Force Diagrams - Revision history

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	[[File:Haer PBG erection force Diagram s.png\|400px\|right\|Haer PBG erection force Diagram s\|link=~~https~~://commons.wikimedia.org/wiki/File:Haer_PBG_erection_force_Diagram_s.png]]		[[File:Haer PBG erection force Diagram s.png\|400px\|right\|Haer PBG erection force Diagram s\|link=http://commons.wikimedia.org/wiki/File:Haer_PBG_erection_force_Diagram_s.png]]
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	~~<big><big>'''Parallelogram, Triangle or Polygon?'''</big></big>~~		__TOC__


	~~__TOC__~~		<big><big>'''Parallelogram, Triangle or Polygon?'''</big></big>


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	=====Calculating Resultants=====		=====Calculating Resultants=====

	[[File:TrigCalcs.jpg\|400px\|left\|Trigonometric Ratios]]		[[File:TrigCalcs.jpg\|400px\|left\|Trigonometric Ratios]]

	The magnitude of the '''Resultant''' may be measured if drawn to scale but may also be calculated.		The magnitude of the '''Resultant''' may be measured if drawn to scale but may also be calculated.

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	For non-right-angled triangles use the '''[[Trigonometry\|Sine Rule]]''' ''(hence the magnitudes of forces in a '''Force Polygon''' can be calculated by dividing into triangles)''		For non-right-angled triangles use the '''[[Trigonometry\|Sine Rule]]''' ''(hence the magnitudes of forces in a '''Force Polygon''' can be calculated by dividing into triangles)''
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			{{Construction Kits Buyers Guide}}

	[[Category:Frames]]		[[Category:Frames]]
	[[Category:Terminology]]		[[Category:Terminology]]

DT Online at 09:58, 27 December 2015

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	[[File:Haer PBG erection force Diagram s.png\|400px\|right\|Haer PBG erection force Diagram s\|link=https://commons.wikimedia.org/wiki/File:Haer_PBG_erection_force_Diagram_s.png]]		[[File:Haer PBG erection force Diagram s.png\|400px\|right\|Haer PBG erection force Diagram s\|link=https://commons.wikimedia.org/wiki/File:Haer_PBG_erection_force_Diagram_s.png]]
	~~<big>~~<big><big>'''Parallelogram, Triangle or Polygon?'''~~</big>~~</big></big>		<big><big>'''Parallelogram, Triangle or Polygon?'''</big></big>

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			<big><big><big>'''Parallelogram, Triangle or Polygon?'''</big></big></big>


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	=====~~Parallelogram, Triangle or Polygon?~~=====		=====Vectors=====

	Forces can be represented by lines drawn to scale - known as '''[[Vector\|Vectors]]''' provided we know:		Forces can be represented by lines drawn to scale - known as '''[[Vector\|Vectors]]''' provided we know:

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	Common sense tells us in such a simple example that these two opposing forces could be replaced by a single '''1 [[newton]]'''. force pointing upwards from point '''O''' ''(i.e. the '''[[~~Resultant and Equilibrant~~\|Resultant]]''')'' and the point '''O''' could be held still by a downwards pointing '''1 [[newton]]''' force ''(i.e. the '''[[~~Resultant and Equilibrant~~\|Equilibrant]]''')''.		Common sense tells us in such a simple example that these two opposing forces could be replaced by a single '''1 [[newton]]'''. force pointing upwards from point '''O''' ''(i.e. the '''[[Vector\|Resultant]]''')'' and the point '''O''' could be held still by a downwards pointing '''1 [[newton]]''' force ''(i.e. the '''[[Vector\|Equilibrant]]''')''.

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			[[File:Haer PBG erection force Diagram s.png\|400px\|right\|Haer PBG erection force Diagram s\|link=https://commons.wikimedia.org/wiki/File:Haer_PBG_erection_force_Diagram_s.png]]

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	For non-right-angled triangles use the '''[[Trigonometry\|Sine Rule]]''' ''(hence the magnitudes of forces in a '''Force Polygon''' can be calculated by dividing into triangles)''		For non-right-angled triangles use the '''[[Trigonometry\|Sine Rule]]''' ''(hence the magnitudes of forces in a '''Force Polygon''' can be calculated by dividing into triangles)''

	[[Category:~~Bridges and Beams~~]]		[[Category:Frames]]
	[[Category:Terminology]]		[[Category:Terminology]]

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=====Parallelogram, Triangle or Polygon?=====

Forces can be represented by lines drawn to scale - known as '''[[Vector|Vectors]]''' provided we know:
[[File:Resultant.png|300px|right|Resultant and Equilibrant]]
*the magnitude
*the point of application
*the line of action
*the direction

From any point '''O''' a line can be draw to a scale representing an upward pointing force of '''2 [[newton|newtons]]''' and similarly from the same point, a shorter line to represent the downward pointing force of '''1 [[newton]]'''.

