Statics/Print version

Statics

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Introduction

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Introduction

Statics is the branch of mechanics concerned with the study of forces and the effect of forces on a non-deformable, or rigid, system when the system is in a state of equilibrium.

This course is a crucial prerequisite for later areas, such as Dynamics and Properties of Materials. It utilizes principles of physics and calculus. It is fundamental in many different branches of engineering, from mechanical to civil engineering, and the principles of equilibrium, moment of inertia, and center of gravity will be revisited in more advanced fields. It is because an understanding of these topics is so crucial that statics does not cover a wide range of topics. Every problem will deal with some combination of two equations: the net forces being equal to zero, and/or net moments being equal to zero.

Statics in the taxonomy of physical sciences.

Who is This Book For?

This book is for undergraduate students pursuing, or thinking about pursuing, a degree in engineering. As specified early, statics is needed in almost every field of engineering. It is not an introductory course, however. Statics does not attempt to reinvent the wheel, and is built firmly on the foundation of Physics. Newton's Laws are essential to statics. Mathematically, one should have a firm grip on vector algebra, though it is possible to solve many simple problems without the use of vectors. While it would benefit a student to understand calculus concepts, like the ability to calculate an area or a volume, most statics problems will involve simple geometries.

How is this book organized

This book is organized to introduce concepts that will be later used in Dynamics and Properties of Materials. As such, it will begin by reviewing over vector mathematics and then introducing simple concepts, like Newton's Second Law, and then increasing the ways with which those concepts work together. Mathematically, this will involve increasing the number of forces, increasing the number of dimensions, and then expanding particle equilibrium to Rigid Body Equilibrium and Structural Equilibrium.

Finally, we will go over Moments of Inertia and Center of Mass. These concepts will be critical in higher level disciplines and a firm understanding will allow you to analyze more complex problems when the concepts of stress, strain, and acceleration are introduced.

Measurement and Units

In statics and mechanics, units can be expressed in terms of three basic dimensions: length, mass, and time. All other units are created from combinations of these three basic units.

Force can be considered a fourth basic unit. Known as a derived measurement, it comes from Newton's 2nd Law:

$\ \ \!\mathrm {F} =\mathrm {m} \ \mathrm {a} \,\!$

Here, force is defined as the amount of mass multiplied by the acceleration (length per second squared) that the mass achieves.

International System of Units (SI Units)

In the SI system of units, the three specified base units are the units of length, mass and time. A fourth unit, that of force, is derived from the base units.

• The unit of length is the meter (m).
• The unit of mass is the kilogram (kg).
• The unit of time is the second (s).
• The unit of force is the newton (N), where:

$\ \!\mathrm {N} =1\ \!\mathrm {kg} \ {\frac {\mathrm {m} }{\mathrm {s} ^{2}}}\,\!$

When working with units that are either large multiples or small fractions of these units, prefixes are often used in order to keep the numbers manageable. For example,

$1000\ \!\mathrm {m} =1\ \!\mathrm {km}$

The following table gives a more detailed description of prefixes.

Prefix	Abbrev.	Factor
peta-	P	1015
tera-	T	1012
giga-	G	109
mega-	M	106
kilo-	k	103
hecto-	h	102
deca-	da	101
deci-	d	10-1
centi-	c	10-2
milli-	m	10-3
micro-	μ	10-6
nano-	n	10-9
pico-	p	10-12
femto-	f	10-15
atto-	a	10-18

All other measurements are derived using variations of these four basic units and the listed prefixes.

Common SI units are listed in the following table.

Value	Units (long)	Units (abbrev.)
Acceleration	meter per second squared	${\frac {\mathrm {m} }{\mathrm {s} ^{2}}}$
Angle	radian	$\!\mathrm {rad}$
Angular Acceleration	radian per second squared	${\frac {\mathrm {rad} }{\mathrm {s} ^{2}}}$
Angular Velocity	radian per second	${\frac {\mathrm {rad} }{\mathrm {s} }}$
Area	square meter	$\!\mathrm {m} ^{2}$
Density	kilogram per cubic meter	${\frac {\mathrm {kg} }{\mathrm {m} ^{3}}}$
Energy	joule	$\!\mathrm {J}$ or $\!\mathrm {Nm}$
Force	newton	$\!\mathrm {N}$
Frequency	hertz	$\!\mathrm {Hz}$ or ${\frac {1}{\mathrm {s} }}$
Impulse	newton-second	$\!\mathrm {Ns}$ or $\!\mathrm {kg} {\frac {\mathrm {m} }{\mathrm {s} }}$
Length	meter	$\!\mathrm {m}$
Mass	kilogram	$\!\mathrm {kg}$
Force Moment	newton-meter	$\!\mathrm {Nm}$
Power	watt	$\!\mathrm {W}$ or ${\frac {\mathrm {J} }{\mathrm {s} }}$
Pressure	pascal	$\!\mathrm {Pa}$ or ${\frac {\mathrm {N} }{\mathrm {m} ^{2}}}$
Stress	pascal	$\!\mathrm {Pa}$ or ${\frac {\mathrm {N} }{\mathrm {m} ^{2}}}$
Time	second	$\!\mathrm {s}$
Velocity	meter per second	${\frac {\mathrm {m} }{\mathrm {s} }}$
Volume (solids)	cubic meter	$\!\mathrm {m} ^{3}$
Volume (liquids)	litre	$\!\mathrm {L}$ or $\!\mathrm {dm} ^{3}$
Work	joule	$\!\mathrm {J}$ or $\!\mathrm {Nm}$

British and American Customary Units

While the International System of units is in common use throughout much of the world, engineers may still encounter British or American units. Therefore, it is a good idea to have some familiarity with them.

While the basic units in International System of units are length, mass, and time--with the unit of force defined in terms of these--in the British and American units, the base units are length, force and time, with mass being defined in terms of these.

• The unit of length is the foot (ft).
• The unit of force is the pound (lb), which is occasionally called pound-force (lbf).
• The unit of time is the second (s).

The unit of mass in British and American units is the slug. It is defined as the amount of mass accelerated at a rate of 1 ft/s^2 when 1 pound of force is applied.

$1\ \!\mathrm {slug} =1\ \!\mathrm {lb} _{f}\ {\frac {\mathrm {ft} }{\mathrm {s} ^{2}}}$

Occasionally, mass is described as a pound-mass. It is equal to the mass required to move one lb of weight when acted upon by the standard acceleration of gravity. On Earth, the standard acceleration of gravity is about 32.2 ft/s^2, this means that one slug is 32.2 lb(mass).

