Archive:Force equilibrium - Revision history

wikademia>Wikademia: moved Force equilibrium to Archive:Force equilibrium

2009-05-31T04:44:16Z

moved Force equilibrium to Archive:Force equilibrium

← Older revision	Revision as of 04:44, 31 May 2009
(No difference)

wikademia>Wikademia at 03:42, 10 May 2009

2009-05-10T03:42:46Z

← Older revision		Revision as of 03:42, 10 May 2009
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	'''Part of the [[~~Topic:~~Statics\|Statics]] course offered by the ''[[~~Topic:~~Applied ~~Mechanics~~\|Division of Applied Mechanics]]'', ''[[~~School:~~Engineering\|School of Engineering]]'' and the ''[[Portal:Engineering and Technology\|Engineering and Technology Portal]]'''''		'''Part of the [[Statics\|Statics]] course offered by the ''[[Applied mechanics\|Division of Applied Mechanics]]'', ''[[Engineering\|School of Engineering]]'' and the ''[[Portal:Engineering and Technology\|Engineering and Technology Portal]]'''''

	==Lecture==		==Lecture==

wikademia>Wikademia at 02:28, 10 May 2009

2009-05-10T02:28:26Z

← Older revision		Revision as of 02:28, 10 May 2009
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	==Lecture==		==Lecture==
	Static equilibrium defines the state in which the sum of the forces, and torque, on each particle of the system is zero. A particle in mechanical equilibrium is undergoing neither linear nor rotational acceleration; however it could be translating or rotating at a constant velocity.		Static equilibrium defines the state in which the sum of the forces, and torque, on each particle of the system is zero. A particle in mechanical equilibrium is undergoing neither linear nor rotational acceleration; however it could be translating or rotating at a constant velocity.
			{{?}}
	===Irrotational Equilibrium===		===Irrotational Equilibrium===

wikademia>Wikademia: Text replace - 'w:' to 'Wikipedia:'

2009-04-11T18:14:15Z

Text replace - 'w:' to 'Wikipedia:'

← Older revision		Revision as of 18:14, 11 April 2009
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	'''Study guide:'''		'''Study guide:'''
	# Wikipedia article:[[w:Force System\|Force System]]		# Wikipedia article:[[Wikipedia:Force System\|Force System]]
	# Wikipedia article:[[w:Mechanical_equilibrium\|Equilibrium]]		# Wikipedia article:[[Wikipedia:Mechanical_equilibrium\|Equilibrium]]
	# Wikipedia article:[[w:Free_body_diagram\|Free Body Diagram]]		# Wikipedia article:[[Wikipedia:Free_body_diagram\|Free Body Diagram]]
	# Wikipedia article:[[w:Gravitational_acceleration\|Gravitational Acceleration]]		# Wikipedia article:[[Wikipedia:Gravitational_acceleration\|Gravitational Acceleration]]
	# Wikipedia article:[[w:Newton\|Newton]]		# Wikipedia article:[[Wikipedia:Newton\|Newton]]
	# Wikipedia article:[[w:Torque\|Torque]]		# Wikipedia article:[[Wikipedia:Torque\|Torque]]
	# Wikipedia article:[[w:Moment of inertia\|Moment of inertia]]		# Wikipedia article:[[Wikipedia:Moment of inertia\|Moment of inertia]]
	# Wikipedia article:[[w:Angular acceleration\|Angular acceleration]]		# Wikipedia article:[[Wikipedia:Angular acceleration\|Angular acceleration]]

