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For a National Board Exam Review:

A uniform area jet travels to $600 \frac{ft}{s}$ and a $100\frac{ft^3}{s}$. What is the horizontal force acts on the water jet if it undergoes a $180^\circ$ turn?

I would normally provide an answer but even review instructor cannot answer question...

So I try:

$${ F_x = \frac{ \gamma Q }{ g } \left(\nu_2 \cos(\theta) - \nu_1\right)}$$ $${ F_x = \frac{ 62.4 \frac{lb_m}{ft^3} ( 100 \frac{ft^3}{s} ) }{ 32.2 \frac{ft}{s^2} } \left(600 \frac{ft}{s^2} \cos(180^\circ)- 0\right) = -116273.2919\,lb_f }$$ Is the negative sign logical? All choices are from the range of ${1.35*10^5}$ to ${ 5.32*10^5 }$ lbf

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    $\begingroup$ May I assume that "with a flow of" is missing in the description? I can not explain the 100 otherwise... $\endgroup$
    – arc_lupus
    Aug 30, 2015 at 11:36

2 Answers 2

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This can be solved easily by calculating momentum. Momentum is mass x velocity. Force is momentum / time.

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Try considering the force required to stop the jet, then the same force again to accelerate the jet back to the same velocity, but in the opposite direction.

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