$1\ \!\mathrm {slug} =32.2\ \!\mathrm {lb_{m}}$

Conversion from one System of Units to Another

While we can do all our calculations in one set of units or the other, as long as we are consistent, there are times we will want to convert from one system to the other.

• Unit of Length 1 ft = 0.3048 m
• Unit of Force 1 lb = 4.448 N
• Unit of Mass 1 slug = 1 lb-s^2/ft = 14.59 kg

As mentioned earlier, the second is the same in both systems of units and so no conversion is required.

Common British and American Customary units and their SI equivalents are listed in the table below.

Value	Conversion
Acceleration	$1\ {\frac {\!\mathrm {ft} }{\!\mathrm {s} ^{2}}}=0.3048\ {\frac {\!\mathrm {m} }{\!\mathrm {s} ^{2}}}$
Area	$1\ \!\mathrm {ft} ^{2}=0.0929\ \!\mathrm {m} ^{2}$
Energy	$1\ \!\mathrm {ft} \ \!\mathrm {lb} =1.356\ \!\mathrm {J}$
Force	$1\ \!\mathrm {lb} =4.448\ \!\mathrm {N}$
Impulse	$1\ \!\mathrm {lb} \ \!\mathrm {s} =4.448\ \!\mathrm {N} \ \!\mathrm {s}$
Length	$1\ \!\mathrm {ft} =0.3048\ \!\mathrm {m}$
	$1\ \!\mathrm {mile} =1.609\ \!\mathrm {km}$
Mass	$1\ \!\mathrm {lb} \ \!\mathrm {mass} =0.4536\ \!\mathrm {kg}$
	$1\ \!\mathrm {slug} =14.59\ \!\mathrm {kg}$
	$1\ \!\mathrm {ton} =907.2\ \!\mathrm {kg}$
Moment	$1\ \!\mathrm {lb} \ \!\mathrm {ft} =1.356\ \!\mathrm {N} \ \!\mathrm {m}$
Power	$1\ \!\mathrm {ft} \ {\frac {\!\mathrm {lb} }{\!\mathrm {s} }}=1.356\ \!\mathrm {W}$
	$1\ \!\mathrm {hp} =745.7\ \!\mathrm {W}$
Pressure	$1\ \!\mathrm {psi} =6.895\ \!\mathrm {kPa}$
Stress	$1\ \!\mathrm {psi} =6.895\ \!\mathrm {kPa}$
Velocity	$1\ {\frac {\!\mathrm {ft} }{\!\mathrm {s} }}=0.3048\ {\frac {\!\mathrm {m} }{\!\mathrm {s} }}$
	$1\ \!\mathrm {mph} =1.609\ {\frac {\!\mathrm {km} }{\!\mathrm {hr} }}$
Volume (solids)	$1\ \!\mathrm {ft} ^{3}=0.02832\ \!\mathrm {m} ^{3}$
Volume (liquids)	$1\ \!\mathrm {gal} =3.785\ \!\mathrm {L}$
Work	$1\ \!\mathrm {ftlb} =1.356\ \!\mathrm {J}$

Example

According to the official National Hockey League rulebook, "The official size of the (hockey) rink shall be two hundred feet (200') long and eighty-five feet (85') wide." What are the dimensions in SI units?

Solution

From the above table:

$1\ \mathrm {ft} =0.3048\ \mathrm {m}$

Using dimensional analysis we find the length and width in meters.

$l\ =\ 200\ \mathrm {ft} \ \cdot \ {\frac {0.3048\ \mathrm {m} }{1\ \mathrm {ft} }}=\ 60.96\ \mathrm {m}$

$w\ =\ 85\ \mathrm {ft} \ \cdot \ {\frac {0.3048\ \mathrm {m} }{1\ \mathrm {ft} }}=\ 25.96\ \mathrm {m}$

Significant Digits

When we talk about measurements and calculations, we need to understand the degree of accuracy involved.

The accuracy of our calculations cannot be more precise than the accuracy of our measurements.

Suppose we are provided with a distance to an accuracy of one decimal place, say 9.8 m. We are told an object travels this distance in 0.81 seconds. It does not make sense to say the object is traveling at a velocity of 12.11111111 m/s, that is, to eight decimal places.

This is because neither the distance nor the time taken to travel this distance is specified to this degree of precision. In fact, they are both specified to an accuracy of only two significant digits.

For reasons we will discuss shortly, we can say the object is traveling at a velocity of 12.1 m/s.

For many calculations in statics, we work to at most three significant digits.

Rules for Finding the Correct Number of Significant Digits

In general, when making a calculation, the answer can not have more significant digits than any of the numbers used in calculating it. The number of significant digits in an answer is equal to the minimum number of significant digits used in the calculation.

Here are rules that will help outline whether or not a digit is significant or not.

1. Non-zero numbers are always significant.
2. Zeros placed in between two other digits are significant.
3. Zeros placed at the end of a number, after a decimal, are significant.

References

1 - Both the principal SI units used in mechanics and the US Customary units and their SI equivalents are taken from Beer, Ferdinand P. and El Russell Johnston, Jr. "Vector Mechanics For Engineers, Statics" 3rd edition, McGraw Hill c 1977. It should be possible to find similar tables in other texts on this subject.

2 - Rules for taking significant digits are taken from www.physics.uoguelph.ca

• Advanced Expression Evaluator
• Unit Converter and Calculator

Vector Math

Adding Vectors

Let's say you have a box on the ground, and the box is being pulled in two directions with a certain force. You can predict the motion of the box by finding the net force acting on the box. If each force vector (where the magnitude is the tension in the rope, and the direction is the direction that the rope is "pointing") can be measured, you can add these vectors to get the net force. There are two methods for adding vectors:

Parallelogram Method

Addition of two vectors

This is a graphical method for adding vectors. First, a little terminology:

• The tail of a vector is where it originates.
• The head of a vector is where it goes. The head is the end with the arrowhead.

This method is most easily executed using graph paper. Establish a rectangular coordinate system, and draw the first vector to scale with the tail at the origin. Then, draw the second vector (again, to scale) with its tail coincident with the head of the first vector. Then, the properties of the sum vector are as follows:

• The length of the sum vector is the distance measured from the origin to the head of the second vector.
• The direction of the sum vector is the angle.

Example

In the image at the right, the vectors (10, 53°07'48") and (10, 36°52'12") are being added graphically. The result is (19.80, 45°00'00"). (How did I measure out those angles so precisely? I did that on purpose.)

The native vector format for the parallelogram method is the 'polar form'.

