Difference between revisions of "Physical Applications"

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<strong>Work Done by a Variable Force</strong>
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==Work==
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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.
 +
 
 +
Work is the result of a force on a point that follows a curve '''X''', with a velocity '''v''', at each instant.  The small amount of work ''δW'' that occurs over an instant of time ''dt'' is calculated as
 +
:<math> \delta W = \mathbf{F}\cdot d\mathbf{s} = \mathbf{F}\cdot\mathbf{v}dt </math>
 +
where the '''F''' ⋅ '''v''' is the power over the instant ''dt''.  The sum of these small amounts of work over the trajectory of the point yields the work,
 +
:<math> W =  \int_{t_1}^{t_2}\mathbf{F} \cdot \mathbf{v}dt =  \int_{t_1}^{t_2}\mathbf{F} \cdot {\tfrac{d\mathbf{s}}{dt}}dt =\int_C \mathbf{F} \cdot d\mathbf{s},</math>
 +
where ''C'' is the trajectory from '''x'''(''t''<sub>1</sub>) to '''x'''(''t''<sub>2</sub>).  This integral is computed along the trajectory of the particle, and is therefore said to be ''path dependent''.
 +
 
 +
If the force is always directed along this line, and the magnitude of the force is ''F'', then this integral simplifies to
 +
:<math> W = \int_C F\,ds</math>
 +
where ''s'' is displacement along the line. If '''F''' is constant, in addition to being directed along the line, then the integral simplifies further to
 +
:<math> W = \int_C F\,ds = F\int_C ds = Fs</math>
 +
where ''s'' is the displacement of the point along the line.
 +
 
 +
This calculation can be generalized for a constant force that is not directed along the line, followed by the particle.  In this case the dot product '''F''' ⋅ ''d'''''s''' = ''F'' cos ''θ'' ''ds'', where ''θ'' is the angle between the force vector and the direction of movement, that is
 +
:<math>W = \int_C \mathbf{F} \cdot d\mathbf{s} = Fs\cos\theta.</math>
 +
 
 +
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===
 +
 
 +
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:
 +
:''Work of a force is the line integral of its scalar tangential component along the path of its application point.''
 +
:If the force varies (e.g. compressing a spring) we need to use calculus to find the work done.  If the force is given by {{math|''F''(''x'')}} (a function of {{mvar|x}}) then the work done by the force along the x-axis from {{mvar|a}} to {{mvar|b}} is:
 +
 
 +
:<math> W = \int_{a}^{b} \mathbf{F(s)} \cdot d\mathbf{s}</math>
 +
 
 +
===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, ''C'' = '''x'''(''t''), defines the work input to the system by the force.
 +
 
 +
====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>
 +
where ''dx''(''t'') defines the trajectory ''C'' and '''v''' is the velocity along this trajectory.  In general this integral requires the path along which the velocity is defined, so the evaluation of work is said to be path dependent.
  
[https://youtu.be/KRKWJre--6U Work Done by a Variable Force] by Krista King
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The time derivative of the integral for work yields the instantaneous power,
 +
:<math>\frac{dW}{dt} = P(t) = \mathbf{F}\cdot \mathbf{v} .</math>
  
[https://youtu.be/LbAxcMQ7J6c Work Done By a Variable Force Physics Problems, Force Displacement Graphs - Calculus] by The Organic Chemistry Tutor
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====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)).
 +
</math>
  
[https://youtu.be/TLw8xbmnY3c Work Problems - Calculus] by The Organic Chemistry Tutor
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The function ''U''('''x''') is called the potential energy associated with the applied force. The force derived from such a potential function is said to be conservative. Examples of forces that have potential energies are gravity and spring forces.
  
 +
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."
  
 +
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
 +
:<math qid=Q25342>P(t) = -\nabla U \cdot \mathbf{v} = \mathbf{F}\cdot\mathbf{v}.</math>
  
<strong>Work (Rope/Cable Problems)</strong>
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===Work by gravity===
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[[File:Work of gravity F dot d equals mgh.JPG|right|thumb|Gravity ''F'' = ''mg'' does work ''W'' = ''mgh'' along any descending path]]
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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.
  
