3.5 Force Exerted by Water Striking a Wall

When a stream of water from a pipe strikes a surface like a wall, it exerts a continuous force. This phenomenon is a direct application of Newton's laws of motion. The force arises because the wall must change the momentum of the moving water, and according to Newton's third law, the water exerts an equal and opposite force back on the wall.

Key Concepts and Derivation

To understand the force, we analyze the change in momentum of the water. This concept is closely related to Scalar and Vector Quantities→ as velocity and force are vector quantities.

1. Change in Momentum of the Water

Consider a small mass of water, $m$, moving with an initial velocity, $V$, that strikes a wall and comes to a complete stop.

Initial Momentum ($p_i$): The momentum of the mass $m$ just before it hits the wall is: $p_i=mV$

Final Momentum ($p_f$): Since the water comes to rest, its final velocity is 0. $p_f=m(0)=0$

Change in Momentum ($\Delta p$): The change in momentum of the water is: $\Delta p=p_f-p_i=0-mV=-mV$

The negative sign indicates that the momentum has decreased in the direction of the initial motion.

2. Force Exerted by the Wall on the Water

According to Newton's Second Law of Motion, the force applied is equal to the rate of change of momentum. The force that the wall exerts on the water to stop it is: $F_{\text{on water}}=\frac{\Delta p}{\Delta t}=\frac{-mV}{\Delta t}$

We can group the terms as: $F_{\text{on water}}=-\left(\frac{m}{\Delta t}\right)V$

Here, the term $\left(\frac{m}{\Delta t}\right)$ represents the mass flow rate—the mass of water striking the wall per unit of time (e.g., in kg/s).

3. Force Exerted by the Water on the Wall

According to Newton's Third Law of Motion, for every action, there is an equal and opposite reaction. The force exerted by the water on the wall is the reaction force to the force exerted by the wall on the water. $F_{\text{on wall}}=-F_{\text{on water}}$

Substituting the expression from the previous step: $F_{\text{on wall}}=-\left(-\left(\frac{m}{\Delta t}\right)V\right)=\left(\frac{m}{\Delta t}\right)V$

Final Formula:

The force exerted by the water on the wall is the product of the mass flow rate and the velocity of the water. $\text{Force}=(\text{Mass Flow Rate})\times (\text{Velocity})$

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Example Calculation

Problem: Water flows from a fire hose at a rate of 3 kg/s and strikes a wall with a velocity of 5 m/s. The water comes to a complete stop upon impact. What is the force exerted on the wall?

Solution:

1. Identify the given values:

   • Mass Flow Rate $\left(\frac{m}{\Delta t}\right)=3$ kg/s
   • Velocity ($V$) = 5 m/s

2. Apply the formula: $F=\left(\frac{m}{\Delta t}\right)V$ $F=(3\text{ kg/s})\times (5\text{ m/s})$ $F=15\text{ kg}\cdot {\text{m/s}}^2=15\text{ N}$

The force exerted on the wall by the water is 15 Newtons.

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Possible Questions and Answers

Q: Why is the mass flow rate used instead of just the mass?

A: The force is continuous because water is continuously striking the wall. Therefore, we need to consider how much mass is hitting the wall per second, which is the mass flow rate.

Q: What factors would increase the force of the water on the wall?

A: The force would increase if either the mass flow rate (more water per second) or the velocity of the water (faster stream) is increased.

Q: What if the water rebounds with the same speed?

A: If the water rebounds with velocity $-V$, the change in momentum is $\Delta p=-mV-mV=-2mV$. Thus, the force exerted would be $F=2\left(\frac{m}{\Delta t}\right)V$.