Computational Method

When you use the computational method, you must resolve each vector into its x- and y-components. Then, simply add the respective components.

Converting Polar Vectors to Rectangular Vectors

If a vector is given by (r, θ), where r is the length and θ is the direction,

• x = r cos θ
• y = r sin θ

Converting Rectangular Vectors to Polar Vectors

If a vector is given by $\left\langle x,y\right\rangle$,

• $r={\sqrt {x^{2}+y^{2}}}$
• $\theta =\arctan {\frac {y}{x}}$

Remember that the arctan() function only returns values in the range [-π/2, π/2]; therefore, if your vector is in the second or third quadrant, you will have to add π to whatever angle is returned from the arctan() function.

Example

Again referring to the image at the right, notice that the first vector can be expressed as $\left\langle 6,8\right\rangle$, and the second is equivalent to $\left\langle 8,6\right\rangle$. (Verify this.) Then, you simply add the components:

$\left\langle 6,8\right\rangle +\left\langle 8,6\right\rangle =\left\langle 6+8,8+6\right\rangle =\left\langle 14,14\right\rangle$

You should verify that $\left\langle 14,14\right\rangle$ is equal to (19.80, 45°00'00").

Multiplying Vectors

There are two ways to multiply vectors. I will not get into specific applications here; you will see many of those as you progress through the book.

The Dot Product

The dot product of two vectors results in a scalar. The dot product is the sum of the product of the components. For example:

   < 1 , 2 >
 ∙ < 3 , 4 >
 -----------
     |   +-----> 2 x 4 = 8
     +---------> 1 x 3 = 3
                      ------
                        11

Calculating bond angles of a symmetrical tetrahedral molecule such as methane using a dot product

A useful relation between vectors, their lengths, and the angle between them is given by the definition of the dot product:

${\vec {a}}\cdot {\vec {b}}=ab\cos \theta$

• ${\vec {a}}$ and ${\vec {b}}$ are the vectors.
• $a$ and $b$ are the vectors' magnitude.
• $\theta$ is the angle between the vectors.

The Cross Product

The cross product of two vectors results in another vector. The cross product is only applicable to 3-space vectors. Remember the three unit vectors:

• ${\hat {i}}$ is the unit vector along the x-axis
• ${\hat {j}}$ is the unit vector along the y-axis
• ${\hat {k}}$ is the unit vector along the z-axis

Now if you have two vectors ${\vec {a}}=\left\langle x_{1},y_{1},z_{1}\right\rangle$ and ${\vec {b}}=\left\langle x_{2},y_{2},z_{2}\right\rangle$, the cross product is given by solving a determinant as follows:

${\vec {a}}\times {\vec {b}}={\begin{vmatrix}{\hat {i}}&{\hat {j}}&{\hat {k}}\\x_{1}&y_{1}&z_{1}\\x_{2}&y_{2}&z_{2}\end{vmatrix}}={\begin{vmatrix}y_{1}&z_{1}\\y_{2}&z_{2}\end{vmatrix}}{\hat {i}}+{\begin{vmatrix}z_{1}&x_{1}\\z_{2}&x_{2}\end{vmatrix}}{\hat {j}}+{\begin{vmatrix}x_{1}&y_{1}\\x_{2}&y_{2}\end{vmatrix}}{\hat {k}}$

For more information on how to solve a determinant, consult the external links below.

The cross product of two vectors, the lengths of those vectors, and the short angle between the vectors is given by the following relation:

${\vec {a}}\times {\vec {b}}=ab\sin \theta$

The Right-Hand Rule

Geometrically, the cross product gives a vector that is perpendicular to the two arguments. Notice the reference to a vector, not the vector. This is because there are infinitely many vectors that are normal to two non-zero vectors. The direction of the cross product can be determined using the right-hand rule: Extend the fingers of your right hand, lay your straightened hand along the first vector, pointing your finger tips in the same direction as the vector. Curl your fingers through the short angle from the first vector to the second vector. Your thumb will point in the direction of the product vector.

Dots and Crosses of the Unit Vectors

Dot Products

• A unit vector dotted into itself gives one.
• A unit vector dotted into a different unit vector gives zero.

Cross Products

Order the unit vectors in this order: ${\hat {i}},{\hat {j}},{\hat {k}}$. Start at the first vector, move to the second vector, and keep going to the cross-product. If you moved immediately to the right first, the answer is positive. If you moved to the left first, the answer is negative. For example:

• ${\hat {i}}\times {\hat {j}}={\hat {k}}$
• ${\hat {j}}\times {\hat {i}}=-{\hat {k}}$

Vector Rules

Given vectors ${\vec {a}},{\vec {b}},{\vec {c}},$ and scalar r:

• ${\vec {a}}\cdot {\vec {b}}={\vec {b}}\cdot {\vec {a}}$
• $r({\vec {a}}\cdot {\vec {b}})=(r{\vec {a}})\cdot {\vec {b}}={\vec {a}}\cdot (r{\vec {b}})$
• ${\vec {a}}\times {\vec {b}}=-{\vec {b}}\times {\vec {a}}$
• $r({\vec {a}}\times {\vec {b}})=(r{\vec {a}})\times {\vec {b}}={\vec {a}}\times (r{\vec {b}})$
• $({\vec {a}}\times {\vec {b}})\cdot {\vec {c}}={\vec {a}}\cdot ({\vec {b}}\times {\vec {c}})$

External Links

Determinants on Wikipedia

Newton's Laws and Equilibrium

They are not the law pertaining to Statics The laws are some bit different then this, They Are the sources of Statics

Newton's First Law

Newton's first law states that an object that remains in uniform motion will remain in uniform motion unless it is acted upon by an external force. This also includes that an object at rest will remain at rest unless it is acted upon by an external force. When more than one force acts upon an object, the vector sum of these forces is the resultant force.

When the resultant force on an object is zero, it will remain at rest if it is at rest, or continue to move in a straight line at a constant velocity if it is in motion. There is no change in either the magnitude or direction of its velocity. That is, there is zero acceleration. This concept can also be applied to motion in any selected direction.

Consider an object moving along the x-axis. If no net force is applied to the object along the x-axis, it will continue to move along the x-axis at a constant velocity with no acceleration.We can extend this to the y- and z- axes.

In any system, unless the applied forces cancel each other out, that is, the resultant force is zero, there will be acceleration in the direction of the resultant force.The net force on such an object is zero

Force Equilibrium

In static systems, where motion does not occur, the sum of the forces in all directions must always equal zero (otherwise, it's a dynamics problem). This concept can be represented mathematically with the following equations.