	[[Category:Statics]]		[[Category:Statics]]

wikademia>Wikademia: 17 revisions

2009-04-10T19:59:54Z

17 revisions

← Older revision	Revision as of 19:59, 10 April 2009
(No difference)

wikademia>Gustable at 10:50, 4 July 2008

2008-07-04T10:50:25Z

← Older revision		Revision as of 10:50, 4 July 2008
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	[[Category:Statics]]		[[Category:Statics]]
	~~[[Category:Advanced Classical Mechanics]]~~
	~~[[Category:Applied Mechanics]]~~
	~~[[Category:Mechanical Engineering]]~~
	~~[[Category:Engineering]]~~

wikademia>Emesee: Force Equilibrium moved to Force equilibrium: casing

2008-06-10T02:51:24Z

Force Equilibrium moved to Force equilibrium: casing

← Older revision	Revision as of 02:51, 10 June 2008
(No difference)

wikademia>Mu301: /* Assignments */ name

2008-01-08T03:20:06Z

Assignments: name

← Older revision		Revision as of 03:20, 8 January 2008
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	==Assignments==		==Assignments==
	'''Activities:'''		'''Activities:'''
	* Create an [[activity]]		* Create an [[Force Equilibrium/activity\|activity]]

	'''Readings:'''		'''Readings:'''

wikademia>Gustable: Finally finished

2007-10-20T03:24:29Z

Finally finished

← Older revision		Revision as of 03:24, 20 October 2007
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	The force upward on the legs is found by balancing these forces in the equation given by Newton's Law. Gravity is acting downward on the 200 kg object and the 100 kg table. Therefore, we may substitute the acceleration due to gravity on Earth <math> \vec g </math> which is <math> 9.81 \frac{m}{s^2} </math>, for <math> \vec a </math>.		The force upward on the legs is found by balancing these forces in the equation given by Newton's Law. Gravity is acting downward on the 200 kg object and the 100 kg table. Therefore, we may substitute the acceleration due to gravity on Earth <math> \vec g </math> which is <math> 9.81 \frac{m}{s^2} </math>, for <math> \vec a </math>.
			<BR>
			<BR>
	Substitute in known values...		Substitute in known values...

	<math> \sum \vec F_y = M * \vec a = 0 = - (200 kg * 9.81 \frac{m}{s^2}) - (100 kg * 9.81 \frac{m}{s^2}) + (F_L * 4) </math>		<math> \sum \vec F_y = M * \vec a = 0 = - (200 kg * 9.81 \frac{m}{s^2}) - (100 kg * 9.81 \frac{m}{s^2}) + (F_L * 4) </math>
			<BR>
	<BR>		<BR>
	And simplifying...		And simplifying...
	<BR><BR>		<BR>
			<BR>
	<math> \vec F_L = \frac{200 kg * 9.81 \frac{m}{s^2} + 100 kg * 9.81 \frac{m}{s^2}}{4} </math>		<math> \vec F_L = \frac{200 kg * 9.81 \frac{m}{s^2} + 100 kg * 9.81 \frac{m}{s^2}}{4} </math>

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	Notice that it is critical that a consistent sign convention be followed throughout the entire analysis effort. The sign convention is typically chosen in more complex problems to ease the total amount of algebra necessary to analyze the equations in the <math> x </math>, <math> y </math> and <math> z </math> axis. We chose positive values to mean upward forces and negative values to mean downward forces, but we could have also used the opposite, provided we were consistent.		Notice that it is critical that a consistent sign convention be followed throughout the entire analysis effort. The sign convention is typically chosen in more complex problems to ease the total amount of algebra necessary to analyze the equations in the <math> x </math>, <math> y </math> and <math> z </math> axis. We chose positive values to mean upward forces and negative values to mean downward forces, but we could have also used the opposite, provided we were consistent.


	===Rotational Equilibrium===		===Rotational Equilibrium===
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	[[image:LeverExample.png\|thumb\|200px\|left]]		[[image:LeverExample.png\|thumb\|200px\|left]]
	Consider a massless lever with two weights attached and a single massless support:		Consider a massless lever with two weights attached and a single massless support:
			<BR>
			<BR>
	''Weight of Object 1, <math>\vec F_1</math> = 10 lb''<br>		''Weight of Object 1, <math>\vec F_1</math> = 10 lb''<br>
	''Distance of Object 1 from fulcrum = 10 ft''<br>		''Distance of Object 1 from fulcrum = 10 ft''<br>
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	''Weight of Object 2, <math>\vec F_2</math> = 80 lb''<br>		''Weight of Object 2, <math>\vec F_2</math> = 80 lb''<br>
	''Distance of Object 2 from fulcrum = 4 ft''<br>		''Distance of Object 2 from fulcrum = 4 ft''<br>
			<BR>
	What forces act upon the massless lever?''		What forces act upon the massless lever?''
			<BR>
			<BR>
	First let's write the static torque equation for the system. Forces 1 and 2 are actually the weights of the two objects...		First let's write the static torque equation for the system. Forces 1 and 2 are actually the weights of the two objects...

