Problem 3 - Applied Mathematics

Question

Water issues from a pipe. whose cross section is \(c \mathrm{~m}^{2}\), in a horizontal jet with velocity \(v \mathrm{~ms}^{-1}\). What force must be exerted by a shield placed perpendicular to the jet to bring the water to a horizontal stop? (The mass of \(1 \mathrm{~m}^{3}\) of water is \(10^{3} \mathrm{~kg}\) ).

Step-by-Step Solution

Answer

The force required is \(10^3 \cdot c \cdot v^2\).

Step 1: Find the mass flow rate of the water

The mass flow rate (denoted as \(\text{m\text{\_dot}}\)) is the mass of water flowing per second. It can be calculated by multiplying the density of water (\(\rho\)), the cross-sectional area of the pipe (\(c\)), and the velocity of the water (\(v\)). Since the density of water is given as 1000 kg/m³, the formula becomes: \[ \dot{m} = \rho \cdot c \cdot v = 10^3 \cdot c \cdot v \]

Step 2: Determine the change in momentum per second

The force required to stop the water jet can be found using Newton's second law, which states that force is the rate of change of momentum. Since the shield brings the water to a stop, the final velocity is 0 m/s, and the change in velocity is \(-v\). The change in momentum per second is given by: \[ F = \dot{m} \cdot \text{change in velocity} = \dot{m} \cdot (-v) \]

Step 3: Calculate the force

Substituting \(\dot{m}\) from Step 1 into the equation from Step 2: \[ F = (10^3 \cdot c \cdot v) \cdot (-v) = -10^3 \(c \cdot v^2 \) \] Since force is a vector quantity and we are interested in its magnitude, we take the positive value: \[ F = 10^3 \cdot c \cdot v^2 \]

Key Concepts

Mass Flow Rate

Mass flow rate is an important concept in fluid dynamics that tells us how much mass of a fluid flows through a given cross-sectional area per unit of time. To find the mass flow rate, we use the formula \[ \text{mass flow rate} (\text{m\text{\textunderscore dot}}) = \rho \times c \times v \], where

• \( \rho \) is the density of the fluid
• \( c \) is the cross-sectional area
• \( v \) is the velocity of the fluid

For water, the density \( \rho \) is typically taken as 1000 kg/m³. So, for a given area and velocity, you can multiply these three values together to find the mass flow rate. In this exercise, knowing the mass flow rate of the water helps us understand how much mass is encountering the shield per second.

Newton's Second Law

Newton's Second Law of Motion states that the force acting on an object is equal to the rate of change of its momentum. Mathematically, it is represented as \[ F = \frac{dp}{dt} \] where \( F \) is the force, and \( \frac{dp}{dt} \) is the rate of change of momentum. Momentum itself is the product of mass and velocity (\[ p = m \times v \]). When a force is applied, it changes the velocity, thereby changing the momentum. In the exercise, the water jet hits the shield and stops, which means its final velocity is zero. Therefore, the change in momentum per second translates to a force applied to the shield. This helps us determine the force needed to stop the water jet using the mass flow rate and velocity.

Momentum Change

Momentum change is the difference in an object's momentum before and after an event. For a fluid, momentum change can be computed using the formula \[ \text{momentum change} = \text{mass flow rate} \times \text{change in velocity} \]. In this example, the change in velocity is from \( v \) (where \( v \) is the initial velocity) to 0 (since the water comes to a stop). Thus, the change in velocity is \( \text{change in velocity} = 0 - v = -v \). So, if we plug in our values, we get \[ F = \text{mass flow rate} \times (-v) \]. This formula can be expanded using our previously found mass flow rate to give us the force. This tells us how strong the water's impact is on the shield and allows us to calculate the forces acting on objects in fluid flow scenarios.