Rotational Equilibrium

The concept also applies to rotational motion.

If the resultant moment about an axis is zero, the object will have no rotational acceleration about the axis. If the object is not spinning, it will not start to spin. If the object is spinning, it will continue to spin at the same constant angular velocity. Again, we can extend this to moments about the y-axis and the z-axis. This is represented mathematically with the following. ..............................................................................................................................................

What happens if the sum of forces or moments is not zero?

If the object has a net moment not equal to zero, it will spin about whichever axes are not zero. If the object is acted upon by a net force not equal to zero, it will accelerate in whatever direction the net force is. Once the object moves it cannot be analyzed with statics; instead, the rules of dynamics take over. The acceleration of the object is governed by Newton's second law,

$\sum F=m\cdot a$

You will never need this equation in Statics.

Forces As Vectors

Reading(s):

Vectors, Chapter 1.1 - 1.7

Scalars and Vectors and force

Scalar

A scalar is a quantity possessing only a magnitude. Examples include mass, volume, and length. In this book, scalars are represented by letters in italic type: $A$. Scalar quantities may be manipulated following the rules of simple algebra.

Vector

A vector is a quantity that has both a magnitude and a direction. Examples include velocity, position, and force. In this book, vectors are represented by letters with arrows over them: ${\vec {A}}$. Vector quantities are manipulated using vector mathematics, which is described in some detail in the following section.

Examples of Vector vs. Scalar Quantities

Velocity vs. Speed

Consider a car traveling South at a speed of 110 km per hour.

We can describe the motion of the car as a velocity with a magnitude of 110 km per hour and a direction of south. The velocity is a vector because it indicates magnitude and direction.

The motion of the car could also be described as a scalar by saying the speed is 110km per hour and ignoring the direction. Speed is a scalar because it consists of only a magnitude.

Force

An applied force is a vector quantity having both a magnitude and a direction.

Consider a hot air balloon suspended over a farmer's field at a constant altitude. The buoyant force on the ballon pushes the balloon up. At the same time, gravity exerts a force on the balloon pulling the balloon down.

The buoyant force and the force due to gravity act in opposite directions. If they have the same magnitude, the balloon will remain suspended at the same altitude. If the buoyant force is greater than the force due to gravity, the balloon will rise. If the force due to gravity is greater than the buoyant force, than the balloon will fall.

Vector Representation (Two Dimensions)

Forces, and any other vector, may be represented in a number of ways.

Graphically

A vector may be represented graphically by an arrow. The magnitude of the vector corresponds to the length of the arrow, and the direction of the vector corresponds to the angle between the arrow and a coordinate axis. The sense of the direction is indicated by the arrowhead. But when a vector changes direction the unit will change 10x

Polar Notation

In polar notation, the vector is represented by the magnitude of the vector, $r$ and its angle from the coordinate axis, $\theta$ in the form:

${\vec {A}}=r\angle \theta$

or

${\vec {A}}=(r,\theta )$

Component Notation

In component notation, a vector is represented by the magnitude of the components of the vector along the coordinate axes.

Scalar Notation

The components of the vector are represented as scalar values which are positive if their sense is in the same direction as the coordinate axis, and negative if their sense is opposite of the coordinate axis. For a vector with a positive x-component and a negative y-component:

${\vec {A}}=A_{x}+(-A_{y})=A_{x}-A_{y}$

Cartesian Notation

The components of the vector are represented as positive scalar values multiplied by cartesian unit vectors. Cartesian unit vectors are vectors with a magnitude of one that represent the direction of the coordinate axes. The unit vector ${\hat {i}}$ represents the x-axis, and the unit vector ${\hat {j}}$ represents the y-axis. The vectors sense is indicated by the sign of the unit vector. For a vector with a positive x-component and a negative y-component:

${\vec {A}}=A_{x}{\hat {i}}+A_{y}(-{\hat {j}})=A_{x}{\hat {i}}-A_{y}{\hat {j}}$

or

${\vec {A}}=\langle A_{x},-A_{y}\rangle$

Forces as Vectors

In engineering statics, we often convert forces into to component notation. Replacing a force with its components makes it easier to compute the resultant of a group of forces acting on a body. Conversion of a force from polar to component notation is accomplished by the following transformations:

For force ${\vec {F}}=F\angle \alpha =F_{x}+F_{y}$

$F_{x}\ =F\cos \alpha$

$F_{y}\ =F\sin \alpha$

$F\ ={\sqrt {F_{x}^{2}+F_{y}^{2}}}$

$\alpha \ =\arctan {\frac {F_{y}}{F_{x}}}$

Example 1

The graph to the left shows the 30 degree force, the plot to the right shows how the 30 degree force can be separated into X and Y components

Consider a force, ${\vec {F}}$, with a magnitude, $F$, of 100 N acting in the x-y plane. This force acts at an angle, $\alpha$, of 30 degrees to x-axis. What is this force in component notation?

We can replace this force with a pair of forces acting along the x-axis and the y-axis as follows.

$F_{x}=F\cos \alpha =100cos(30)\ =86.6N$

$F_{y}=F\sin \alpha =100sin(30)\ =50.0N$

Example 2

Two forces acting in the x-y plane are acting on a point. The first force is 100 N at an angle of 0 degrees. The second force is 50 N acting at an angle of 60 degrees. What is the resultant?

First, resolve the forces into their x and y components.

$F_{1x}=F_{1}\cos(A)\ =100cos(0)=100$

$F_{1y}=F_{1}\sin(A)\ =100sin(0)=0$

$F_{2x}=F_{2}\cos(B)\ =50cos(60)=25$

$F_{2y}=F_{2}\sin(B)\ =50sin(60)=43.3$

Sum all the forces in the x-direction.

$F_{x}=F_{1x}+F_{2x}\ =100+25=125$

Sum all the forces in the y-direction.

$F_{y}=F_{1y}+F_{2y}\ =0+43.3=43.3$

Finally, convert the resultant components back into polar notation.

${\sqrt {\left(\vert F_{x}\vert ^{2}+\vert F_{y}\vert ^{2}\right)}}=132.3$

$\theta \ =\arctan \left({\frac {F_{y}}{F_{x}}}\right)=19.1^{\circ }$

The resultant force is 132.3 N at an angle of 19.1 degrees.

coming soon!

Quick links:Statics, Mechanical Engineering

Resultants of Force Systems (contents)

A force system is a collection of forces acting at specified locations (may also include couples). Thus the set of forces shown on any free body diagram make up a force system. Force system is simply a term used to describe a group of forces.

