Physical Applications: Fresh Take

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Physical Applications: Fresh Take

Calculate the mass of linear and circular objects using their density distributions
Compute the work required in various situations, such as pumping fluids or moving objects along a path
Determine the force exerted by water on a vertical surface underwater

Mass and Density

The Main Idea

Mass-Density Formula for One-Dimensional Objects:
- For a thin rod along the $x$ -axis from $a$ to $b$ : $m = \int_a^b \rho(x) dx$
Mass-Density Formula for Two-Dimensional Disks:
- For a disk of radius $r$ with radial density: $m = \int_0^r 2\pi x \rho(x) dx$
Concept of Density Functions:
- Linear density: mass per unit length
- Area density: mass per unit area
- Radial density: density varying along the radius

Problem-Solving Strategies

One-Dimensional Case:
- Partition the rod into segments
- Approximate mass of each segment: $m_i \approx \rho(x_i^*) \Delta x$
- Sum and take the limit as $n$ approaches infinity
Two-Dimensional Case:
- Partition the disk into concentric washers
- Approximate area of each washer: $A_i \approx 2\pi x_i^* \Delta x$
- Approximate mass of each washer: $m_i \approx 2\pi x_i^* \rho(x_i^*) \Delta x$
- Sum and take the limit as $n$ approaches infinity

Find the mass of a rod with density $\rho(x) = \sin x$ over the interval $[\frac{\pi}{2}, \pi]$ .

$\begin{array}{rcl} m &=& \int_{\frac{\pi}{2}}^{\pi} \rho(x) dx \\ &=& \int_{\frac{\pi}{2}}^{\pi} \sin x dx \\ &=& [-\cos x]_{\frac{\pi}{2}}^{\pi} \\ &=& (-\cos \pi) - (-\cos \frac{\pi}{2}) \\ &=& 1 - 0 = 1 \end{array}$

Consider a thin rod oriented on the $x$ -axis over the interval $\left[1,3\right].$ If the density of the rod is given by $\rho (x)=2{x}^{2}+3,$ what is the mass of the rod?

$\frac{70}{3}$

Watch the following video to see the worked solution to this example.

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Calculate the mass of a disk with radius $4$ and radial density $\rho(x) = \sqrt{x}$ .

$\begin{array}{rcl} m &=& \int_0^4 2\pi x \rho(x) dx \\ &=& \int_0^4 2\pi x \sqrt{x} dx \\ &=& 2\pi \int_0^4 x^{3/2} dx \\ &=& 2\pi [\frac{2}{5}x^{5/2}]_0^4 \\ &=& 2\pi \cdot \frac{2}{5} \cdot 32 \\ &=& \frac{128\pi}{5} \end{array}$

Let $\rho (x)=3x+2$ represent the radial density of a disk. Calculate the mass of a disk of radius $2$ .

$24\pi$

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “2.5 Physical Applications” here (opens in new window).

Work Done by a Force

The Main Idea

Definition of Work:
- Work is the energy transfer when a force moves an object
- Units: Joules (J); Force in Newtons (N), Distance in meters (m)
Work Formula for Constant Force: $W = F \cdot d$ (Force × Distance)
Work Formula for Variable Force: $W = \int_a^b F(x) dx$
Application to Springs:
- Hooke’s Law: $F(x) = kx$
- $k$ is the spring constant (N/m)

Problem-Solving Strategy

Partition the interval $[a,b]$ into $n$ segments
Approximate work in each segment: $W_i \approx F(x_i^*) \Delta x$
Sum the work over all segments: $W \approx \sum_{i=1}^n F(x_i^*) \Delta x$
Take the limit as $n \to \infty$ to get the exact work: $W = \lim_{n \to \infty} \sum_{i=1}^n F(x_i^*) \Delta x = \int_a^b F(x) dx$

Suppose it takes a force of $8$ lb to stretch a spring $6$ in. from the equilibrium position. How much work is done to stretch the spring $1$ ft from the equilibrium position?

$8$ ft-lb

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “2.5 Physical Applications” here (opens in new window).

A spring requires a $10$ N force to compress it $0.2$ m. How much work is done to stretch it $0.5$ m?

Find spring constant $k$ :

$10 = k(0.2)$ , so $k = 50$ N/m

Set up the work integral:

$W = \int_0^{0.5} kx dx = \int_0^{0.5} 50x dx$

Evaluate the integral:

$\begin{array}{rcl} W &=& 50 \int_0^{0.5} x dx \\ &=& 50 [\frac{1}{2}x^2]_0^{0.5} \\ &=& 25 [(0.5)^2 - 0] \\ &=& 6.25 \text{ J} \end{array}$

Work Done in Pumping

The Main Idea

Work in Pumping:
- Calculating work required to pump water out of tanks
- Varies based on tank shape and water level
Key Principle:
- Work = Force (Weight of water) × Distance (Height lifted)
General Work Formula: $W = \int_a^b F(x) dx$ , where $F(x)$ depends on tank shape
Weight-Density of Water:
- $9800$ N/m³ (metric) or $62.4$ lb/ft³ (imperial)

Problem-Solving Strategy

Sketch the tank and choose a coordinate system
Calculate volume of a representative water layer
Determine force using weight-density
Calculate lifting distance for the layer
Estimate work for the layer: Force × Distance
Sum work for all layers (Riemann sum)
Take limit and evaluate integral for exact work

Calculate work to pump water from a full inverted conical tank (height $12$ ft, base radius $4$ ft) until $4$ ft of water remains.

Set up coordinate system: $x$ -axis vertical, origin at top

Partition interval $[0, 8]$ ( $8$ ft of water pumped)

Use similar triangles to find radius at each layer:

$r_i = 4 - \frac{x_i^*}{3}$

Volume of layer: $V_i = \pi (4-\frac{x_i^*}{3})^2 \Delta x$

Force: $F_i \approx 62.4\pi (4-\frac{x_i^*}{3})^2 \Delta x$

Distance lifted: approximately $x_i^*$

Work integral:

$W = \int_0^8 62.4\pi x(4-\frac{x}{3})^2 dx$

Evaluate integral:

$W \approx 33,450$ ft-lb

A tank is in the shape of an inverted cone, with height $10$ ft and base radius $6$ ft. The tank is filled to a depth of $8$ ft to start with, and water is pumped over the upper edge of the tank until $3$ ft of water remain in the tank. How much work is required to pump out that amount of water?

Approximately $43,255.2$ ft-lb

Hydrostatic Force and Pressure

The Main Idea

Hydrostatic Pressure:
- Pressure exerted by a fluid at rest due to its weight
- $p = \rho s$ , where $\rho$ is weight density and $s$ is depth
Force on Horizontal Surface:
- $F = \rho As$ , where $A$ is area
Force on Vertical or Angled Surface:
- $F = \int_a^b \rho w(x)s(x)dx$
- $w(x)$ : width function
- $s(x)$ : depth function
Pascal’s Principle:
- Pressure at a given depth is the same in all directions

Problem-Solving Strategy

Sketch the situation and choose a coordinate system
Determine depth $s(x)$ and width $w(x)$ functions
Identify the weight density $\rho$ of the liquid
Set up and evaluate the force integral: $F = \int_a^b \rho w(x)s(x)dx$

Calculate the force on one end of a water trough $15$ ft long, with ends shaped like inverted isosceles triangles (base $8$ ft, height $3$ ft).

Set coordinate system: $x$ -axis vertical, $x = 0$ at top

Depth function: $s(x) = x$

Width function (similar triangles): $w(x) = 8 - \frac{8}{3}x$

Force integral:

$\begin{array}{rcl} F &=& \int_0^3 62.4(8 - \frac{8}{3}x)x dx \\ &=& 62.4 \int_0^3 (8x - \frac{8}{3}x^2) dx \\ &=& 62.4 [4x^2 - \frac{8}{9}x^3]_0^3 \\ &=& 748.8 \text{ lb} \end{array}$

When the reservoir is at its average level, the surface of the water is about 50 ft below where it would be if the reservoir were full. What is the force on the face of the dam under these circumstances?

Approximately $7,164,520,000$ lb or $3,582,260$ t