[https://youtu.be/YMaR5KJj2zQ Ex 1: Integration Application - Work Lifting an Object] by James Sousa
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If an object with weight ''mg'' is displaced upwards or downwards a vertical distance ''y''<sub>2</sub> − ''y''<sub>1</sub>, the work ''W'' done on the object is:
 +
:<math>W = F_g (y_2 - y_1) = F_g\Delta y = mg\Delta y</math>
 +
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.
  
[https://youtu.be/pKZzfYIgq1Q Ex 2: Integration Application - Work Lifting an Object and Cable] by James Sousa
+
===Work by gravity in space===
 +
The force of gravity exerted by a mass ''M'' on another mass ''m'' is given by
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:<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.-->
 +
where '''r''' is the position vector from ''M'' to ''m''.
  
[https://youtu.be/XEwdZqz93z4 Ex: Find the Work Lifting a Leaking Bucket of Sand and Rope Given Mass] by James Sousa
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Let the mass ''m'' move at the velocity '''v'''; then the work of gravity on this mass as it moves from position '''r'''(''t''<sub>1</sub>) to  '''r'''(''t''<sub>2</sub>) is given by
 +
:<math> W=-\int^{\mathbf{r}(t_2)}_{\mathbf{r}(t_1)}\frac{GMm}{r^3}\mathbf{r}\cdot d\mathbf{r}=-\int^{t_2}_{t_1}\frac{GMm}{r^3}\mathbf{r}\cdot\mathbf{v}dt.</math>
 +
Notice that the position and velocity of the mass ''m'' are given by
 +
:<math> \mathbf{r} = r\mathbf{e}_r, \qquad\mathbf{v} =  \frac{d\mathbf{r}}{dt} = \dot{r}\mathbf{e}_r + r\dot{\theta}\mathbf{e}_t,</math>
 +
where '''e'''<sub>''r''</sub> and '''e'''<sub>''t''</sub> are the radial and tangential unit vectors directed relative to the vector from ''M'' to ''m'', and we use the fact that <math>  d \mathbf{e}_r/dt  = \dot{\theta}\mathbf{e}_t. </math>  Use this to simplify the formula for work of gravity to,
 +
:<math> W=-\int^{t_2}_{t_1}\frac{GmM}{r^3}(r\mathbf{e}_r)\cdot(\dot{r}\mathbf{e}_r + r\dot{\theta}\mathbf{e}_t)dt = -\int^{t_2}_{t_1}\frac{GmM}{r^3}r\dot{r}dt = \frac{GMm}{r(t_2)}-\frac{GMm}{r(t_1)}.</math>
 +
This calculation uses the fact that
 +
:<math> \frac{d}{dt}r^{-1}=-r^{-2}\dot{r}=-\frac{\dot{r}}{r^2}.</math>
 +
The function
 +
:<math> U=-\frac{GMm}{r}, </math>
 +
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.
  
[https://youtu.be/oWcN96kcbkM Finding Work using Calculus - The Cable/Rope Problem - Part b] by patrickJMT
+
===Work by a spring===
 +
[[File:Analogie ressorts contrainte.svg|upright|right|thumb|Forces in springs assembled in parallel]]
  
[https://youtu.be/UJF_34TSAwQ Work done using a rope to lift a weight] by patrickJMT
+
Consider a spring that exerts a horizontal force '''F''' = (−''kx'', 0, 0) that is proportional to its deflection in the ''x'' direction independent of how a body moves.  The work of this spring on a body moving along the space with the curve '''X'''(''t'') = (''x''(''t''), ''y''(''t''), ''z''(''t'')), is calculated using its velocity, '''v''' = (''v''<sub>x</sub>, ''v''<sub>y</sub>, ''v''<sub>z</sub>), to obtain
 +
:<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.
  
[https://youtu.be/TLw8xbmnY3c Work Problems - Calculus] by The Organic Chemistry Tutor
+
====Hooke's law for linear springs====
 +
Consider a simple helical spring that has one end attached to some fixed object, while the free end is being pulled by a force whose magnitude is {{mvar|F<sub>s</sub>}}. Suppose that the spring has reached a state of equilibrium, where its length is not changing anymore. Let {{mvar|x}} be the amount by which the free end of the spring was displaced from its "relaxed" position (when it is not being stretched). Hooke's law states that
 +
:<math>F_s = kx</math>
  
 +
or, equivalently,
 +
:<math>x = \frac{F_s}{k}</math>
  
 +
where {{mvar|k}} is a positive real number, characteristic of the spring. Moreover, the same formula holds when the spring is compressed, with {{mvar|F<sub>s</sub>}} and {{mvar|x}} both negative in that case. According to this formula, the graph of the applied force {{mvar|F<sub>s</sub>}} as a function of the displacement {{mvar|x}} will be a straight line passing through the origin, whose slope is {{mvar|k}}.
  