Resultants

When two or more forces or moments are combined, the combination is called a resultant. Any force system can be simplified about a point to a resultant consisting of one force and one couple (either or both of which may be zero). The force is applied to the point and is the vector sum of all of the forces in the system. The couple is the vector sum of all of the couples in the force system plus the vector sum of all of the moments about the point of all of the forces in the force system. One immediate consequence of this definition of resultant is that the resultant about any point of the external force system acting on any system in equilibrium must be zero. The statement that the sum of the external forces and the sum of the external moments acting on a system in equilibrium is zero is often replaced by the statement that the resultant acting on a system in equilibrium is zero. An important attribute of the resultant of a force system is that if you apply the resultant of a force system to a rigid body, the effect on that rigid body is exactly the same as the effect of the original force system. It is in this sense that the resultant is "equivalent" to the original force system. Thus as we study the behavior of rigid bodies under the action of forces, the resultant of the external force system will be of vital interest. Care must be taken in replacing force systems with their resultant when dealing with deformable bodies as the effects of the resultant acting on a non-rigid system will be different than the effects of the actual force system.

Equipollence

The "equivalence" of effect on a rigid body of a force system and its resultant is the motivation for the term equipollence. Two force systems are said to be equipollent if they have the same resultant about a point. Thus two equipollent force systems have the same effect on a rigid body. Thus the precise way to state what we have learned about the toe is that the upper clamping surface exerts a force system on the toe that is equipollent to a force of 540 lb in the -Y direction acting at a point at the top, middle of the toe. Typically we are not this precise and merely state that the clamping surface is exerting a force of 540 lb in the -Y direction on the top, middle point of the toe. It is important for you to recognize that when you hear such statements, what they mean is that the force system acting on the object is equipollent to the specified force and that the particular distribution or combination of forces involved may be quite complicated.

Rigid Body Equilibrium (contents)

Rigid body equilibrium exists when a rigid body under the influence of a force system is in a stable equilibrium.

What is a rigid body?

It is a fictional object that is a useful approximation. The rigid body cannot be deformed, which means it cannot be squished or broken. Mathematically, we would say that every point on the rigid body will retain their distance and position relative to one another not matter the force. If we were to subject it to two forces approaching infinity, it would not be harmed.

How are they useful?

Rigid bodies are useful approximations because we will often be faced with materials that are not under enormous amounts of strain- the force is very small relative to the material's modulus of elasticity. We will go over how to see whether or not a rigid body is in equilibrium in two dimensions, and then be able to generalize that to three dimensions.

Two-Dimensional Equilibrium

There are six equations expressing the equilibrium of a rigid body in 3 dimensions.

Sum of Forces: $\sum _{}^{}F_{x}=0$, $\sum _{}^{}F_{y}=0$, $\sum _{}^{}F_{z}=0$

Sum of Moments: $\sum _{}^{}M_{x}=0$, $\sum _{}^{}M_{y}=0$, $\sum _{}^{}M_{z}=0$

In two dimensions one direction of force and two directions of moments can be ignored. When forces exist only in the x and y directions, there cannot be a moment in any direction except z. The equations of concern when forces only exist in the x and y directions are shown below.

Sum of Forces: $\sum _{}^{}F_{x}=0$, $\sum _{}^{}F_{y}=0$

Sum of Moments: $\sum _{}^{}M_{z}=0$

To solve two dimensional statics problems:

• draw a free body diagram
• Write equations for force equilibrium
• Write equations for moment equilibrium
• Once you have the same number of equations as unknowns the problem can be solved, you may have to strategically pick points to write moment equations etc.

Couple

F1 and F2 are equal

A couple exerts the same moment at every point as demonstrated in this section. After looking at the picture to the right the following equations can be written. Counter clockwise moments are considered positive and clockwise negative. F1 = F2

$\sum M_{A}=F_{1}*a-F_{2}*(b+a)=F_{1}*a-F_{1}*(b+a)=-F_{1}*b=-F*b$

$\sum M_{B}=-F_{1}*(b+c)+F_{2}*(c)=-F*b$

$\sum M_{C}=-F_{1}*d-F_{2}*e=-F*(d+e)=-F*b$

The moment about all points is the force multiplied by the distance between the forces. If you found the moment about D, or any other point you would continue to find the same moment. All points have the same moment, even points that aren't in the x-y plane.

Example 1

Question

The picture to the right shows the forces acting on a parked car. If the weight of the car acts exactly halfway between the two wheels and the weight is 1000 lbs how much force is exerted on the rear wheel? What about the front wheel?

Answer

Writing the force equations

$\sum F_{x}=0$

There are no forces in the x direction

$\sum F_{y}=0=R_{f}+R_{B}-W=R_{f}+R_{B}-1000lbs$

Writing moment equation about front wheel

$\sum M_{f}=0=-5ft*W+10*R_{B}=-5000lbft+10*R_{B}$

$\ R_{B}=500lb$

subbing $R_{B}$ back into the $sumF_{y}$

$\sum F_{y}=0=R_{f}+R_{B}-W=R_{f}+500lbs-1000lbs$

$\ R_{f}=500lbs$

Please note that $R_{f}$ and $R_{B}$ are distributed over two wheels. Each front wheel supports half of $R_{f}$ and each back wheel supports half of $R_{B}$.

Example 2

Question:

A uniform ring of mass $m\,\!$ and radius $r\,\!$ carries an eccentric mass $m_{0}\,\!$ at a radius $b\,\!$ and is in equilibrium position on an incline, which makes an angle $\alpha \,\!$ with the horizontal. If the contacting surfaces are rough enough to prevent slipping, write the expression for the angle $\theta \,\!$ which defines the equilibrium position.

Answer: $\theta =sin^{-1}[{\frac {r}{b}}(1+{\frac {m}{m_{0}}})sin\alpha ]\,\!$

Three-Dimensional Equilibrium

There are six equations expressing the equilibrium of a rigid body in 3 dimensions.

Sum of Forces: $\sum _{}^{}F_{x}=0$, $\sum _{}^{}F_{y}=0$, $\sum _{}^{}F_{z}=0$

Sum of Moments: $\sum _{}^{}M_{x}=0$, $\sum _{}^{}M_{y}=0$, $\sum _{}^{}M_{z}=0$

To solve three dimensional statics problems:

Im mad bc why r there only 5 problems here i need the six specific equilibruim equations for a 3 dimensional system. HELP BRO.

• Resolve force components
• Resolve moments about a point
• Resolve moments about a line or axis. See Scalar Triple Product

Knowing how to find the Determinate of a matrix can help a lot here.

