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| {{Other uses}}
| | Airconditioning and Refrigeration Mechanic Andy Sergeant from Maple Ridge, likes to spend time wargames, swiss watches replica and compose music. Always enjoys taking a trip to places such as Rice Terraces of the Philippine Cordilleras.<br><br>my website :: http://mdmototrials.com/ ([http://event-ology.net/profile/zaiwn http://event-ology.net]) |
| In [[physics]], '''net force''' is ''the overall force'' acting on an object. In order to calculate the net force, the body is isolated and interactions with the [[Environment (systems)|environment]] or constraints are introduced as [[forces]] and torques forming a [[free-body diagram]].
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| The net force does not have the same effect on the movement of the object as the original system forces, unless the point of application of the net force and an associated torque are determined so that they form the [[resultant force]] and torque. It is always possible to determine the torque associated with a point of application of a net force so that it maintains the movement of the object under the original system of forces.
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| With its associated torque, the net force becomes the ''[[resultant force]]'' and has the same effect on the [[rotation]]al motion of the object as all actual forces taken together.<ref>Symon, Keith R. (1964), Mechanics, Addison-Wesley, ISBN 60-5164</ref> It is possible for a system of forces to define a torque-free resultant force. In this case, the net force when applied at the proper line of action has the same effect on the body as all of the forces at their points of application. It is not always possible to find a torque-free resultant force.
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| <!--This discussion of forces on a particle and rigid body is confusing--A force acting on an object may cause changes in the motion or in the shape (configuration) of the object. When two or more forces are acting on an object, the concepts of net force and resultant force are intended to simplify description of their effect on its motion.
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| If the forces are acting on a [[particle]] (the size of the object is so small that it can be approximated by a point), they can only change its [[velocity]]. In that case, there is no difference between the net force and the resultant force because no rotation is associated with such objects.
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| If the object is an extended but [[rigid body]] (no change in shape), the forces can change its velocity (i.e. the velocity of its [[center of mass]], usually called its linear velocity) as well as its [[angular velocity]]. In that case, it may be useful to distinguish the resultant force from the net force. And even in the case of non-rigid objects (deformable bodies or systems), the concepts of net and resultant force are equally applicable to description of their overall motion.
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| Some authors use "resultant force" and "net force" as synonyms, even when the forces act on extended bodies. However, this is generally not the case (see the note on usage at the end of the article) in [[mechanics]] and in those technical applications where full understanding and actual calculations of the rotating body [[dynamics (mechanics)|dynamics]] (or even of the [[static equilibrium]]) are required.
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| ==Total force==
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| The sum of forces acting on a particle is called the total force or the net force. The net force is a single force that replaces the effect of the original forces on the particle's motion. It gives the particle the same [[acceleration]] as all those actual forces together as described by the [[Newton's laws of motion|Newton's second law of motion]].
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| [[Image:Addition of forces.JPG|thumb|350px|Another method for diagramming addition of forces]]
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| Force is a [[Euclidean vector|vector]] quantity, which means that it has a magnitude and a direction, and it is usually denoted using boldface such as '''F''' or by using an arrow over the symbol, such as <math>\scriptstyle \vec F</math>.
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| Graphically a force is represented as line segment from its point of application ''A'' to a point ''B'' which defines its direction and magnitude. The length of the segment ''AB'' represents the magnitude of the force.
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| [[Vector calculus]] was developed in the late 1800s and early 1900s, however, the [[parallelogram rule]] for addition of forces is said to date from the ancient times, and it is explicitly noted by Galileo and Newton.<ref>Michael J. Crowe (1967). A History of Vector Analysis : The Evolution of the Idea of a Vectorial System. Dover Publications; Reprint edition. ISBN 0-486-67910-1</ref>
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| The diagram shows the addition of the forces <math>\scriptstyle \vec{F}_{1}</math> and <math>\scriptstyle \vec{F}_{2}</math>. The sum <math>\scriptstyle \vec F</math> of the two forces is drawn as the diagonal of a parallelogram defined by the two forces.