 +
Hooke's law for a spring is sometimes, but rarely, stated under the convention that {{mvar|F<sub>s</sub>}} is the restoring force exerted by the spring on whatever is pulling its free end. In that case, the equation becomes
 +
:<math>F_s = -kx</math>
 +
since the direction of the restoring force is opposite to that of the displacement.
  
<strong>Work (Spring Problem)</strong>
+
===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.
  
[https://youtu.be/Fd65ZxqpktE Ex: Find the Force Required to Stretch a Spring (Integration App)] by James Sousa
+
==Hydrostatic pressure==
  
[https://youtu.be/x_0YWeHXZFE Work and Hooke's Law - Ex 1] by patrickJMT
+
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.
  
[https://youtu.be/_YXA8JeGLjo Work and Hooke's Law - Ex 2] by patrickJMT
+
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:
  
[https://youtu.be/ZKfYUxzlLx8 Work Done on Elastic Springs] by Krista King
+
:<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>
  
[https://youtu.be/D5vnqohxEOI Hooke's Law Physics, Basic Introduction, Restoring Force, Spring Constant, Practice Problems] by The Organic Chemistry Tutor
+
where:
[https://youtu.be/TLw8xbmnY3c Work Problems - Calculus] by The Organic Chemistry Tutor
+
* {{mvar|p}} is the hydrostatic pressure (Pa),
 +
* {{mvar|ρ}} is the fluid density (kg/m<sup>3</sup>),
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* {{mvar|g}} is gravitational acceleration (m/s<sup>2</sup>),
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* {{mvar|A}} is the test area (m<sup>2</sup>),
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* {{mvar|z}} is the height (parallel to the direction of gravity) of the test area (m),
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* {{math|''z''<sub>0</sub>}} is the height of the zero reference point of the pressure (m).
  
 +
For water and other liquids, this integral can be simplified significantly for many practical applications, based on the following two assumptions: Since many liquids can be considered incompressible, a reasonable good estimation can be made from assuming a constant density throughout the liquid. (The same assumption cannot be made within a gaseous environment.) Also, since the height {{mvar|h}} of the fluid column between {{mvar|z}} and {{math|''z''<sub>0</sub>}} is often reasonably small compared to the radius of the Earth, one can neglect the variation of {{mvar|g}}. Under these circumstances, the integral is simplified into the formula:
 +
:<math>p - p_0 = \rho g h,</math>
  
<strong>Work (Pumping Fluid Out of a Tank)</strong>
+
where {{mvar|h}} is the height {{math|''z'' − ''z''<sub>0</sub>}} of the liquid column between the test volume and the zero reference point of the pressure. This formula is often called Stevin's law. Note that this reference point should lie at or below the surface of the liquid. Otherwise, one has to split the integral into two (or more) terms with the constant {{math|''ρ''<sub>liquid</sub>}} and {{math|''ρ''(''z''′)<sub>above</sub>}}. For example, the absolute pressure compared to vacuum is:
  
[https://youtu.be/1x-xMf4TFNM Ex: Determine the Work Required to Pump Water Out of Trough (Isosceles Triangle)] by James Sousa
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:<math>p = \rho g H + p_\mathrm{atm},</math>
  
[https://youtu.be/1x-xMf4TFNM Ex: Determine the Work Required to Pump Water Out of a Circular Cylinder] by James Sousa
+
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.
  
