1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Firetruck going down a slope

  1. Mar 28, 2009 #1
    1. The problem statement, all variables and given/known data
    A fire department's tanker truck has a total mass of 21000kg, including 15000kg of water. Its brakes fail at the top of a long 3 degree slope and it begins to roll downward, starting from rest. In an attempt to stop the truck, firefighters direct a stream of water parallel to the slope beginning as soon as the truck starts to roll. The water leaves the 6cm diameter hose nozzle at 50m/s. Will the truck stop before it runs out of water? If so, when? If not, what is the minimum speed reached?

    2. Relevant equations



    3. The attempt at a solution
    The rate of loss of mass per second is
    [tex]\frac{dm}{dt}=50\pi r^2 m^3/s[/tex]

    [tex]\frac{dm}{dt}=141 kg/s[/tex]
    So after 15000/141=106 seconds the fire truck will have run out of water.
    The only other force acting is gravity. (not sure why friction isn't included)
    [tex]F_{net}=m(t)g\sin\theta-141 \times 50=(21000-141t)g\sin\theta-7050[/tex]

    for [tex]0\leq t\leq 106[/tex]

    Fire truck will stop next when a=0 which occurs at


    51.5 < 106

    So i say yes the fire truck will stop before it runs out of water
    Does my argument sound ok ?
  2. jcsd
  3. Mar 29, 2009 #2
    Because this is an exercise :smile:
    You mean v=0.

    m(t) dv/dt = m(t)g' - vw dm/dt

    where g' = g sin(theta)

    You have to integrate to find v(t).
  4. Mar 29, 2009 #3


    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi danb! Welcome to PF! :smile:
    Nice one! :biggrin:
  5. Mar 29, 2009 #4
    Re: Welcome to PF!

    Hi tiny-tim, thanks for the welcome :smile:
  6. Mar 30, 2009 #5
    after having a second look at this problem

    [tex]F_{net} = \frac{dp}{dt} = ma =m\frac{dv}{dt}=m\frac{dv}{dt}+v\frac{dm}{dt}[/tex]
    this sounds pretty silly but doesnt the m dv/dt cancel from both sides?

    moving on
    let F= flow rate
    M=initial mass of truck and water=21000kg
    v_w = velocity of water
    initial velocity of truck = 0

    so using newton we have
    [tex]m(t) \frac{dv}{dt} = m(t)g\sin\theta-F v_w[/tex]
    [tex]\int dv=g\sin\theta\int dt-Fv_w\int \frac{dt}{m(t)}=gt\sin\theta-Fv_w\int \frac{dt}{M-Ft}[/tex]
    [tex]v(t)=gt\sin\theta+Fv_w \ln(M-Ft) + C[/tex]
    i can find the constant C by using
    but with the ln(t) and t in the same equation it seems a bit tricky to solve for t when v=0 unless using maple or mathematica
    did i go overkill somewhere?
  7. Mar 30, 2009 #6
    No, it's part of [tex]\frac{d}{dt}(mv)[/tex]. [tex]F_{net} = \frac{dp}{dt}[/tex] turns into [tex]mg sin \theta = m\frac{dv}{dt}+v\frac{dm}{dt}[/tex]
  8. Mar 30, 2009 #7
    Yea, I'm not sure how the problem can be solved analytically if v goes to zero. I guess you can find v when the water runs out and hope it's positive :smile:
  9. Mar 30, 2009 #8


    User Avatar
    Science Advisor
    Homework Helper

    Hi vladimir69! :smile:

    Since the question asks …
    … solve for dv/dt = 0 first, and find the speed v then …

    if it's positive, and if you've not run out of water, then that's your answer

    (only if it's negative do you have a problem with ln(t) :wink:)
  10. Apr 2, 2009 #9
    ok thanks for the effort guys
    greatly appreciated
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook