Physical Applications - Revision history

Khanh: /* Work and potential energy */

2022-01-15T23:14:00Z

Work and potential energy

Lila: /* Resources */

2021-10-29T16:59:05Z

Resources

Lila: /* Resources */

2021-10-15T19:07:28Z

Resources

Lila at 19:06, 15 October 2021

2021-10-15T19:06:53Z

Lila: /* Work by a spring */

2021-10-15T19:06:12Z

Work by a spring

Lila: /* Hydrostatic force on submerged surfaces */

2021-10-15T19:03:49Z

Hydrostatic force on submerged surfaces

Lila at 19:03, 15 October 2021

2021-10-15T19:03:07Z

Lila: /* Hydrostatic pressure */

2021-10-15T19:01:27Z

Hydrostatic pressure

Lila: /* Work by a spring */

2021-10-15T19:00:47Z

Work by a spring

Lila: /* Hydrostatic pressure */

2021-10-15T19:00:10Z

Hydrostatic pressure

@@ Line 28: / Line 28: @@
 ===Work and potential energy===
-The scalar product of a force '''F''' and the velocity '''v''' of its point of application defines the power input to a system at an instant of time.  Integration of this power over the trajectory of the point of application, {{nowrap|1=''C'' = '''x'''(''t'')}}, defines the work input to the system by the force.
+The scalar product of a force '''F''' and the velocity '''v''' of its point of application defines the power input to a system at an instant of time.  Integration of this power over the trajectory of the point of application, ''C'' = '''x'''(''t''), defines the work input to the system by the force.
 ====Path dependence====

← Older revision		Revision as of 16:59, 29 October 2021
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	* [https://youtu.be/kkq8ruV8_Jw Introduction to Pressure & Fluids] by The Organic Chemistry Tutor		* [https://youtu.be/kkq8ruV8_Jw Introduction to Pressure & Fluids] by The Organic Chemistry Tutor
	* [https://youtu.be/3jG-hWgUJko Hydrostatic Force Problems] by The Organic Chemistry Tutor		* [https://youtu.be/3jG-hWgUJko Hydrostatic Force Problems] by The Organic Chemistry Tutor
		+
		+	==Licensing==
		+	Content obtained and/or adapted from:
		+	* [https://en.wikipedia.org/wiki/Work_(physics) Work (physics), Wikipedia] under a CC BY-SA license
		+	* [https://en.wikipedia.org/wiki/Hydrostatics Hydrostatics, Wikipedia] under a CC BY-SA license
		+	* [https://en.wikipedia.org/wiki/Hooke%27s_law Hooke's Law, Wikipedia] under a CC BY-SA license

@@ Line 142: / Line 142: @@
 * [https://en.wikipedia.org/wiki/Work_(physics) Work (physics)], Wikipedia
 * [https://en.wikipedia.org/wiki/Hydrostatics Hydrostatics], Wikipedia
 <strong>Work Done by a Variable Force</strong>

@@ Line 47: / Line 47: @@
 In this case, the gradient of work yields
 :<math qid=Q11402> \nabla W =  -\nabla U= -\left(\frac{\partial U}{\partial x}, \frac{\partial U}{\partial y}, \frac{\partial U}{\partial z}\right) = \mathbf{F},</math>
-and the force '''F''' is said to be "derivable from a potential."<ref>J. R. Taylor, [https://books.google.com/books?id=P1kCtNr-pJsC&pg=PA117&lpg=PA117#v=onepage&q&f=false ''Classical Mechanics,''] University Science Books, 2005.</ref>
+and the force '''F''' is said to be "derivable from a potential."
 Because the potential ''U'' defines a force '''F''' at every point '''x''' in space, the set of forces is called  a force field.  The power applied to a body by a force field is obtained from the gradient of the work, or potential, in the direction of the velocity '''V''' of the body, that is
@@ Line 127: / Line 127: @@
 ===Hydrostatic force on submerged surfaces===
-The horizontal and vertical components of the hydrostatic force acting on a submerged surface are given by the following:<ref name=" F-M" />
+The horizontal and vertical components of the hydrostatic force acting on a submerged surface are given by the following:
 :<math>\begin{align} F_\mathrm{h} &= p_\mathrm{c}A \\ F_\mathrm{v} &= \rho g V \end{align}</math>

@@ Line 83: / Line 83: @@
 :<math> W=\int_0^t\mathbf{F}\cdot\mathbf{v}dt =-\int_0^tkx v_x dt = -\frac{1}{2}kx^2. </math>
 For convenience, consider contact with the spring occurs at ''t'' = 0, then the integral of the product of the distance ''x'' and the x-velocity, ''xv''<sub>x</sub>''dt'', over time ''t'' is (1/2)''x''<sup>2</sup>. The work is the product of the distance times the spring force, which is also dependent on distance; hence the ''x''<sup>2</sup> result.
 ===Work by a gas===

@@ Line 1: / Line 1: @@
-=Work=
+==Work==
 For moving objects, the quantity of work/time (power) is integrated along the trajectory of the point of application of the force. Thus, at any instant, the rate of the work done by a force (measured in joules/second, or '''watts''') is the scalar product of the force (a vector), and the velocity vector of the point of application. This scalar product of force and velocity is known as instantaneous power. Just as velocities may be integrated over time to obtain a total distance, by the fundamental theorem of calculus, the total work along a path is similarly the time-integral of instantaneous power applied along the trajectory of the point of application.
@@ Line 19: / Line 19: @@
 When a force component is perpendicular to the displacement of the object (such as when a body moves in a circular path under a central force), no work is done, since the cosine of 90° is zero. Thus, no work can be performed by gravity on a planet with a circular orbit (this is ideal, as all orbits are slightly elliptical). Also, no work is done on a body moving circularly at a constant speed while constrained by mechanical force, such as moving at constant speed in a frictionless ideal centrifuge.
-==Work done by a variable force==
+===Work done by a variable force===
 Calculating the work as "force times straight path segment" would only apply in the most simple of circumstances, as noted above. If force is changing, or if the body is moving along a curved path, possibly rotating and not necessarily rigid, then only the path of the application point of the force is relevant for the work done, and only the component of the force parallel to the application point velocity is doing work (positive work when in the same direction, and negative when in the opposite direction of the velocity). This component of force can be described by the scalar quantity called ''scalar tangential component'' ({{math|''F'' cos(''θ'')}}, where {{mvar|θ}} is the angle between the force and the velocity). And then the most general definition of work can be formulated as follows:
@@ Line 27: / Line 27: @@
 :<math> W = \int_{a}^{b} \mathbf{F(s)} \cdot d\mathbf{s}</math>
-==Work and potential energy==
+===Work and potential energy===
 The scalar product of a force '''F''' and the velocity '''v''' of its point of application defines the power input to a system at an instant of time.  Integration of this power over the trajectory of the point of application, {{nowrap|1=''C'' = '''x'''(''t'')}}, defines the work input to the system by the force.
-===Path dependence===
+====Path dependence====
 Therefore, the work done by a force '''F''' on an object that travels along a curve ''C'' is given by the line integral:
 :<math> W = \int_C \mathbf{F} \cdot d\mathbf{x} =  \int_{t_1}^{t_2}\mathbf{F}\cdot \mathbf{v}dt,</math>
@@ Line 38: / Line 38: @@
 :<math>\frac{dW}{dt} = P(t) = \mathbf{F}\cdot \mathbf{v} .</math>
-===Path independence===
+====Path independence====
 If the work for an applied force is independent of the path, then the work done by the force, by the gradient theorem, defines a potential function which is evaluated at the start and end of the trajectory of the point of application. This means that there is a potential function ''U''('''x'''), that can be evaluated at the two points '''x'''(''t''<sub>1</sub>) and '''x'''(''t''<sub>2</sub>) to obtain the work over any trajectory between these two points. It is tradition to define this function with a negative sign so that positive work is a reduction in the potential, that is
 :<math> W = \int_C \mathbf{F} \cdot \mathrm{d}\mathbf{x} =  \int_{\mathbf{x}(t_1)}^{\mathbf{x}(t_2)} \mathbf{F} \cdot \mathrm{d}\mathbf{x} =  U(\mathbf{x}(t_1))-U(\mathbf{x}(t_2)).
@@ Line 52: / Line 52: @@
 :<math qid=Q25342>P(t) = -\nabla U \cdot \mathbf{v} = \mathbf{F}\cdot\mathbf{v}.</math>
-==Work by gravity==
+===Work by gravity===
 [[File:Work of gravity F dot d equals mgh.JPG|right|thumb|Gravity ''F'' = ''mg'' does work ''W'' = ''mgh'' along any descending path]]
 In the absence of other forces, gravity results in a constant downward acceleration of every freely moving object. Near Earth's surface the acceleration due to gravity is ''g'' = 9.8&nbsp;m⋅s<sup>−2</sup> and the gravitational force on an object of mass ''m'' is ''F'''<sub>g</sub> = ''mg''. It is convenient to imagine this gravitational force concentrated at the center of mass of the object.
@@ Line 60: / Line 60: @@
 where ''F<sub>g</sub>'' is weight (pounds in imperial units, and newtons in SI units), and Δ''y'' is the change in height ''y''. Notice that the work done by gravity depends only on the vertical movement of the object. The presence of friction does not affect the work done on the object by its weight.
-==Work by gravity in space==
+===Work by gravity in space===
 The force of gravity exerted by a mass ''M'' on another mass ''m'' is given by
 :<math> \mathbf{F}=-\frac{GMm}{r^3}\mathbf{r},</math><!-- please do not replace r^3 with r^2: the vector r is not a unit vector.-->
@@ Line 77: / Line 77: @@
 is the gravitational potential function, also known as gravitational potential energy.  The negative sign follows the convention that work is gained from a loss of potential energy.
-==Work by a spring==
+===Work by a spring===
 [[File:Analogie ressorts contrainte.svg|upright|right|thumb|Forces in springs assembled in parallel]]
@@ Line 84: / Line 84: @@
 For convenience, consider contact with the spring occurs at ''t'' = 0, then the integral of the product of the distance ''x'' and the x-velocity, ''xv''<sub>x</sub>''dt'', over time ''t'' is (1/2)''x''<sup>2</sup>. The work is the product of the distance times the spring force, which is also dependent on distance; hence the ''x''<sup>2</sup> result.
-==Work by a gas==
+===Work by a gas===
 :<math> W=\int_a^b{P}dV  </math>
 Where ''P'' is pressure, ''V'' is volume, and ''a'' and ''b'' are initial and final volumes.
@@ Line 112: / Line 112: @@
 where {{mvar|H}} is the total height of the liquid column above the test area to the surface, and {{math|''p''<sub>atm</sub>}} is the atmospheric pressure, i.e., the pressure calculated from the remaining integral over the air column from the liquid surface to infinity. This can easily be visualized using a pressure prism.
 ==Resources==