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| <!---This intuitive description is not intuitive---This can be grasped intuitively: if the total force <math>\scriptstyle \vec F</math> should describe the joint effect of the two forces on a particle (which is the intrinsic meaning of addition), its direction should be closer to the direction of the stronger force <math>\scriptstyle \vec{F}_{2}</math>, and its amount should be greater than the amount of <math>\scriptstyle \vec{F}_{2}</math> because <math>\scriptstyle \vec{F}_{1}</math> also helps in pulling the particle in that "general" direction (for the forces shown in the diagram).
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| Independent of this approximate intuitive judgment, the rule of parallelogram gives the exact result, which is easily verified by measuring the effects of the forces. The result can be approximately evaluated from the diagram or, based on the diagram, precisely calculated using elementary [[trigonometry]].
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| <!---This is not an alternative to the parallelogram rule, it is the parallelogram rule---Instead of using the parallelogram rule, the same result can be obtained by a simpler procedure (shown on the right side of the diagram). The line segments representing the original forces can be translated (in any order) so that one begins where the other ends. The same result for the vector sum is the line drawn from the beginning of the first segment to the end of the second – or to the end of the last one – which enables simple addition of more than two vectors. At the bottom of the diagram, this procedure is applied to the addition of two parallel and antiparallel forces, leading to the intuitively expected result: for parallel forces the amounts add up, whereas for the forces in opposite directions (antiparallel) the amount of the smaller force is subtracted from the bigger one.
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| Forces applied to an extended body can have different points of application. Forces are bound vectors and can be added only if they are applied at the same point. The net force obtained from all the forces acting on a body will not preserve its motion unless they are applied at the same point and the appropriate torque associated with the new point of application is determined. The net force on a body applied at a single point with the appropriate torque is known as the [[resultant force]] and torque.
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| ==Parallelogram rule for the addition of forces==
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| [[File:Parallelogram1.svg|right|Parallelogram ABCD]]
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| A force is known as a bound vector which means it has a direction and magnitude and a point of application. A convenient way to define a force is by a line segment from a point ''A'' to a point ''B''. If we denote the coordinates of these points as '''A'''=(A<sub>x</sub>, A<sub>y</sub>, A<sub>z</sub>) and '''B'''=(B<sub>x</sub>, B<sub>y</sub>, B<sub>z</sub>), then the force vector applied at ''A'' is given by
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| :<math> \mathbf{F}= \mathbf{B}-\mathbf{A} = (B_x-A_x, B_y-A_y, B_z-A_z). </math>
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| The length of the vector '''B'''-'''A''' defines the magnitude of '''F''', and is given by
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| :<math> |\mathbf{F}| = \sqrt{(B_x-A_x)^2+(B_y-A_y)^2+(B_z-A_z)^2}. </math>
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| The sum of two forces '''F'''<sub>1</sub> and '''F'''<sub>2</sub> applied at ''A'' can be computed from the sum of the segments that define them. Let '''F'''<sub>1</sub>='''B'''-'''A''' and '''F'''<sub>2</sub>='''D'''-'''A''', then the sum of these two vectors is
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| :<math> \mathbf{F}=\mathbf{F}_1+\mathbf{F}_2 = \mathbf{B}-\mathbf{A} + \mathbf{D}-\mathbf{A},</math>
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| which can be written as
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| :<math> \mathbf{F}=\mathbf{F}_1+\mathbf{F}_2 = 2(\frac{\mathbf{B}+\mathbf{D}}{2}-\mathbf{A})=2(\mathbf{E}-\mathbf{A}),</math>
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| where '''E''' is the midpoint of the segment '''BD''' that joins the points ''B'' and ''D''.
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| Thus, the sum of the forces '''F'''<sub>1</sub> and '''F'''<sub>2</sub> is twice the segment joining ''A'' to the midpoint ''E'' of the segment joining the endpoints ''B'' and ''D'' of the two forces. The doubling of this length is easily achieved by defining a segments '''BC''' and '''DC''' parallel to '''AD''' and '''AB''', respectively, to complete the parallelogram ''ABCD''. The diagonal '''AC''' of this parallelogram is the sum of the two force vectors. This is known as the parallelogram rule for the addition of forces.
