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Force on a fire truck

  1. Sep 26, 2016 #1
    1. The problem statement, all variables and given/known data
    A fire truck pumps a stream of water on a burning building at a rate K kg/s. The stream leaves the truck at an angle ##\theta## with respect to the horizontal and strikes the building horizontally at height h above the nozzle. What is the magnitude of the force on the truck due to the ejection of the water stream?

    2. Relevant equations


    3. The attempt at a solution
    We know that the rate of change of momentum of the water coming out of the nozzle must be equal and opposite to the force on the truck, from Newton's third law, then ##\vec{F_{truck}} = -\dot{\vec{P}}##. Now, we know that ##\dot{\vec{P}} = K \vec{v_0}##, so now all we must do is find the magnitude of the initial velocity. In the y-direction, we know that the velocity of the stream is zero when y = h (since it hits the building horizontally. So we can use the equation ##v_y^2 = v_{0y}^2 - 2gh##. Since the final velocity at h is zero, we can solve for the initial velocity in the y-direction, which is just ##v_{0y} = \sqrt{2gh}##. And since ##v_{0y} = v_0 \sin \theta##, we have that ##v_0 = \frac{\sqrt{2gh}}{\sin \theta}##. Thus, the magnitude of ##\dot{P}## and thus of ##F_{truck}## is ##K \frac{\sqrt{2gh}}{\sin \theta}##

    I feel like I am doing something wrong...
     
  2. jcsd
  3. Sep 27, 2016 #2

    Simon Bridge

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    What makes you think you are doing something wrong?
     
  4. Sep 27, 2016 #3
    I've just never seen an expression in physics where the sin function is in the denominator...
     
  5. Sep 27, 2016 #4

    Simon Bridge

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    It is an aesthetic to put trig functions in the numerator ... so you may prefer to use the cosecant instead: ##F=K\sqrt{2gh}\csc\theta## ... better?

    There are lots of occasion in physics to divide by the sine of an angle - especially with vectors. Your result has the right dimensions.
     
  6. Sep 27, 2016 #5
    Alright... I'll just take it that my solution is correct then. Thanks! Also, one more thing. If I were asked to then find the direction of the force, how would I do that? Would I just say that it is in the direction ##- \hat{v_0}##?
     
  7. Sep 27, 2016 #6

    haruspex

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    You are, but not what you thought.
    You have an equation relating a momentum vector P to initial velocity vector v0. What are you taking P to be exactly there? What is its direction?
     
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