# Calculating the Force on a Fire Truck Pumping Water onto a Burning Building

• Mr Davis 97
In summary, the magnitude of the force on the truck due to the ejection of the water stream is equal to K times the square root of 2gh divided by the sine of the angle θ. The direction of the force can be determined by taking the negative direction of the initial velocity vector, which is equal to the negative direction of the momentum vector.
Mr Davis 97

## Homework Statement

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?

## 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...

What makes you think you are doing something wrong?

I've just never seen an expression in physics where the sin function is in the denominator...

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.

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}##?

Mr Davis 97 said:
I feel like I am doing something wrong
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?

## 1. What is the purpose of the force on a fire truck?

The force on a fire truck is used to propel the vehicle forward, allowing it to reach the scene of a fire quickly and efficiently. It also helps the truck carry heavy equipment and water to extinguish the fire.

## 2. How is the force on a fire truck generated?

The force on a fire truck is generated by the vehicle's engine, which powers the pump and creates pressure in the water lines. This pressure then propels the water out of the fire hose, creating a force that can be directed at the fire.

## 3. Can the force on a fire truck be adjusted?

Yes, the force on a fire truck can be adjusted by controlling the amount of pressure in the water lines. This allows firefighters to determine the appropriate force needed to extinguish the fire without causing damage to the surrounding area.

## 4. How does the force on a fire truck affect the handling of the vehicle?

The force on a fire truck can significantly affect the handling of the vehicle, especially at higher speeds. The weight and force from the water inside the truck's tank can make it more difficult to maneuver, and the force from the water being expelled can cause the truck to push or pull to one side.

## 5. Are there safety precautions in place to prevent accidents caused by the force on a fire truck?

Yes, there are safety measures in place to prevent accidents caused by the force on a fire truck. Firefighters are trained to handle the force of the water and to be aware of its effects on the truck's handling. Additionally, the force can be controlled and adjusted as needed to ensure the safety of both firefighters and the public.

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