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* '''[[Lesson:In Statics the Sum of the Forces is Equal to Zero]]'''
'''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]]'''''
 
'''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]]'''''


==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.   
 
{{?}}
===Free Body Diagrams===
===Irrotational Equilibrium===


Newton's Second Law for a static system (Equilibrium):
Newton's Second Law for a static system (Equilibrium):
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The sum of the forces is equal to the mass times the acceleration. The 2nd Law tells us that if the object or system is motionless, the acceleration is equal to zero. Therefore the sum of the vector forces must be equal to zero.
The sum of the forces is equal to the mass times the acceleration. The 2nd Law tells us that if the object or system is motionless, the acceleration is equal to zero. Therefore the sum of the vector forces must be equal to zero.


'''Example:'''
'''Example:'''
 
[[Image:Equilibrium.JPG|left|250px|thumb]]
[[Image:Equilibrium.JPG|left|300px|thumb]]
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>.
<BR>
<BR>
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>
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, and we may solve for <math> \vec F_L </math>...
 
<BR> 
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.
<BR>
 
<math> \vec F_L = \frac{1962  N + 981  N}{4} = 735.75  N </math>
  <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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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.


===Corollary for torque===
===Rotational Equilibrium===


Lesson: In Statics the Sum of the Torques is Equal to Zero.
Lesson: In Statics the Sum of the Torques (Moments) is Equal to Zero.


The sum of all rotational forces, or torques, denoted by the capital Greek letter tau (<math>\Tau</math>), is also zero.  Commonly used units for torque are foot•pounds (ft•lb) and Newton•meters (N•m).
The sum of all rotational forces, or torques, denoted by the capital Greek letter <i>tau ('''<math>\Tau </math>''')</i>, is also zero.  Commonly used units for torque are <i>Foot•pounds ('''ft•lb''')</i> and <i>Newton•meters ('''N•m''')</i>.


Newton's Second Law (applied to torques):
Newton's Second Law (applied to torques):


Σ<math>\Tau</math> = Σ(ωI) = ω<sub>1</sub>I<sub>1</sub> + ω<sub>2</sub>I<sub>2</sub> + ω<sub>3</sub>I<sub>3</sub> + • • •  + ω<sub>n</sub>I<sub>n</sub>
<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 moment of inertia, I) times the angular acceleration (denoted by the lower case Greek letter omega, ω). 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.


Torques may also be calculated as forces times distances:
Torques may also be calculated as forces times distances:


Σ<math>\Tau</math> = Σ(Fd) = F<sub>1</sub>d<sub>1</sub> + F<sub>2</sub>d<sub>2</sub> + F<sub>3</sub>d<sub>3</sub> +  • • •  + F<sub>n</sub>d<sub>n</sub>
  <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: Consider a massless lever with two weights attached and a single massless support:
 
 
  |          +  Weight of object 1    = 10 lb
  |        /|  Distance from fulcrum = 10 ft
  |        / | 
  |      R/  |  Distance of support
W|    E/|S w1  from fulcrum        =  7 ft
A|    V/ |U   
  L|  E/  |P    Weight of object 2    = 80 lb
L|  L/|  |P    Distance from fulcrum =  4 ft   
  |  / w2 |O   
  | /    |R    What forces act upon the
  <u>|/      |T  </u>    massless lever ?
    FLOOR
 
First let's write the static torque equation for the system:
 
Σ<math>\Tau</math> = Σ(Fd) = F<sub>1</sub>d<sub>1</sub> + F<sub>S</sub>d<sub>S</sub> + F<sub>2</sub>d<sub>2</sub> = 0
 
Forces 1 and 2 are actually the weights of the two objects:
 
Σ<math>\Tau</math> = Σ(Fd) = w<sub>1</sub>d<sub>1</sub> + F<sub>S</sub>d<sub>S</sub> + w<sub>2</sub>d<sub>2</sub> = 0
 
Substitute in known values:
 
Σ<math>\Tau</math> = Σ(Fd) = (-10 lb)(10 ft) + F<sub>S</sub>(7 ft) + (-80 lb)(4 ft) = 0
 
Solve for F<sub>S</sub>:
 
Σ<math>\Tau</math> = Σ(Fd) = (-100 ft•lb) + F<sub>S</sub>(7 ft) + (-320 ft•lb) = 0
 
F<sub>S</sub>(7 ft) = (420 ft•lb)
 
F<sub>S</sub> = (420 ft•lb)/(7 ft) = 60 lb
 
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.
 
Note that we didn't need to use the fulcrum as the point from which all distances are measured, we could have chosen any point along the lever.  However, as with forces, a consistent sign convention must be used.  In this case, positive may be used for clockwise (CW) torques measured from one point of view, with negative torques used for counterclockwise (CCW) torques measured from the same POV.  The opposite system could also be used, so long as we were consistent.
 