[https://youtu.be/fJtxJv5sdqo Calculating the Work Required to Drain a Tank] by patrickJMT
+
===Hydrostatic force on submerged surfaces===
 +
The horizontal and vertical components of the hydrostatic force acting on a submerged surface are given by the following:
  
[https://youtu.be/TLw8xbmnY3c Work Problems - Calculus] by The Organic Chemistry Tutor
+
:<math>\begin{align} F_\mathrm{h} &= p_\mathrm{c}A \\ F_\mathrm{v} &= \rho g V \end{align}</math>
  
 +
where:
 +
*{{math|''p''<sub>c</sub>}} is the pressure at the centroid of the vertical projection of the submerged surface
 +
*{{mvar|A}} is the area of the same vertical projection of the surface
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*{{mvar|ρ}} is the density of the fluid
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*{{mvar|g}} is the acceleration due to gravity
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*{{mvar|V}} is the volume of fluid directly above the curved surface
  
<strong>Hydrostatic Pressure and Force</strong>
 
  
[https://youtu.be/zSyh3IzQMdc Ex: Find the Hydrostatic Force on a Horizontal Plate (No Calculus)] by James Sousa
+
==Resources==
 +
* [https://en.wikipedia.org/wiki/Work_(physics) Work (physics)], Wikipedia
 +
* [https://en.wikipedia.org/wiki/Hydrostatics Hydrostatics], Wikipedia
 +
* [https://en.wikipedia.org/wiki/Hooke%27s_law Hooke's Law], Wikipedia
  
[https://youtu.be/Ntk9RpCCSeM Ex: Find the Hydrostatic Force on a Semicircle Window Submerged in Water] by James Sousa
+
<strong>Work Done by a Variable Force</strong>
 +
* [https://youtu.be/KRKWJre--6U Work Done by a Variable Force] by Krista King
 +
* [https://youtu.be/LbAxcMQ7J6c Work Done By a Variable Force Physics Problems] by The Organic Chemistry Tutor
 +
* [https://youtu.be/TLw8xbmnY3c Work Problems - Calculus] by The Organic Chemistry Tutor
  
[https://youtu.be/8i7xOtAo6ws Ex: Find the Hydrostatic Force on a Dam in the Shape of a Degree 4 Polynomial] by James Sousa
 
  
[https://youtu.be/12MhraQo0TY Hydrostatic Force - Basic Idea / Deriving the Formula] by patrickJMT
+
<strong>Work (Rope/Cable Problems)</strong>
 +
* [https://youtu.be/YMaR5KJj2zQ Ex 1: Integration Application - Work Lifting an Object] by James Sousa, Math is Power 4U
 +
* [https://youtu.be/pKZzfYIgq1Q Ex 2: Integration Application - Work Lifting an Object and Cable] by James Sousa, Math is Power 4U
 +
* [https://youtu.be/XEwdZqz93z4 Ex: Find the Work Lifting a Leaking Bucket of Sand Given Mass] by James Sousa, Math is Power 4U
 +
* [https://youtu.be/XEwdZqz93z4 Ex: Find the Work Lifting a Leaking Bucket of Sand and Rope Given Mass] by James Sousa, Math is Power 4U
 +
* [https://youtu.be/oWcN96kcbkM Finding Work using Calculus - The Cable/Rope Problem - Part b] by patrickJMT
 +
* [https://youtu.be/UJF_34TSAwQ Work done using a rope to lift a weight] by patrickJMT
 +
* [https://youtu.be/TLw8xbmnY3c Work Problems - Calculus] by The Organic Chemistry Tutor
  
[https://youtu.be/cXNmCaTod58 Hydrostatic Force - Complete Example #1] by patrickJMT
 
  
[https://youtu.be/h8kMaW2q9EM Hydrostatic Force - Complete Example #2, Part 1 of 2] by patrickJMT
+
<strong>Work (Spring Problem)</strong>
 +
* [https://youtu.be/3_-7IJO6EVc Ex: Find the Work Required to Stretch a Spring] by James Sousa, Math is Power 4U
 +
* [https://youtu.be/Fd65ZxqpktE Ex: Find the Force Required to Stretch a Spring] by James Sousa, Math is Power 4U
 +
* [https://youtu.be/x_0YWeHXZFE Work and Hooke's Law - Ex 1] by patrickJMT
 +
* [https://youtu.be/_YXA8JeGLjo Work and Hooke's Law - Ex 2] by patrickJMT
 +
* [https://youtu.be/ZKfYUxzlLx8 Work Done on Elastic Springs] by Krista King
 +
* [https://youtu.be/D5vnqohxEOI Hooke's Law Physics, Basic Introduction, Restoring Force, Spring Constant] by The Organic Chemistry Tutor
 +
* [https://youtu.be/TLw8xbmnY3c Work Problems] by The Organic Chemistry Tutor
  
[https://youtu.be/F2poHPZZBhE Hydrostatic Force - Complete Example #2, Part 2 of 2]  by patrickJMT
 