	<math>\sum \tau = \sum (\vec F * d) = \vec F_1 * d_1 + \vec F_2 * d_2 + \vec F_S * d_S = 0 </math>		<math>\sum \tau = \sum (\vec F * d) = \vec F_1 * d_1 + \vec F_2 * d_2 + \vec F_S * d_S = 0 </math>
			<BR>
			<BR>
	Substitute in known values...		Substitute in known values...

	<math>\sum \tau = \sum (\vec F * d) = (-10 lb)(10 ft) + (-80 lb)(4 ft) + \vec F_S * (7 ft) = 0 </math>		<math>\sum \tau = \sum (\vec F * d) = (-10 lb)(10 ft) + (-80 lb)(4 ft) + \vec F_S * (7 ft) = 0 </math>
			<BR>
	~~Solve for~~ <~~math~~>~~\vec F_S~~<~~/math~~> ...		<BR>
			Simplify...

	<math>\sum \tau = \sum (\vec F * d) = (-100 ft.lb.) + \vec F_S * (7 ft) + (-320 ft.lb.) = 0 </math>		<math>\sum \tau = \sum (\vec F * d) = (-100 ft.lb.) + \vec F_S * (7 ft) + (-320 ft.lb.) = 0 </math>
	<BR>		<BR>
	<BR>		<BR>
			And solve for <math>\vec F_S</math> ...

	<math>\vec F_S * (7 ft) \ = (420 ft.lb.) </math> and therefore <math>\vec F_S = (420 ft.lb.) / (7 ft) \ = 60 lb.</math>		<math>\vec F_S * (7 ft) \ = (420 ft.lb.) </math> and therefore <math>\vec F_S = (420 ft.lb.) / (7 ft) \ = 60 lb.</math>
			<BR>
			<BR>
	Thus, a force of 60 pounds acts upward on the lever from the support in order to balance the torques on the lever. Since the total weight on the lever is 90 pounds and 60 of those pounds are countered by the support, the remaining 30 pounds must act upward on the lever at the fulcrum in the lower, left corner.		Thus, a force of 60 pounds acts upward on the lever from the support in order to balance the torques on the lever. Since the total weight on the lever is 90 pounds and 60 of those pounds are countered by the support, the remaining 30 pounds must act upward on the lever at the fulcrum in the lower, left corner.

wikademia>Gustable: Finished

2007-10-20T03:15:02Z

Finished

← Older revision		Revision as of 03:15, 20 October 2007
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	Consider a table with four legs and a 200 kg object resting motionless in the center of the table. What force acts upon the bottom of each table leg <math> \vec F_L </math>?		Consider a table with four legs and a 200 kg object resting motionless in the center of the table. What force acts upon the bottom of each table leg <math> \vec F_L </math>?

	The force upward on the legs is found by balancing these forces in the equation given by Newton's Law. Gravity is acting downward on the 200 kg object and the 100 kg table. Therefore, we may substitute the acceleration due to gravity on Earth <math> \vec g </math> (<math> 9.81 \frac{m}{s^2} </math>), for <math> \vec a </math>.		The force upward on the legs is found by balancing these forces in the equation given by Newton's Law. Gravity is acting downward on the 200 kg object and the 100 kg table. Therefore, we may substitute the acceleration due to gravity on Earth <math> \vec g </math> which is <math> 9.81 \frac{m}{s^2} </math>, for <math> \vec a </math>.