Trusses

Trusses are series of trianglular supports with no intraspan loads. Classified as two force members,the beams have applied forces only at the ends. The resultant forces at the ends must be equal in magnitude and opposite in direction, along the line of the joints of the member. These forces are called axial forces. The member is said to be in compression if T is negative (ie, the forces at each end are toward each other) or in tension if T is positive.

The main strategies for analyzing trusses are the method of joints and the method of sections.

Engineering Trusses

Blue lines are members of the truss, the blue circle is the joint which all members are attached to

The following picture shows a three member truss. Angles, and one force are given in the picture. Find the force in the vertical and diagonal member by using what you know about equilibrium.

Then assign a coordinate system. In this problem a two dimensional X,Y plane can be used with the positive X axis being horizontal running from left to right(pointing to the right). The Y axis will be vertical and pointing upwards.

Since this is a statics problem, the joint cannot move. The only way the joint won't move is if the forces acting on it are in equilibrium (the net force must be equal to zero in all directions). An intelligent step now would be to sum X and Y forces at the joint.

Assign variable names so you don't get members confused with each other, let's call the vertical member's tension $T_{1}$ and the diagonal member's tension $T_{2}$.

$\sum F_{x}=0=-10-T_{2}\cos(30)+T_{1}\cos(90)$

$\sum F_{y}=0=T_{1}+T_{2}\sin(30)$

The cosine of 90 degrees is zero so $T_{1}$ drops out. You could also drop out $T_{1}$ simply by noticing that $T_{1}$ does not have an X component.

$\sum F_{x}=0=-10-T_{2}\cos(30)$

$\ T_{2}=-11.547N$

The negative sign indicates that the force is in the opposite direction as in the picture. This means the member is in compression instead of tension. Now we can solve for $T_{1}$.

$\sum F_{y}=0=T_{1}+T_{2}sin(30)$

$\sum F_{y}=0=T_{1}+-11.547sin(30)$

$\ T_{1}=5.774N$

This answer came out positive, meaning the direction indicated in the picture was correct and the member is in tension.

The problem has been solved, we know the force in every member of the truss and whether the member is in tension or compression. If you are having problems visualizing the problem see if the image below helps.

This picture displays every force acting on the joint in the correct orientation, you can see that the forces will be add to zero so the joint will not move.

Quick links:Statics, Civil Engineering, Mechanical Engineering

Method of Joints

The method of joints is a way to find unknown forces in a truss structure. The principle behind this method is that all forces acting on a joint must add to zero. If there were a net force, the joint would move.

Example 1

Question

Find the force in member BC of the truss pictured to the right.

Answer

Using the method of joints, the force could be found by isolating the joint at either end of the member (joint B or C). Neither joint can be solved without further analysis; however, joint B can be solved if the force in member $AB$ and $BH$ is found.

To find force $AB$ analyze joint A. This joint has an external vertical force of 300N which must be countered by the members attached to the joint. Member $AE$ cannot possibly support any vertical load, otherwise it would not be loaded axially and the entire structure would no longer be a truss. If $AE$ has no load then member $AB$ is in 300N of tension.

When joint H is analyzed it is found that the force in members $BH$ and $HC$ must be zero. The reason why neither member can carry any load is that member $BH$ can only take a vertical load and member $HC$ can only take a horizontal load. In a real world application this structure might be useful if there was a load applied at joint $H$. Now joint $B$ can be analyzed.

The picture to the left shows the forces affecting joint B.

$\sum F_{y}=0=BH-BA+BC\cos(50)-BE\cos(50)$

$\sum F_{x}=0=BC\sin(50)+BE\sin(50)$

Substitution

From analysis of joint $A$

$\ BA=300N(Tension)$

From analysis of joint $H$

$\ BH=0N$

Put values for $BA$ and $BH$ into the equilibrium equations for joint B.

$\sum F_{y}=0=0-300N+BC\cos(50)-BE\cos(50)$

$\sum F_{x}=0=BC\sin(50)+BE\sin(50)$

$\ BC\sin(50)=-BE\sin(50)$

$\ BC=-BE$

Now $BC$ can be inserted in place of $-BE$ in $\sum F_{y}$, which gives:

$2BC\cos(50)-300N=0$

Finally, $BC$ can be solved for as follows:

$BC=233.4N$

Method of Sections

The method of sections is another method to determine forces in members of a truss structure. In order to find unknown forces in using the method of sections, sections of the truss structure must be isolated. The net force on the entire isolated section must be zero since the isolated section does not move (if it did move it wouldn't be a statics problem). This method is often faster because instead of moving from joint to joint with the method of joints, the members of interest can be immediately isolated.

Example 1

Question

The truss pictured to the right is secured with pin mounts to the brown concrete block. A 100kg mass is hanging from the rightmost joint. For simplicity the weight of the trusses in the structure can be ignored. Calculate the compression in member A-B.

Answer

This problem could be solved using the method of joints, but you would have to start at the joint above the weight then solve many other joints between the first joint and member A-B. The fifth joint solved would give member A-B's loading. It is easier using the method of sections.

The method of joints isolates a joint to find unknown forces. The method of sections is the same except an entire section is isolated. It should be obvious at this point that there cannot be any net force or moment on the section, if there was the section would move.

In order to find the stress in member A-B it is convenient to isolate the section of the truss as indicated by the image to the left. In problems where the method of joints was used, a joint was isolated by cutting surrounding members. Cutting is also used to isolate a section, three of the members of the truss were cut to isolate the section in this example. Because all cut members were part of a truss, all members supported an axial force of tension or compression. After being cut, all members must have the same axial force applied to support the section.

Equilibrium equations can now be written.

$\sum F_{x}=0=-T_{1}-T_{2}\cos(60)-T_{3}$

$\sum F_{y}=0=T_{2}\sin(60)-981N$

The equation summing forces in the Y direction only has one unknown because all cut members except A-B are horizontal.

$\ 981N=T_{2}\sin(60)$

$\ T_{2}=1130N$

Because $T_{2}$ is positive, member A-B is in 1130N of tension.

Quick links:Statics

Multiforce Members

There are many structural problems in statics that do not involve a truss. For example, a beam is not loaded axially. A beam is analyzed in terms of bending moments and shear. Many simple tools can be analyzed with statics.

Example 1

Question

The picture to the right shows a tool used to compress springs. In the tool's current state a person is squeezing the pliers with 40 Newtons applied to each handle to hold the spring(shown in orange) in a steady partially compressed state. The black circle is a pin (a pin can transfer forces, but cannot resist turning). The forces used to squeeze the pliers are vertical in addition to the forces applied on the spring.