  
[https://youtu.be/xTDA3H6-FAY Hydrostatic pressure] by Krista King
+
<strong>Work (Pumping Fluid Out of a Tank)</strong>
 +
* [https://youtu.be/1x-xMf4TFNM Ex: Determine the Work Required to Pump Water Out of a Circular Cylinder] by James Sousa, Math is Power 4U
 +
* [https://youtu.be/1x-xMf4TFNM Ex: Determine the Work Required to Pump Water Out of Trough (Isosceles Triangle)] by James Sousa, Math is Power 4U
 +
* [https://youtu.be/RPXnZl3przM Ex: Determine the Work Required to Pump Water Out of Trough (Quadratic Cross Section)] by James Sousa, Math is Power 4U
 +
* [https://youtu.be/fJtxJv5sdqo Calculating the Work Required to Drain a Tank] by patrickJMT
 +
* [https://youtu.be/TLw8xbmnY3c Work Problems - Calculus] by The Organic Chemistry Tutor
  
[https://youtu.be/FgwWPlYiZgY Hydrostatic force] by Krista King
 
  
[https://youtu.be/kkq8ruV8_Jw Introduction to Pressure & Fluids - Physics Practice Problems] by The Organic Chemistry Tutor
+
<strong>Hydrostatic Pressure and Force</strong>
 +
* [https://youtu.be/zSyh3IzQMdc Ex: Find the Hydrostatic Force on a Horizontal Plate (No Calculus)] by James Sousa, Math is Power 4U
 +
* [https://youtu.be/7exmlrEbNXI Ex: Find the Hydrostatic Force on a Vertical Plane in the Shape of a Isosceles Triangle] by James Sousa, Math is Power 4U
 +
* [https://youtu.be/Ntk9RpCCSeM Ex: Find the Hydrostatic Force on a Semicircle Window Submerged in Water] by James Sousa, Math is Power 4U
 +
* [https://youtu.be/8i7xOtAo6ws Ex: Find the Hydrostatic Force on a Dam in the Shape of a Degree 4 Polynomial] by James Sousa, Math is Power 4U
 +
* [https://youtu.be/12MhraQo0TY Hydrostatic Force - Basic Idea / Deriving the Formula] by patrickJMT
 +
* [https://youtu.be/cXNmCaTod58 Hydrostatic Force - Complete Example #1] by patrickJMT
 +
* [https://youtu.be/h8kMaW2q9EM Hydrostatic Force - Complete Example #2, Part 1 of 2] by patrickJMT
 +
* [https://youtu.be/F2poHPZZBhE Hydrostatic Force - Complete Example #2, Part 2 of 2] by patrickJMT
 +
* [https://youtu.be/xTDA3H6-FAY Hydrostatic Pressure] by Krista King
 +
* [https://youtu.be/FgwWPlYiZgY Hydrostatic Force] by Krista King
 +
* [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 - Calculus 2] 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

Latest revision as of 17:14, 15 January 2022

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.

Work is the result of a force on a point that follows a curve X, with a velocity v, at each instant. The small amount of work δW that occurs over an instant of time dt is calculated as

where the Fv is the power over the instant dt. The sum of these small amounts of work over the trajectory of the point yields the work,

where C is the trajectory from x(t1) to x(t2). This integral is computed along the trajectory of the particle, and is therefore said to be path dependent.

If the force is always directed along this line, and the magnitude of the force is F, then this integral simplifies to

where s is displacement along the line. If F is constant, in addition to being directed along the line, then the integral simplifies further to

where s is the displacement of the point along the line.

This calculation can be generalized for a constant force that is not directed along the line, followed by the particle. In this case the dot product Fds = F cos θ ds, where θ is the angle between the force vector and the direction of movement, that is

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

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 (F cos(θ), where θ is the angle between the force and the velocity). And then the most general definition of work can be formulated as follows:

Work of a force is the line integral of its scalar tangential component along the path of its application point.
If the force varies (e.g. compressing a spring) we need to use calculus to find the work done. If the force is given by F(x) (a function of x) then the work done by the force along the x-axis from a to b is:

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, C = x(t), defines the work input to the system by the force.

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:

where dx(t) defines the trajectory C and v is the velocity along this trajectory. In general this integral requires the path along which the velocity is defined, so the evaluation of work is said to be path dependent.