Also note that the table example used previously stated that the object was in the center of the table.  If not, then balancing the torques (in two directions) would result in more force supported by some table legs and less by others.  If the object was directly over one of the legs, for example, then it's entire weight would be supported by that leg, in addition to one-fourth of the table's weight.  The other legs would then only support one-fourth of the table's weight each.
 
 
******************
===Force Vectors===
[[Image:Vector_components.png|right|200px|thumb]]
A force vector is a force defined in two or more dimensions with a component vector in each dimension which may all be summed to equal the force vector.  Similarly, the magnitude of each component vector, which is a scalar quantity, may be multiplied by the [[w:unit vector|unit vector]] in that dimension to equal the component vector.
 
<math>\vec F(x,y,z) = \vec F_x + \vec F_y + \vec F_z = F_x\hat{i} + F_y\hat{j} + F_z\hat{k}</math>
 
<br>
 
===Moment===
 
For a system wherein a rigid body experiences a force '''F''' at a distance '''L''' from a fixed point, the ''moment'' '''M''' is the quantity (oddly enough of the same units as energy) defined by the force multiplied by the length of distance between the fixed point and the point where the force is applied. The direction of the moment is perpendicular to the force ecotro and the length, using the [[w:right hand rule|right hand rule]].
 
 
<math> \vec M \ = \vec F * L </math>
 
 
In the event that a force impacts the rigid body at an angle other than a right angle <math>\vec F = F\angle\alpha = F_x + F_y</math>, the moment is determined by the component of the force vector <math>\vec F</math> that is orthogonal to the length '''L'''.


<br>


[[Image:Moment.PNG|left|400px|thumb]]
'''Example:'''
'''Example:'''
 
[[image:LeverExample.png|thumb|200px|left]]
M = Force * Length = 100 Newtons * 10 Meters = 1,000 Newton-meters (N-m)
Consider a massless lever with two weights attached and a single massless support:
 
<BR>
 
<BR>
'''Example:'''
''Weight of Object 1, <math>\vec F_1</math> = 10 lb''<br>
Force '''F''' is incident on the end of a rigid body of length '''L''' at an angle '''A''' degrees from the central axis of the body '''x'''
''Distance of Object 1 from fulcrum = 10 ft''<br>
(Hint: draw a free body diagram).
''Distance of support from fulcrum  = 7 ft''<br>
 
''Weight of Object 2, <math>\vec F_2</math> = 80 lb''<br>
Then  <math>F_y \ = \vec F \sin (A)</math>   and  <math>M \ = F_y \ * L </math>
''Distance of Object 2 from fulcrum = 4 ft''<br>
 
<BR>
What forces act upon the massless lever?''
<BR>
<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...


===Couple===
<math>\sum \tau = \sum (\vec F * d) = \vec F_1 * d_1 + \vec F_2 * d_2 + \vec F_S * d_S = 0 </math>
A couple is a pair of equal and opposite force vectors that are some distance apart and that act upon the same body, thus causing a rotation.
Imagine that force <math> F_1 \ </math> and force <math> F_2 \ </math> are incident at two locations along a rigid body of total length <math> L \ </math> at positions <math> a \ </math> and <math> b \ </math>, where <math> a \ + b = L </math>.
(Hint: draw a free body diagram)
 
Then  <math> \vec M = F(a+b) - (Fa)</math>  
 
 
===Resultant===
[[Image:Resultant.JPG|left|400px|thumb]]
<BR>
<BR>
<BR>
<BR>
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>
<BR>
<BR>
<BR>
<BR>
Simplify...
<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>
<BR>
<BR>
<BR>
<BR>
Any system of forces may be reduced to a system of components and a resulting moment.
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.


That is to say, <math>\vec R = \sum \vec F </math>  and  <math>\vec M_o \ = \sum M </math> about the point <math> O \ </math>
Note that we didn't need to use the fulcrum as the point from which all distances are measured, we could have chosen any point along the lever.  However, as with forces, a consistent sign convention must be used.  In this case, positive may be used for clockwise (CW) torques measured from one point of view, with negative torques used for counterclockwise (CCW) torques measured from the same POV.  The opposite system could also be used, so long as we were consistent.


  <math>\vec R_x = \sum \vec F_x </math> and <math>\vec R_y = \sum \vec F_y </math> and <math>\vec R_z = \sum \vec F_z </math> and then <math>R = \sqrt{\vec R_x^2 + \vec R_y^2 + \vec R_z^2}</math> with <math>\theta = \arctan \frac{F_y}{F_x} </math>
Also note that the table example used previously stated that the object was in the center of the table. If not, then balancing the torques (in two directions) would result in more force supported by some table legs and less by others. If the object was directly over one of the legs, for example, then it's entire weight would be supported by that leg, in addition to one-fourth of the table's weight. The other legs would then only support one-fourth of the table's weight each.
 
