Question

A garden hose is used to fill a 20 - \(L\) bucket in 1 min. The speed of the water leaving the hose is \(1.05 \mathrm{~m} / \mathrm{s}\). What force acts on the person holding the hose? a) \(0.35 \mathrm{~N}\) b) \(2.1 \mathrm{~N}\) c) \(9.8 \mathrm{~N}\) d) \(12 \mathrm{~N}\) e) \(21 \mathrm{~N}\)

Short answer

a) 0.35 N b) 0.50 N c) 0.75 N d) 1.05 N Answer: a) 0.35 N

Step-by-step solution

Find the mass flow rate of the water

To find the mass flow rate, we need to first find the volume flow rate, which can be found by dividing the volume of the bucket by the time it takes to fill it. The volume flow rate (V') is given by: \(V' = \frac{V_{bucket}}{time} = \frac{20L}{60\, seconds} = \frac{1}{3} L/s\) Next, we'll convert the volume flow rate from liters per second to cubic meters per second (m^3/s) by dividing by 1000. The density of water (\(\rho\)) is 1000 kg/m^3. Multiply volume flow rate by the density of water to obtain the mass flow rate (M): \(M = V' \times \rho = \frac{1}{3} × \frac{1}{1000} m^3/s × 1000 kg/m^3 = \frac{1}{3} kg/s\)

Calculate the momentum of the water

Now we need to calculate the momentum change of the water per second, which will give us the force exerted by the water on the hose. To find the momentum, we will multiply the mass flow rate by the speed of the water: \(p = M \times v = \frac{1}{3} kg/s \times 1.05 m/s = \frac{1.05}{3} kg \cdot m/s\)

Determine the force exerted on the person holding the hose

Finally, we can determine the force exerted on the person holding the hose by using the formula that relates force, mass, and acceleration: \(F = \frac{dp}{dt} = \frac{dp}{dt} = \frac{1.05}{3} kg \cdot m/s\) \(F = 0.35 N\) So, the force acting on the person holding the hose is \(0.35~N\). The correct answer is (a).

Key concepts

Understanding Mass Flow Rate in Fluid Dynamics

Mass flow rate is a crucial concept in fluid dynamics and plays a significant role in various applications such as engineering, meteorology, and hydraulics. It quantifies the amount of mass passing through a given surface per unit time. Its typical unit is kilograms per second (kg/s).

In our problem, we compute the mass flow rate to understand the quantity of water being delivered by the garden hose. Determining the mass flow rate is a two-step process. First, we find the volume flow rate by dividing the total volume of the water by the time it took to fill the 20-liter bucket. Then, we convert this rate into mass flow rate by using the density of water, which tells us how much mass of water flows through per second.

It is imperative to note that the mass flow rate is central to the calculation of force in our example because it directly impacts the amount of momentum transferred by the water leaving the hose. Understanding mass flow rate not only helps us in solving this problem but also forms the foundation for many other fluid dynamics equations and principles.

Conservation of Momentum in Fluid Streams

Momentum conservation is a fundamental principle in physics that states that the total amount of momentum in an isolated system remains constant if no external forces are applied. In fluid dynamics, this principle helps in analyzing the behavior of fluids in motion and predicting the forces involved.

For the garden hose scenario, we apply momentum conservation to find the force the water exerts. We calculate the change in momentum per second, which is equal to the mass flow rate multiplied by the velocity of the water. This figure provides a clear understanding of how the water, once leaving the hose, imparts a force on any object it encounters, including the person holding the hose.

The usefulness of momentum conservation in fluid dynamics extends beyond garden hoses. It informs the design of jet engines, water turbines, and more by allowing us to predict the resultant forces when fluids change speed or direction.

Calculating Force from Fluid Motion

Force calculation in fluid dynamics is integral to understanding how fluids interact with their surroundings. According to Newton’s second law of motion, force is the product of mass and acceleration. However, in a flowing fluid, we often measure force as a change in momentum over time.

In our textbook problem, we need to determine the force action on the person holding the hose based on the water's motion being emitted. By calculating the change in momentum per unit of time (momentum per second from the previous step), we can directly find the force exerted on the hose and consequently, on the person. This practical example illuminates how fluid movement can translate into physical forces, offering insights into phenomena such as hydraulic systems and propulsion forces in aerospace applications.

It’s essential for students to grasp the relationship between the flow of fluid and the forces generated to predict how fluids will behave in real-world situations, embodying the underlying power of fluid dynamics.