ft/s] and the cross-sectional area is 0.8 [ft2].

    2. Relevant equations


    3. The attempt at a solution
    I'm pretty confused about how to start this problem. I tried starting out with drawing a free body diagram (attached) and then found the mass flow rate:
    m=ρAv=(62.4lb/ft3)(0.8ft2)(68.5ft/s)
    m=3419.5 lb/s

    I was thinking somehow I could multiply that mass rate to get the weight of the tank or something, but now I'm thinking that wouldn't really make sense... Did I even need to find the mass flow rate?

    Then I summed the forces in the x and y direction:

    ΣFx=(FH2O)cos(30°)=T

    ∑Fy=FN+FA-Wtank-Wcart-(FH2O)sin(30°) +FB

    Is there someway I could calculate the pressure at the nozzle outlet? And then multiply that pressure by the area to get the force?

    Sorry if this seems a little scattered!
     

    Attached Files:

  2. jcsd
  3. Feb 6, 2016 #2
    The rate of water momentum entering the combination of tank and cart is zero. What is the horizontal component of the rate of water momentum exiting the combination of tank and cart? This is equal to the horizontal force exerted by the combination of tank and cart on the water (i.e., the rate of change of momentum of the water). How does that relate to the horizontal force exerted by the water on the combination of tank and cart?

    Let me guess. In your course, you are learning about macroscopic momentum balances on fluid flows.

    Chet
     
  4. Feb 6, 2016 #3
    Haha I think that's what we're learning about! I haven't seen an example like this yet though, so maybe we'll be doing problems like this next class!

    So for the horizontal component of the momentum exiting should be this?: v*m*cos(30)

    So does that equal the tension in the rope? T=v*m*cos(30)=(68.5 ft/s)*(3419.5 lb/s)*cos(30) T=202854.11 (ft*lb)/(s2)

    I hope I understood you correctly! I'm not very comfortable with english units, do you know how to convert this answer to lbf? Or do the above units already condense down to lbf?


     
  5. Feb 6, 2016 #4
    Assuming you did the rest of the arithmetic correctly, you gotta divide by gc.
     
  6. Feb 6, 2016 #5
    What is gc?
     
  7. Feb 7, 2016 #6
    $$g_c=32.2 \frac{lb_mft}{lb_fsec^2}$$
     
  8. Feb 7, 2016 #7
    Hahah oh, the subscript threw me off a little. Thanks!
     
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