			Substitute in known values...

	<math> \sum \vec F_y = M * \vec a = 0 = - (200 kg * 9.81 \frac{m}{s^2}) - (100 kg * 9.81 \frac{m}{s^2}) + (F_L * 4) </math>		<math> \sum \vec F_y = M * \vec a = 0 = - (200 kg * 9.81 \frac{m}{s^2}) - (100 kg * 9.81 \frac{m}{s^2}) + (F_L * 4) </math>
	<br>		<BR>
	<br>		And simplifying...
			<BR><BR>
	<math> \vec F_L = \frac{200 kg * 9.81 \frac{m}{s^2} + 100 kg * 9.81 \frac{m}{s^2}}{4} </math>		<math> \vec F_L = \frac{200 kg * 9.81 \frac{m}{s^2} + 100 kg * 9.81 \frac{m}{s^2}}{4} </math>

	The unit <math> \frac{kg-m}{s^2} </math> is equivalent to the unit of force called a Newton <i>'''N'''</i>. Thus when multiplying through, the kilograms cancel out: <math> \vec F_L = \frac{1962 N + 981 N}{4} = 735.75 N </math>		The unit <math> \frac{kg.m}{s^2} </math> is equivalent to the unit of force called a Newton <i>'''N'''</i>. Thus when multiplying through, the kilograms cancel out, and we may solve for <math> \vec F_L </math>...
			<BR>
			<BR>
			<math> \vec F_L = \frac{1962 N + 981 N}{4} = 735.75 N </math>

	This means that each leg exerts a downward force of 735.75 Newtons on the floor, and the floor simultaneously exerts the same force upward on each table leg.		This means that each leg exerts a downward force of 735.75 Newtons on the floor, and the floor simultaneously exerts the same force upward on each table leg.
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	Newton's Second Law (applied to torques):		Newton's Second Law (applied to torques):

	<math>\sum \tau = \sum (\vec \omega * I) = (\vec \omega_1 * I_1) + (\vec \omega_2 * I_2) + (\vec \omega_3 * I_3) + ... + (\vec \omega_n * I_n)</math>		<math>\sum \tau = \sum (\vec \omega * I) = (\vec \omega_1 * I_1) + (\vec \omega_2 * I_2) + (\vec \omega_3 * I_3) + ... + (\vec \omega_n * I_n)</math>

	The sum of the torques is equal to the rotational mass or <i> moment of inertia ('''I''')</i> times the angular acceleration, denoted by the lower case Greek letter <i>omega ('''<math>\omega</math>''')</i>. The 2nd Law tells us that if the object or system is motionless, the angular acceleration is equal to zero. Therefore the sum of the vector torques must be equal to zero.		The sum of the torques is equal to the rotational mass or <i> moment of inertia ('''I''')</i> times the angular acceleration, denoted by the lower case Greek letter <i>omega ('''<math>\omega</math>''')</i>. The 2nd Law tells us that if the object or system is motionless, the angular acceleration is equal to zero. Therefore the sum of the vector torques must be equal to zero.
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	Torques may also be calculated as forces times distances:		Torques may also be calculated as forces times distances:

	<math>\sum \vec \tau = \sum (\vec F * d) = \vec F_1 * d_1 + \vec F_2 * d_2 + \vec F_3 * d_3 + ... + \vec F_n * d_n </math>		<math>\sum \vec \tau = \sum (\vec F * d) = \vec F_1 * d_1 + \vec F_2 * d_2 + \vec F_3 * d_3 + ... + \vec F_n * d_n </math>


	'''Example:'''		'''Example:'''
			[[image:LeverExample.png\|thumb\|200px\|left]]
	Consider a massless lever with two weights attached and a single massless support:		Consider a massless lever with two weights attached and a single massless support:

	~~[[image:LeverExample.png\|thumb\|200px\|left]]~~

	''Weight of Object 1, <math>\vec F_1</math> = 10 lb''<br>		''Weight of Object 1, <math>\vec F_1</math> = 10 lb''<br>
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	What forces act upon the massless lever?''		What forces act upon the massless lever?''