@@ Line 92: / Line 92: @@
 In a fluid at rest, all frictional and inertial stresses vanish and the state of stress of the system is called ''hydrostatic''. When this condition of {{math|''V'' = 0}} is applied to the Navier–Stokes equations, the gradient of pressure becomes a function of body forces only. For a barotropic fluid in a conservative force field like a gravitational force field, the pressure exerted by a fluid at equilibrium becomes a function of force exerted by gravity.
-The hydrostatic pressure can be determined from a control volume analysis of an infinitesimally small cube of fluid. Since pressure is defined as the force exerted on a test area ({{math|''p'' = <math>\frac{F}{A}</math>, with {{mvar|p}}: pressure, {{mvar|F}}: force normal to area {{mvar|A}}, {{mvar|A}}: area), and the only force acting on any such small cube of fluid is the weight of the fluid column above it, hydrostatic pressure can be calculated according to the following formula:
+The hydrostatic pressure can be determined from a control volume analysis of an infinitesimally small cube of fluid. Since pressure is defined as the force exerted on a test area (<math>p = \frac{F}{A}</math>, with {{mvar|p}}: pressure, {{mvar|F}}: force normal to area {{mvar|A}}, {{mvar|A}}: area), and the only force acting on any such small cube of fluid is the weight of the fluid column above it, hydrostatic pressure can be calculated according to the following formula:
 :<math>p(z)-p(z_0)=\frac{1}{A}\int_{z_0}^z dz' \iint_A dx' dy'\, \rho (z') g(z') = \int_{z_0}^z dz'\, \rho (z') g(z') ,</math>