Find the compressive force on the spring and the force on the pin.

Hint

Think of how the plier components can be disassembled and analyzed for equilibrium.

Answer

If the pin in the question's picture was welded to the hole it fits in, the force from the spring could be any value; however, the pin is free to rotate (it can't resist a turning moment) so the picture to the left can be used for further analysis.

If moments are summed about the black pin (let's designate it "pin A"), the spring force(designated "F") can be solved for.

$\sum M_{A}=0=-6in*40N+2in*F$

$\ F=120N$

This means the spring is being compressed with 120N.

The force in the pin can be solved for with force equilibrium. Let's call the pin's $x$ force $P_{x}$ and the $y$ force $P_{y}$.

$\sum F_{x}=0=P_{x}$

$\sum F_{y}=0=40N+P_{y}+120N$

$\ P_{y}=-160N$

Since the equations were written in terms of the assumed force directions pictured and since $P_{y}$ is negative, the force acts in the opposite direction as shown in the picture.

The pin exerts a downward force on the handle shown in the left picture, but it exerts an upward force of the same magnitude on the other handle. This statement is very important and can be further explained by the picture to the right. The picture shows a dark gray and light gray rectangle. The light and dark gray corresponds to the previous pictures. The black rectangle represents the pin which goes through both handles as indicated with dashed lines. The dark handle is exerting an upward force of 160N on the pin and the light handle is exerting a downward force of 160N on the pin. Because the forces are equal and opposite the pin does not travel vertically. The force from the dark handle is transmitted through the pin to the light handle and the force from the light handle is transmitted through the pin to the dark handle. The pin is in 160N of shear.

Moment of Inertia (contents)

The moment of inertia can be defined as the second moment about an axis and is usually designated the symbol I. The moment of inertia is very useful in solving a number of problems in mechanics. For example, the moment of inertia can be used to calculate angular momentum, and angular energy. Moment of inertia is also important in beam design.

Shape moment of inertia for flat shapes

The area moment of inertia takes only shape into account, not mass.

It can be used to calculate the moment of inertia of a flat shape about the x or y axis when I is only important at one cross-section.

$I_{x}=\int y^{2}\,dA$

$I_{y}=\int x^{2}\,dA$

$I_{z}=\int r^{2}\,dA$

Because, for flat shapes, $r^{2}=x^{2}+y^{2}$ the following is true $I_{z}=I_{x}+I_{y}$

The shape moment of inertia of the cross-section of a beam is used in Structural Engineering in order to find the stress and deflection of the beam.

Shape moment of inertia for 3D shapes

The moment of inertia I=∫r²dm for a hoop, disk, cylinder, box, plate, rod, and spherical shell or solid can be found from this figure.

For more shapes see Mass Moments Of Inertia Of Common Geometric Shapes.

Mass moment of inertia

The mass moment of inertia takes mass into account. The mass moment of inertia of a point mass about a reference axis is equal to mass multiplied by the square of the distance from that point mass to the reference axis:

$I_{pointmass}=r^{2}m\,$

The metric units are kg*m^2.

The mass moment of inertia of any body of mass rotating around any axis is equal to the sum of the mass moment of inertia of each of the particles of that body:

$I_{m}=\sum r_{i}^{2}m_{i}$

Rather than adding up each particle individually, sometimes we can take a mathematical shortcut by integrating over all the particles:

$I_{m}=\int r^{2}\,dm$

Radius of gyration

The radius of gyration is the radius at which you could concentrate the entire mass to make the moment of inertia equal to the actual moment of inertia. If the mass of an object was 2kg, and the moment of inertia was $18kg*m^{2}$, then the radius of gyration would be 3m. In other words, if all of the mass was concentrated at a distance of 2m from the axis, then the moment of inertia would still be $18kg*m^{2}$. Radius of gyration is represented with a k.

$k={\sqrt {I/m}}\,$

The formula for the area radius of gyration replaces the mass with area.

$k={\sqrt {I/A}}\,$

Parallel axis theorem

x runs through the center of mass, while x' is parallel to x

If the moment of inertia is known about an axis that runs through the center of mass, then the moment of inertia about any parallel axis is given by,

$I_{parallel-axis}=I_{center\,of\,mass}+md^{2}$

where d is the distance between the two axis of rotation.

further reading

• Mechanics/Rigid Bodies

Center of Gravity and Centroid (contents)

To discuss the concept of the center of gravity, center of mass, and the centroid

To show how to determine the location of the center of gravity and centroid for a system of discrete particles and a body of arbitrary shape

To use the theorems of Pappas and Guldinus for finding the area and volume for a surface of revolution

To present a method for finding the resultant of a general distributed loading and show how it applies to finding the resultant of a fluid

---

9.1 - Center of Gravity and Center of Mass for a System of Particles

Center of Gravity The center of gravity G is a point which locates the resultant weight of a system of particles, if a body is in gravitational field; if not, center of gravity is a center of inertia of the body (= a system of particles). To show how to determine this point consider the system of n particles fixed within a region of space. The weights of a particle comprise a system of parallel forces which can be replaced by a single (equivalent) resultant weight having the defined point G of application.

The resultant weight must be equal to the total weight of all n particles; that is:

W_R=ΣW

Center of Gravity and Centroid (contents)

To discuss the concept of the center of gravity, center of mass, and the centroid

To show how to determine the location of the center of gravity and centroid for a system of discrete particles and a body of arbitrary shape

To use the theorems of Pappas and Guldinus for finding the area and volume for a surface of revolution

To present a method for finding the resultant of a general distributed loading and show how it applies to finding the resultant of a fluid

---

9.1 - Center of Gravity and Center of Mass for a System of Particles

Center of Gravity The center of gravity G is a point which locates the resultant weight of a system of particles, if a body is in gravitational field; if not, center of gravity is a center of inertia of the body (= a system of particles). To show how to determine this point consider the system of n particles fixed within a region of space. The weights of a particle comprise a system of parallel forces which can be replaced by a single (equivalent) resultant weight having the defined point G of application.

The resultant weight must be equal to the total weight of all n particles; that is:

W_R=ΣW

Quizzes

What does the moment equal at a cantilever end? Answer: Zero.

When we call system of forces is not equilibrium stable? Answer: When the sum of all forces doesn't equal Zero.

Condition Of Equilibrium

When forces are in equilibrium, that is, there is no net force and the summation of the particle's moment, taken at any point, is equal to 0.

