# Homework Help: Firetruck going down a slope

1. Mar 28, 2009

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
$$F=ma$$

$$F_{net}=\frac{dp}{dt}$$

$$\theta=3$$

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

$$\frac{dm}{dt}=141 kg/s$$
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)
so
$$F_{net}=m(t)g\sin\theta-141 \times 50=(21000-141t)g\sin\theta-7050$$

for $$0\leq t\leq 106$$

Fire truck will stop next when a=0 which occurs at
$$0=(21000-141t)g\sin\theta-7050$$

$$141t=7257$$

$$t=51.5s$$
51.5 < 106

So i say yes the fire truck will stop before it runs out of water
Does my argument sound ok ?

2. Mar 29, 2009

### danb

Because this is an exercise
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).

3. Mar 29, 2009

### tiny-tim

Welcome to PF!

Hi danb! Welcome to PF!
Nice one!

4. Mar 29, 2009

### danb

Re: Welcome to PF!

Hi tiny-tim, thanks for the welcome

5. Mar 30, 2009

after having a second look at this problem

$$F_{net} = \frac{dp}{dt} = ma =m\frac{dv}{dt}=m\frac{dv}{dt}+v\frac{dm}{dt}$$
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
$$F=141kg/s$$
$$m(t)=M-Ft$$
$$v_w=50$$
$$\theta=3$$

so using newton we have
$$m(t) \frac{dv}{dt} = m(t)g\sin\theta-F v_w$$
$$\int dv=g\sin\theta\int dt-Fv_w\int \frac{dt}{m(t)}=gt\sin\theta-Fv_w\int \frac{dt}{M-Ft}$$
$$v(t)=gt\sin\theta+Fv_w \ln(M-Ft) + C$$
i can find the constant C by using
$$v(0)=0$$
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?

6. Mar 30, 2009

### danb

No, it's part of $$\frac{d}{dt}(mv)$$. $$F_{net} = \frac{dp}{dt}$$ turns into $$mg sin \theta = m\frac{dv}{dt}+v\frac{dm}{dt}$$

7. Mar 30, 2009

### danb

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

8. Mar 30, 2009

### tiny-tim

… 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) )

9. Apr 2, 2009