	First let's write the static torque equation for the system. Forces 1 and 2 are actually the weights of the two objects...		First let's write the static torque equation for the system. Forces 1 and 2 are actually the weights of the two objects...

	<math>\sum \tau = \sum (\vec F * d) = \vec F_1 * d_1 + \vec F_2 * d_2 + \vec F_S * d_S = 0 </math>		<math>\sum \tau = \sum (\vec F * d) = \vec F_1 * d_1 + \vec F_2 * d_2 + \vec F_S * d_S = 0 </math>

	Substitute in known values...		Substitute in known values...

	<math>\sum \tau = \sum (\vec F * d) = (-10 lb)(10 ft) + (-80 lb)(4 ft) + \vec F_S * (7 ft) = 0 </math>		<math>\sum \tau = \sum (\vec F * d) = (-10 lb)(10 ft) + (-80 lb)(4 ft) + \vec F_S * (7 ft) = 0 </math>

	Solve for <math>\vec F_S</math> ...		Solve for <math>\vec F_S</math> ...

	<math>\sum \tau = \sum (\vec F * d) = (-100 ft.lb.) + \vec F_S * (7 ft) + (-320 ft.lb.) = 0 </math>		<math>\sum \tau = \sum (\vec F * d) = (-100 ft.lb.) + \vec F_S * (7 ft) + (-320 ft.lb.) = 0 </math>
			<BR>
	<math>\vec F_S * (7 ft) \ = (420 ft.lb.) </math> and therefore <math>\vec F_S = (420 ft.lb.) / (7 ft) \ = 60 lb. </math>		<BR>
			<math>\vec F_S * (7 ft) \ = (420 ft.lb.) </math> and therefore <math>\vec F_S = (420 ft.lb.) / (7 ft) \ = 60 lb.</math>

	Thus, a force of 60 pounds acts upward on the lever from the support in order to balance the torques on the lever. Since the total weight on the lever is 90 pounds and 60 of those pounds are countered by the support, the remaining 30 pounds must act upward on the lever at the fulcrum in the lower, left corner.		Thus, a force of 60 pounds acts upward on the lever from the support in order to balance the torques on the lever. Since the total weight on the lever is 90 pounds and 60 of those pounds are countered by the support, the remaining 30 pounds must act upward on the lever at the fulcrum in the lower, left corner.

← Older revision		Revision as of 18:14, 11 April 2009
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	'''Study guide:'''		'''Study guide:'''
	# Wikipedia article:[[w:Force System\|Force System]]		# Wikipedia article:[[Wikipedia:Force System\|Force System]]
	# Wikipedia article:[[w:Mechanical_equilibrium\|Equilibrium]]		# Wikipedia article:[[Wikipedia:Mechanical_equilibrium\|Equilibrium]]
	# Wikipedia article:[[w:Free_body_diagram\|Free Body Diagram]]		# Wikipedia article:[[Wikipedia:Free_body_diagram\|Free Body Diagram]]
	# Wikipedia article:[[w:Gravitational_acceleration\|Gravitational Acceleration]]		# Wikipedia article:[[Wikipedia:Gravitational_acceleration\|Gravitational Acceleration]]
	# Wikipedia article:[[w:Newton\|Newton]]		# Wikipedia article:[[Wikipedia:Newton\|Newton]]
	# Wikipedia article:[[w:Torque\|Torque]]		# Wikipedia article:[[Wikipedia:Torque\|Torque]]
	# Wikipedia article:[[w:Moment of inertia\|Moment of inertia]]		# Wikipedia article:[[Wikipedia:Moment of inertia\|Moment of inertia]]
	# Wikipedia article:[[w:Angular acceleration\|Angular acceleration]]		# Wikipedia article:[[Wikipedia:Angular acceleration\|Angular acceleration]]

	[[Category:Statics]]		[[Category:Statics]]