Problem 89 - University Physics with Modern Physics

Question

Water stands at a depth \(H\) in a large, open tank whose side walls are vertical (Fig. \(\mathrm{Pl} 2.89 ) .\) A hole is made in one of the walls at a depth \(h\) below the water surface. (a) At what distance \(R\) from the foot of the wall does the emerging stream strike the floor? (b) How far above the bottom of the tank could a second hole be cut so that the stream emerging from it could have the same range as for the first hole?

Step-by-Step Solution

Answer

(a) \( R = 2\sqrt{h(H-h)} \), (b) Second hole is at depth \(H - h\) above the bottom.

Step 1: Identify Known Variables

We have two known depth variables: \(H\), the total depth of the water in the tank, and \(h\), the depth from the water surface to the hole. We'll find the distance \(R\) where the water strikes the ground.

Step 2: Determine the Velocity of Water from the Hole Using Torricelli's Law

To find the velocity \(v\) of water emerging from the hole, use Torricelli's law, which is derived from Bernoulli's equation:\[ v = \sqrt{2g(H-h)} \]where \(g\) is the acceleration due to gravity.

Step 3: Calculate the Time Water Takes to Hit the Floor

Using the vertical motion under gravity, calculate the time \(t\) it takes for water to hit the floor:\[ t = \sqrt{\frac{2h}{g}} \]where \(h\) is the vertical distance the stream falls.

Step 4: Determine the Horizontal Distance (Range)

Calculate the horizontal range \(R\) using the horizontal velocity and time:\[ R = v \times t = \sqrt{2g(H-h)} \times \sqrt{\frac{2h}{g}} = 2\sqrt{h(H-h)} \]

Step 5: Solve for Second Hole Height for Same Range

For the second hole, at depth \(h' = H - h\), the formula for range is symmetric:\[ R = 2\sqrt{h'(H-h')} \]For the same range, solve \(h' = H - h\). Thus, the second hole can be placed at \(h' = H - h\) above the bottom.

Key Concepts

Fluid Dynamics

Fluid dynamics is the study of how liquids and gases move. It's a branch of physics that focuses on the behavior of fluids in motion. In our exercise, fluid dynamics helps us understand how water flows out of a hole in a tank.

When a hole is made in a container filled with liquid, the fluid flows due to the pressure difference. This pressure force is the driving mechanism behind fluid motion.

• The shape and size of the container, as well as the size and placement of the exit hole, affect the flow rate and behavior of the stream.
• Viscosity and density of the fluid play important roles in how it travels.

In open systems like the water tank in the exercise, the flow is influenced by gravitational forces, making analysis using fluid dynamics essential for understanding the water stream's path.

Bernoulli's Equation

Bernoulli's equation is a crucial principle in fluid dynamics. It helps us predict how fluids will behave under varying pressure conditions. In simple terms, it states that the total mechanical energy along a streamline remains constant.

Using Bernoulli's equation, we can deduce Torricelli's law, which is directly applied to the problem in question. Bernoulli's equation is as follows:\[ P + \frac{1}{2} \rho v^2 + \rho gh = \text{constant} \]

• \(P\) is the pressure energy per unit volume.
• \(\frac{1}{2} \rho v^2\) is the kinetic energy per unit volume.
• \(\rho gh\) is the potential energy per unit volume due to height.

In an open tank, pressure energy can change, affecting the velocity of fluid exiting from the hole. By applying this principle, we find the speed of water leaving the container using Torricelli's law: \[ v = \sqrt{2g(H-h)} \] This equation implies the velocity is determined by the depth of water above the hole.

Projectile Motion

Projectile motion is important when considering how water jets travel after exiting the tank. It occurs when an object is thrown or propelled into the air, affected only by gravity.

In our exercise, once the water exits the hole, it becomes a projectile following a parabolic path. This path is determined by its initial velocity and gravitational force.

• Horizontal velocity is constant, as no forces act in this direction (ignoring air resistance).
• Vertical motion is influenced by gravity, making the stream accelerate downwards as it travels.

The range of the projectile, in this case, the distance it travels horizontally before hitting the ground, is calculated using:\[ R = v \times t = 2\sqrt{h(H-h)} \] This formula integrates both the speed of exiting fluid and the effect of gravity.

Acceleration Due to Gravity

Gravity is a fundamental force affecting the motion of objects on Earth. It provides a constant acceleration pulling objects towards the ground. In physics, it's represented as \( g \) and has an approximate value of \( 9.81 \, \text{m/s}^2 \).

In the context of this exercise, gravity directly influences the projectile motion of the water.

• It causes the water stream to fall downward after exit, determining how fast it strikes the ground.
• It impacts both the time taken to reach the floor and thus the horizontal distance traveled.

Gravity is crucial for calculations in fluid dynamics, as it affects flow rates, falling times, and eventual impact locations of fluids. This constant acceleration makes sure that water follows a predictable trajectory after emerging from the tank.