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| ==Translation and rotation due to a force==
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| ===Point forces===
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| When a force acts on a particle, it is applied to a single point (the particle volume is negligible): this is a point force and the particle is its application point. But an external force on an extended body (object) can be applied to a number of its constituent particles, i.e. can be "spread" over some volume or surface of the body. However, in order to determine its rotational effect on the body, it is necessary to specify its point of application (actually, the line of application, as explained below). The problem is usually resolved in the following ways:
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| *Often the volume or surface on which the force acts is relatively small compared to the size of the body, so that it can be approximated by a point. It is usually not difficult to determine whether the error caused by such approximation is acceptable.
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| *If it is not acceptable (obviously e.g. in the case of gravitational force), such "volume/surface" force should be described as a system of forces (components), each acting on a single particle, and then the calculation should be done for each of them separately. Such a calculation is typically simplified by the use of differential elements of the body volume/surface, and the integral calculus. In a number of cases, though, it can be shown that such a system of forces may be replaced by a single point force without the actual calculation (as in the case of uniform gravitational force).
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| In any case, the analysis of the rigid body motion begins with the point force model. And when a force acting on a body is shown graphically, the oriented line segment representing the force is usually drawn so as to "begin" (or "end") at the application point.
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| ===Rigid bodies===
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| [[File:Free body acceleration.JPG|thumb|300px|How a force accelerates a body]]
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| In the example shown on the diagram, a single force <math>\scriptstyle \vec F </math> acts at the application point '''H''' on a free rigid body. The body has the mass <math>\scriptstyle m </math> and its center of mass is the point '''C'''. In the constant mass approximation, the force causes changes in the body motion described by the following expressions:
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| :<math> \vec a = {\vec F \over m} </math> is the center of mass acceleration; and | |
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| :<math> \vec \alpha = {\vec \tau \over I} </math> is the [[angular acceleration]] of the body. | |
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| In the second expression, <math>\scriptstyle \vec \tau </math> is the [[torque]] or moment of force, whereas <math>\scriptstyle I </math> is the [[moment of inertia]] of the body. A torque caused by a force <math>\scriptstyle \vec F </math> is a vector quantity defined with respect to some reference point:
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| :<math> \vec \tau = \vec r \times \vec F </math> is the torque vector, and
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| :<math> \ \tau = Fk </math> is the amount of torque.
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| The vector <math>\scriptstyle \vec r </math> is the [[position vector]] of the force application point, and in this example it is drawn from the center of mass as the reference point (see diagram). The straight line segment <math>\scriptstyle k </math> is the lever arm of the force <math>\scriptstyle \vec F </math> with respect to the center of mass. As the illustration suggests, the torque does not change (the same lever arm) if the application point is moved along the line of the application of the force (dotted black line). More formally, this follows from the properties of the vector product, and shows that rotational effect of the force depends only on the position of its line of application, and not on the particular choice of the point of application along that line.
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| The torque vector is perpendicular to the plane defined by the force and the vector <math>\scriptstyle \vec r </math>, and in this example it is directed towards the observer; the angular acceleration vector has the same direction. The [[right hand rule]] relates this direction to the clockwise or counter-clockwise rotation in the plane of the drawing.
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| The moment of inertia <math>\scriptstyle I </math> is calculated with respect to the axis through the center of mass that is parallel with the torque. If the body shown in the illustration is a homogenous disc, this moment of inertia is <math>\scriptstyle I=m r^2 /2 </math> . If the disc has the mass 0,5 kg and the radius 0,8 m, the moment of inertia is 0,16 kgm<sup>2</sup>. If the amount of force is 2 N, and the lever arm 0,6 m, the amount of torque is 1,2 Nm. At the instant shown, the force gives to the disc the angular acceleration α = {{math|''τ''}}/I = 7,5 rad/s<sup>2</sup>, and to its center of mass it gives the linear acceleration a = F/m = 4 m/s<sup>2</sup>.