Common sense tells us in such a simple example that these two opposing forces could be replaced by a single '''1 [[newton]]'''. force pointing upwards from point '''O''' ''(i.e. the '''[[Resultant and Equilibrant|Resultant]]''')'' and the point '''O''' could be held still by a downwards pointing '''1 [[newton]]''' force ''(i.e. the '''[[Resultant and Equilibrant|Equilibrant]]''')''.

It is less easy to know the effect of two or more forces acting at a point ''(i.e. ''''concurrent'''')'' but are out of line with each other.

=====Parallelogram of Forces=====

[[File:ParallelogramForces.png|300px|right|Parallelogram of Forces]]

To find the '''Resultant''' (and therefore also the '''Equilibrant''') of two non-aligned forces acting at a point a '''Parallelogram of Forces''' can be used.

'''Vectors''' are drawn to represent the two forces from any point '''O''' as before then lines are drawn parallel to them to complete a parallelogram as shown. The diagonal to this parallelogram is the line of action of the missing '''resultant''' or '''equilibrant''' and it has been proved that its length is equal to the magnitude.

It is clear from this example that the '''resultant''' must be acting upwards and the '''equilibrant''' therefore acting downwards if the point '''O''' is to remain static.

=====Triangle of Forces=====

[[File:TriangleForces.png|300px|right|Triangle of Forces]]

Alternatively, the problem of finding the '''resultant''' or '''equilibrant''' to two concurrent forces can be found using a '''Triangle of Forces'''

In this case, '''vectors''' representing the two concurrent forces are drawn to scale end on end as shown. It can be seen that the closing side of the triangle is actually just the same as the diagonal to the parallelogram above and so also provides the line of action and magnitude of the '''resultant''' and '''equilibrant'''

But it is also a feature of a triangle of forces that the vector direction arrows will follow each other round if the point of application is to remain static. In this way, the triangle of forces gives the direction of the '''equilibrant''' - and so the '''resultant''' is equal and opposite to it.

=====Multiple Concurrent Forces=====

[[File: 4concurrent.png|300px|right|Resolving Concurrent Forces]]

These ideas can be extended when there are more than two concurrent forces.

Several forces can be reduced to a single resultant using '''Parallelogram of Forces''' to reduce each pair of them to a single '''resultant''' then creating a second '''parallelogram of force''' to reduce the two '''resultants''' into one.

This process may be repeated as often as is required, depending on the number of concurrent forces in question, until the point is reached when there are just two forces to resolve into one.

=====Polygon of Forces=====

[[File:PolygonForces.png|300px|right|Polygon of Forces]]

Alternatively, the idea of a '''Triangle of Forces''' can be extended to become a '''Polygon of Forces''' to solve the problem.

'''Vectors''' are drawn end on end, as for the '''triangle of forces''', and the closing side provides the line of action and magnitude of the '''equilibrant''' - and the arrows follow round to give the direction.

=====Bow's Notation=====

[[File:Polygon(Bows).png|300px|right|Using Bow's Notation]]
Application of the '''Polygon of Force''' can be helped by using '''[[Bow's Notation]]'''. The position of the missing '''equilibrant''' can be estimated then capital letters used to mark each space. '''Vectors''' are then drawn as before but labelled '''ab''', '''bc''', etc. - all in lower case letters.

<div style="clear: both;"></div>
=====Calculating Resultants=====

[[File:TrigCalcs.jpg|400px|left|Trigonometric Ratios]]

The magnitude of the '''Resultant''' may be measured if drawn to scale but may also be calculated.

For right-angled triangles '''[[Trigonometry|trigonometric ratios]]''' ''(sine, cosine and tangent)'' can be used.

For non-right-angled triangles use the '''[[Trigonometry|Sine Rule]]''' ''(hence the magnitudes of forces in a '''Force Polygon''' can be calculated by dividing into triangles)''

[[Category:Bridges and Beams]]
[[Category:Terminology]]