  ...then <math>\vec R</math> is magnitude <math>R \ </math> in the direction of <math>\theta </math>


==Assignments==
==Assignments==
'''Activities:'''
'''Activities:'''
* Create an [[activity]]
* Create an [[Force Equilibrium/activity|activity]]


'''Readings:'''
'''Readings:'''
Line 161: Line 104:


'''Study guide:'''  
'''Study guide:'''  
# Wikipedia article:[[w:Force System|Force System]]
# Wikipedia article:[[Wikipedia:Force System|Force System]]
# Wikipedia article:[[w:Vector|Vector]]
# Wikipedia article:[[Wikipedia:Mechanical_equilibrium|Equilibrium]]
# Wikipedia article:[[w:Force Vector|Force Vector]]
# Wikipedia article:[[Wikipedia:Free_body_diagram|Free Body Diagram]]
# Wikipedia article:[[w:Moment|Moment]]
# Wikipedia article:[[Wikipedia:Gravitational_acceleration|Gravitational Acceleration]]
# Wikipedia article:[[w:Couple|Couple]]
# Wikipedia article:[[Wikipedia:Newton|Newton]]
# Wikipedia article:[[w:Resultant|Resultant]]
# Wikipedia article:[[Wikipedia:Torque|Torque]]
# Wikipedia article:[[Wikipedia:Moment of inertia|Moment of inertia]]
# Wikipedia article:[[Wikipedia:Angular acceleration|Angular acceleration]]


[[Category:Statics]]
[[Category:Statics]]
[[Category:Advanced Classical Mechanics]]
[[Category:Applied Mechanics]]
[[Category:Mechanical Engineering]]
[[Category:Engineering]]

Latest revision as of 04:44, 31 May 2009

Part of the Statics course offered by the Division of Applied Mechanics, School of Engineering and the Engineering and Technology Portal

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. ?

Irrotational Equilibrium

Newton's Second Law for a static system (Equilibrium): 

<math> \sum \vec F \ = M * \vec a = 0 </math>

The sum of the forces is equal to the mass times the acceleration. The 2nd Law tells us that if the object or system is motionless, the acceleration is equal to zero. Therefore the sum of the vector forces must be equal to zero.


Example:

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> 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>

And simplifying...

<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 N. Thus when multiplying through, the kilograms cancel out, and we may solve for <math> \vec F_L </math>...

<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.

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

Lesson: In Statics the Sum of the Torques (Moments) is Equal to Zero.

The sum of all rotational forces, or torques, denoted by the capital Greek letter tau (<math>\Tau </math>), is also zero. Commonly used units for torque are Foot•pounds (ft•lb) and Newton•meters (N•m).

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>

The sum of the torques is equal to the rotational mass or moment of inertia (I) times the angular acceleration, denoted by the lower case Greek letter omega (<math>\omega</math>). 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.

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>


Example:

File:LeverExample.png

Consider a massless lever with two weights attached and a single massless support:

Weight of Object 1, <math>\vec F_1</math> = 10 lb
Distance of Object 1 from fulcrum = 10 ft
Distance of support from fulcrum = 7 ft
Weight of Object 2, <math>\vec F_2</math> = 80 lb
Distance of Object 2 from fulcrum = 4 ft

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...

<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...

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

Simplify...

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

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>

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.

Note that we didn't need to use the fulcrum as the point from which all distances are measured, we could have chosen any point along the lever. However, as with forces, a consistent sign convention must be used. In this case, positive may be used for clockwise (CW) torques measured from one point of view, with negative torques used for counterclockwise (CCW) torques measured from the same POV. The opposite system could also be used, so long as we were consistent.

Also note that the table example used previously stated that the object was in the center of the table. If not, then balancing the torques (in two directions) would result in more force supported by some table legs and less by others. If the object was directly over one of the legs, for example, then it's entire weight would be supported by that leg, in addition to one-fourth of the table's weight. The other legs would then only support one-fourth of the table's weight each.

Assignments

Activities:

Readings:

Study guide:

  1. Wikipedia article:Force System
  2. Wikipedia article:Equilibrium
  3. Wikipedia article:Free Body Diagram
  4. Wikipedia article:Gravitational Acceleration
  5. Wikipedia article:Newton
  6. Wikipedia article:Torque
  7. Wikipedia article:Moment of inertia
  8. Wikipedia article:Angular acceleration
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