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| ==Resultant force==
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| [[File:Rezultanta.JPG|thumb|500px|Graphical placing of the resultant force]]
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| [[Resultant force]] and torque replaces the effects of a system of forces acting on the movement of a rigid body. An interesting special case is a torque-free resultant which can be found as follows:
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| #First, vector addition is used to find the net force;
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| #Then use the equation to determine the point of application with zero torque:
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| :<math> \vec r \times \vec F_R = \sum_{i=1}^N ( \vec r_i \times \vec F_i ) </math>
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| where <math>\scriptstyle \vec F_R </math> is the net force, <math>\scriptstyle \vec r</math> locates its application point, and individual forces are <math>\scriptstyle \vec F_i </math> with application points <math>\scriptstyle \vec r_i </math>. It may be that there is no point of application that yields a torque-free resultant.
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| <!--This paragraph confused a resultant force and torque with a torque-free resultant---
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| The above equation may have no solution for <math>\scriptstyle \vec r</math>. In that case there is no resultant force, i.e. no single force can replace all actual forces regarding both linear and angular acceleration of the body. And even when <math>\scriptstyle \vec r</math> can be calculated, it is not unique, because the point of application can move along the line of application without affecting the torque.
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| The diagram illustrates simple graphical methods for finding the line of application of the resultant force of simple planar systems.
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| #Lines of application of the actual forces <math>\scriptstyle \vec{F}_{1}</math> and <math>\scriptstyle \vec{F}_{2}</math> on the leftmost illustration intersect. After vector addition is performed "at the location of <math>\scriptstyle \vec{F}_{1}</math>", the net force obtained is translated so that its line of application passes through the common intersection point. With respect to that point all torques are zero, so the torque of the resultant force <math>\scriptstyle \vec{F}_{R}</math> is equal to the sum of the torques of the actual forces.
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| #Illustration in the middle of the diagram shows two parallel actual forces. After vector addition "at the location of <math>\scriptstyle \vec{F}_{2}</math>", the net force is translated to the appropriate line of application, where it becomes the resultant force <math>\scriptstyle \vec{F}_{R}</math>. The procedure is based on decomposition of all forces into components for which the lines of application (pale dotted lines) intersect at one point (the so-called pole, arbitrarily set at the right side of the illustration). Then the arguments from the previous case are applied to the forces and their components to demonstrate the torque relationships.
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| #The rightmost illustration shows a [[couple (mechanics)|couple]], two equal but opposite forces for which the amount of the net force is zero, but they produce the net torque <math> \scriptstyle \tau = Fd </math> where <math> \scriptstyle \ d </math> is the distance between their lines of application. This is "pure" torque, since there is no resultant force.
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| ==Usage==
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| [[Image:Non-parallel net force.svg|thumb|279px|Vector diagram for addition of non-parallel forces]]
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| Generally, a system of forces acting on a rigid body can always be replaced by one force plus one "pure" torque. The force is the net force, but in order to calculate the additional torque, the net force must be assigned the line of action. The line of action can be selected arbitrarily, but the additional "pure" torque will depend on this choice. In a special case it is possible to find such line of action that this additional torque is zero.
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| The [[resultant force]] and torque can be determined for any configuration of forces. However, an interesting special case is a torque-free resultant which it is useful both conceptually and practically, because the body moves without rotating as if it was a particle.
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| <!---These paragraphs seems to be redundant---
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| When the system of forces can be replaced by a resultant force, this can simplify practical calculations (e.g. in many planar systems, or using the center of gravity in homogenous field, etc.). On the conceptual level, definition of the resultant force underlines the fact that the net force does not fully replace the system of forces (so, for example, the [[work physics|work]] of the net force cannot replace the net work in the case of an extended rigid body, e.g. in the work-energy theorem etc.). And the concept is also useful for a full understanding of a more general approach.
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| [[Image:Parallel net force.svg|thumb|185px|Vector diagram for addition of parallel forces]]
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| Some authors do not distinguish the [[resultant force]] from the net force and use the terms as synonyms.<ref>Resnick, Robert and Halliday, David (1966), Physics, (Vol I and II, Combined edition), Wiley International Edition, Library of Congress Catalog Card No. 66-11527</ref>
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| ==See also==
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| *[[Screw theory]]
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| *[[Center of mass]]
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| *[[Centers of gravity in non-uniform fields]]
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| ==References==
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| {{Reflist}}
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| [[Category:Force]]
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| [[Category:Dynamics]]
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