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Water from a fire hose is directed horizontally against a wall at a rate of 50.0 kg/s and a speed of 42.0 m/s. Calculate the force exerted on the wall, assuming the water’s horizontal momentum is reduced to zero.
Water from a fire hose is directed horizontally against a wall at a rate of 50.0 kilograms per second and a speed of 42.0 meters per second. Calculate the force exerted on the wall, assuming the water’s horizontal momentum is reduced to zero.
Let’s start by highlighting some of the important information this problem statement gives. We’re told the rate at which water leaves the fire hose, 50.0 kilograms of water every second. And that the water is moving at a speed of 42.0 meters per second. We’re asked to solve for the force, we’ll call it capital 𝐹, that the water exerts on the wall, assuming that all of the horizontal momentum of the water is reduced to zero. Let’s draw a diagram of the problem to get clarity on how to move forward.
So here we have our wall and our fire hose firing water into the wall. And we’re told that the hose sprays water at a certain rate we’ve called 𝑟, which is 50.0 kilograms per second. And that the speed of the water as it leaves the hose and impacts the wall is 42.0 meters per second. In addition to assuming that the wall entirely reduces the horizontal momentum of our water to zero, we also assume that there is no energy loss between the water leaving the hose and impacting in the wall. When the water does run into the wall, it presses on the wall with a force. We’ve called it capital 𝐹, and that’s what we wanna solve for.
To do that, let’s recall a theorem that relates the impulse of the water with its momentum. The impulse-momentum theorem states that there is a relationship between the impulse, the force times the change in time, and the change in momentum, 𝑚𝛥𝑣, of an object or a mass. Let’s apply this relationship to our particular scenario to see how it fits.
The equation says that force multiplied by change in time is equal to mass multiplied by change in speed. In our situation, we wanna solve for the force. So to do that, let’s isolate 𝐹 from the left-hand side of the equation by dividing both sides by 𝛥𝑡. When we do that, 𝛥𝑡 on the left-hand side of our equation cancels out. And we see a simplified equation that says that the force 𝐹 is equal to mass times change in speed divided by change in time.
Now let’s look over the given information in our problem. We’re given a rate that we’ve called 𝑟. And the units of that rate are kilograms per second. In other words, it’s an amount of mass per time. That means we can replace the 𝑚 divided by 𝛥𝑡 in our equation with 𝑟 for rate. Then as we look to 𝛥𝑣, we’re told in the problem that the water’s horizontal momentum becomes zero after it hits the wall. That means that its horizontal speed must be zero. So the change in speed of the water from start to finish is 𝑣, 42.0 meters per second minus zero meters per second. So we can replace 𝛥𝑣 in our equation with 𝑣 which specifically represents 42.0 meters per second.
Now to solve for force, we just need to plug in for our values of 𝑟 and 𝑣. When we enter these numbers in on our calculator, we find a force of 2.10 times 10 to the third newtons. This is actually quite a lot of force. This is approximately three times the weight force of the average-sized adult. This is the force that the water exerts on the wall.
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