Difference between revisions of "Physical Applications"
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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 | 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> | :<math> \delta W = \mathbf{F}\cdot d\mathbf{s} = \mathbf{F}\cdot\mathbf{v}dt </math> | ||
− | where the | + | 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> | :<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''. | 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''. |
Revision as of 12:47, 15 October 2021
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 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,
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 F ⋅ ds = F cos θ ds, where θ is the angle between the force vector and the direction of movement,[1] 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.[2] 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:
Resources
Work Done by a Variable Force
- Work Done by a Variable Force by Krista King
- Work Done By a Variable Force Physics Problems by The Organic Chemistry Tutor
- Work Problems - Calculus by The Organic Chemistry Tutor
Work (Rope/Cable Problems)
- Ex 1: Integration Application - Work Lifting an Object by James Sousa, Math is Power 4U
- Ex 2: Integration Application - Work Lifting an Object and Cable by James Sousa, Math is Power 4U
- Ex: Find the Work Lifting a Leaking Bucket of Sand Given Mass by James Sousa, Math is Power 4U
- Ex: Find the Work Lifting a Leaking Bucket of Sand and Rope Given Mass by James Sousa, Math is Power 4U
- Finding Work using Calculus - The Cable/Rope Problem - Part b by patrickJMT
- Work done using a rope to lift a weight by patrickJMT
- Work Problems - Calculus by The Organic Chemistry Tutor
Work (Spring Problem)
- Ex: Find the Work Required to Stretch a Spring by James Sousa, Math is Power 4U
- Ex: Find the Force Required to Stretch a Spring by James Sousa, Math is Power 4U
- Work and Hooke's Law - Ex 1 by patrickJMT
- Work and Hooke's Law - Ex 2 by patrickJMT
- Work Done on Elastic Springs by Krista King
- Hooke's Law Physics, Basic Introduction, Restoring Force, Spring Constant by The Organic Chemistry Tutor
- Work Problems by The Organic Chemistry Tutor
Work (Pumping Fluid Out of a Tank)
- Ex: Determine the Work Required to Pump Water Out of a Circular Cylinder by James Sousa, Math is Power 4U
- Ex: Determine the Work Required to Pump Water Out of Trough (Isosceles Triangle) by James Sousa, Math is Power 4U
- Ex: Determine the Work Required to Pump Water Out of Trough (Quadratic Cross Section) by James Sousa, Math is Power 4U
- Calculating the Work Required to Drain a Tank by patrickJMT
- Work Problems - Calculus by The Organic Chemistry Tutor
Hydrostatic Pressure and Force
- Ex: Find the Hydrostatic Force on a Horizontal Plate (No Calculus) by James Sousa, Math is Power 4U
- Ex: Find the Hydrostatic Force on a Vertical Plane in the Shape of a Isosceles Triangle by James Sousa, Math is Power 4U
- Ex: Find the Hydrostatic Force on a Semicircle Window Submerged in Water by James Sousa, Math is Power 4U
- Ex: Find the Hydrostatic Force on a Dam in the Shape of a Degree 4 Polynomial by James Sousa, Math is Power 4U
- Hydrostatic Force - Basic Idea / Deriving the Formula by patrickJMT
- Hydrostatic Force - Complete Example #1 by patrickJMT
- Hydrostatic Force - Complete Example #2, Part 1 of 2 by patrickJMT
- Hydrostatic Force - Complete Example #2, Part 2 of 2 by patrickJMT
- Hydrostatic Pressure by Krista King
- Hydrostatic Force by Krista King
- Introduction to Pressure & Fluids by The Organic Chemistry Tutor
- Hydrostatic Force Problems by The Organic Chemistry Tutor
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