The time derivative of the integral for work yields the instantaneous power,

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(t1) and x(t2) 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

The function U(x) is called the potential energy associated with the applied force. The force derived from such a potential function is said to be conservative. Examples of forces that have potential energies are gravity and spring forces.

In this case, the gradient of work yields

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

Work by gravity

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 m⋅s−2 and the gravitational force on an object of mass m is F'g = mg. It is convenient to imagine this gravitational force concentrated at the center of mass of the object.

If an object with weight mg is displaced upwards or downwards a vertical distance y2y1, the work W done on the object is:

where Fg is weight (pounds in imperial units, and newtons in SI units), and Δy 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

The force of gravity exerted by a mass M on another mass m is given by

where r is the position vector from M to m.

Let the mass m move at the velocity v; then the work of gravity on this mass as it moves from position r(t1) to r(t2) is given by

Notice that the position and velocity of the mass m are given by

where er and et are the radial and tangential unit vectors directed relative to the vector from M to m, and we use the fact that Use this to simplify the formula for work of gravity to,

This calculation uses the fact that

The function

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

Forces in springs assembled in parallel

Consider a spring that exerts a horizontal force F = (−kx, 0, 0) that is proportional to its deflection in the x direction independent of how a body moves. The work of this spring on a body moving along the space with the curve X(t) = (x(t), y(t), z(t)), is calculated using its velocity, v = (vx, vy, vz), to obtain

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, xvxdt, over time t is (1/2)x2. The work is the product of the distance times the spring force, which is also dependent on distance; hence the x2 result.

Hooke's law for linear springs

Consider a simple helical spring that has one end attached to some fixed object, while the free end is being pulled by a force whose magnitude is Fs. Suppose that the spring has reached a state of equilibrium, where its length is not changing anymore. Let x be the amount by which the free end of the spring was displaced from its "relaxed" position (when it is not being stretched). Hooke's law states that

or, equivalently,

where k is a positive real number, characteristic of the spring. Moreover, the same formula holds when the spring is compressed, with Fs and x both negative in that case. According to this formula, the graph of the applied force Fs as a function of the displacement x will be a straight line passing through the origin, whose slope is k.

Hooke's law for a spring is sometimes, but rarely, stated under the convention that Fs is the restoring force exerted by the spring on whatever is pulling its free end. In that case, the equation becomes

since the direction of the restoring force is opposite to that of the displacement.

Work by a gas

Where P is pressure, V is volume, and a and b are initial and final volumes.

Hydrostatic pressure

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 {{{1}}} 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 (, with p: pressure, F: force normal to area A, 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:

where:

  • p is the hydrostatic pressure (Pa),
  • ρ is the fluid density (kg/m3),
  • g is gravitational acceleration (m/s2),
  • A is the test area (m2),
  • z is the height (parallel to the direction of gravity) of the test area (m),
  • z0 is the height of the zero reference point of the pressure (m).

For water and other liquids, this integral can be simplified significantly for many practical applications, based on the following two assumptions: Since many liquids can be considered incompressible, a reasonable good estimation can be made from assuming a constant density throughout the liquid. (The same assumption cannot be made within a gaseous environment.) Also, since the height h of the fluid column between z and z0 is often reasonably small compared to the radius of the Earth, one can neglect the variation of g. Under these circumstances, the integral is simplified into the formula:

where h is the height zz0 of the liquid column between the test volume and the zero reference point of the pressure. This formula is often called Stevin's law. Note that this reference point should lie at or below the surface of the liquid. Otherwise, one has to split the integral into two (or more) terms with the constant ρliquid and ρ(z′)above. For example, the absolute pressure compared to vacuum is:

where H is the total height of the liquid column above the test area to the surface, and patm 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.

Hydrostatic force on submerged surfaces

The horizontal and vertical components of the hydrostatic force acting on a submerged surface are given by the following:

where:

  • pc is the pressure at the centroid of the vertical projection of the submerged surface
  • A is the area of the same vertical projection of the surface
  • ρ is the density of the fluid
  • g is the acceleration due to gravity
  • V is the volume of fluid directly above the curved surface


Resources

Work Done by a Variable Force


Work (Rope/Cable Problems)


Work (Spring Problem)


Work (Pumping Fluid Out of a Tank)


Hydrostatic Pressure and Force

Licensing

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