$\sum F_{x}=0$

$\sum F_{y}=0$

$\sum F_{z}=0$

$\sum M_{o}=0$

Newton's Second Law of Motion

$F=ma$

Trigonometry

Sine Law

${a \over sinA}={b \over sinB}={c \over sinC}$

Cosine Law

$a^{2}=b^{2}+c^{2}-2bc\cos A\$

$b^{2}=a^{2}+c^{2}-2ac\cos B\$

$c^{2}=a^{2}+b^{2}-2ab\cos C\$

Vector Relations

Dot Product

${\vec {a}}\bullet {\vec {b}}=ab\cos \theta$

Cross Product

${\vec {a}}\times {\vec {b}}={\begin{pmatrix}a_{1}\\a_{2}\\a_{3}\end{pmatrix}}\times {\begin{pmatrix}b_{1}\\b_{2}\\b_{3}\end{pmatrix}}={\begin{pmatrix}a_{2}b_{3}-a_{3}b_{2}\\a_{3}b_{1}-a_{1}b_{3}\\a_{1}b_{2}-a_{2}b_{1}\end{pmatrix}}=\left|{\vec {a}}\right|\left|{\vec {b}}\right|\sin(\theta )\cdot {\vec {e}}$

Centroids Of Common Shapes Of Areas And Lines

Triangular Area

$Area={\frac {b*h}{2}}$

Quarter Circular Area

$Area={\frac {\pi \ r^{2}}{4}}$

Semicircular Area

$Area={\frac {\pi \ r^{2}}{2}}$

Semiparabolic Area

$Area={\frac {2ah}{3}}$

Parabolic Area

$Area={\frac {4ah}{3}}$

Parabolic Spandrel

$Area={\frac {ah}{3}}$

Circular Sector

$Area=\alpha \ r^{2}$

Quarter Circular Arc

$Area={\frac {\pi \ }{2}}$

Semi Circular Arc

$Area=\pi \ r$

Arc Of Circle

$Area=2\alpha \ r$

Area Moments Of Inertia of Common Geometric Shapes

Rectangle

$I_{x}={\frac {1}{3}}bh^{3}$

$I_{y}={\frac {1}{3}}hb^{3}$

$I_{x'}={\frac {1}{12}}bh^{3}$

$I_{y'}={\frac {1}{12}}hb^{3}$

Right Triangular Area

$I_{x}={\frac {1}{12}}bh^{3}$

$I_{y}={\frac {1}{4}}hb^{3}$

$I_{x'}={\frac {1}{36}}bh^{3}$

$I_{y'}={\frac {1}{36}}hb^{3}$

Triangular Area

$I_{x}={\frac {1}{12}}bh^{3}$

$I_{x'}={\frac {1}{36}}bh^{3}$

Circular Area

$J_{C}={\frac {\pi \ r^{4}}{2}}$

$I_{x'}=I_{y'}={\frac {\pi \ r^{4}}{4}}$

Hollow circle

This is used for hollow cylinders where there is solid material between the outer and inner radius, but no material between the inner radius and the center, like a pipe's cross-section.

$I={\frac {\pi \ (r_{o}^{4}-r_{i}^{4})}{4}}$

$r_{o}$ is the outer radius $r_{i}$ is the inner radius

Semicircular Area

$I_{x}=I_{y}={\frac {1}{8}}\pi \ r^{4}$

$I_{x'}=({\frac {\pi }{8}}-{\frac {8}{9\pi }})r^{4}$

$I_{y'}={\frac {1}{8}}\pi \ r^{4}$

Quarter Circle

$I_{x}=I_{y}={\frac {1}{16}}\pi \ r^{4}$

$I_{x'}=I_{y'}=({\frac {\pi }{16}}-{\frac {4}{9\pi }})r^{4}$

further reading

Statics/Geometric_Properties_of_Solids

Mass Moments Of Inertia Of Common Geometric Shapes

Slender Rod

$I_{x}=0$

$I_{y}=I_{z}={\frac {1}{12}}ml^{2}$

Thin Quarter-Circular Rod

$I_{x}=I_{z}=mr^{2}({\frac {1}{2}}-{\frac {4}{\pi ^{2}}})$

$I_{y}=mr^{2}(1-{\frac {8}{\pi ^{2}}})$

Thin Ring

$I_{x}=I_{y}={\frac {1}{2}}mr^{2}$

$I_{z}=mr^{2}$

Sphere

$I_{x}=I_{y}=I_{z}={\frac {2}{5}}mr^{2}$

Hemisphere

$I_{x}=I_{y}={\frac {83}{320}}mr^{2}$

$I_{z}={\frac {2}{5}}mr^{2}$

Thin Circular Disk

$I_{x}=I_{y}={\frac {1}{4}}mr^{2}$

$I_{z}={\frac {1}{2}}mr^{2}$

Rectangular Prism

$I_{x}={\frac {1}{12}}m\left(b^{2}+c^{2}\right)$

$I_{y}={\frac {1}{12}}m\left(a^{2}+c^{2}\right)$

$I_{z}={\frac {1}{12}}m\left(a^{2}+b^{2}\right)$

Right Circular Cylinder

$I_{x}=I_{y}={\frac {1}{12}}m(3r^{2}+h^{2})$

$I_{z}={\frac {1}{2}}mr^{2}$

Right Half Cylinder

$I_{x}={\frac {1}{12}}mh^{2}+mr^{2}({\frac {1}{4}}-{\frac {16}{9\pi ^{2}}})$

$I_{y}={\frac {1}{12}}mh^{2}+{\frac {1}{4}}mr^{2}$

$I_{z}=mr^{2}({\frac {1}{2}}-{\frac {16}{9\pi ^{2}}})$

Thin Rectangular Plate

$I_{x}={\frac {1}{12}}mb^{2}$

$I_{y}={\frac {1}{12}}ma^{2}$

$I_{z}={\frac {1}{12}}m(a^{2}+b^{2})$

Right Circular Cone

$I_{x}=I_{y}={\frac {3}{80}}m({4}{r^{2}}+h^{2})$

$I_{z}={\frac {3}{10}}mr^{2}$

Right Tetrahedron

$I_{x}={\frac {3}{80}}m(b^{2}+c^{2})$

$I_{y}={\frac {3}{80}}m(a^{2}+c^{2})$

$I_{z}={\frac {3}{80}}m(a^{2}+b^{2})$

further reading

• Statics/Geometric_Properties_of_Lines_and_Areas
• Mechanics/Rigid Bodies

Acknowledgements

Beer, Ferdinand P. and E Russell Johnston, Jr. "Vector Mechanics For Engineers, Statics" 3rd edition, McGraw Hill c 1977.