MchT 111 - Mechanics For Technology: Statics

MATHEMATICS REVIEW

In any technical course, math plays a vital role in obtaining correct solutions for the problems.  Statics is no exception.  The student must have a solid base in several mathematical disciplines to set-up and solve the equations of statics.  Algebra, trigonemetry, geometry, and calculus are all very important in the study of statics and beyond.  This course is intended for the first year technology student, and will therefore not require calculus.  This section is intended to be a review of the basic mathematical principles that are used exetensively in this course.  It is an overview, and is not intended to be all inclusive.  The principles involved will be presented primarily in the form of examples.

Algebra:  Most of the problems in this course rely heavily on algebra to solve the equations which are used.  The student will be required to solve equations with one variable and sets of equations with several variables.  In addition, the student should be familiar with the quadratic formula, and with natural logarithms.

Equations With One Variable:  A very common problem is an equation with just one variable.  The variable may appear once or several times throughout the equation.  The equation may be linear or non-linear, and may contain trig functions or logarithmic functions.  

The basic approach used to solve an equation with one variable is to manipulate the equation until the variable is isolated on one side of the equation, and everything esle is on the other side.  In the case of a linear equation, this can be done by adding or subtracting equals to each side, multiplying both sides by equals, or dividing both sides by equals.  Often it will require a series of operations to arrive at the final solution.  In the case of non-linear equations, other operations may have to be performed, such as taking the square root of each side or the tangent of each side.  Examples 2-1 thru 2-4 are all examples taken from statics problems, showing step by step manipulations to determine the final solution.  For several reasons, it is recommended that the student does not skip steps while solving a problem.  Skipping steps can lead to errors, and will always making checking the work more difficult.  

Example 2-1 - Linear Equation With 1 Variable Solve for X in the equation: Multiply both sides by 3: Multiply thru by 2: Subtract 24 from both sides: Add 18X to both sides: Do the arithmetic on the left side: Divide both sides by 18: Substitute 2 into the original equation to check:

Example 2-2 - Non-Linear Equation With 1 Variable Using the Quadratic Formula Solve for X in the equation: X (5X + 1.5) - 2.6 = 1.6X (2.5X+.5) Multiply thru by X on the left and 1.6X on the right: 5X2 + 1.5X - 2.6 = 4X2+.8X Subtract 4X2+.8X from both sides: X2 + .7X - 2.6 = 0 Apply the quadratic formula: Substitute values into the quadratic formula: Solve for X:     X1 = 1.3 X2 = -2 Notice that there are 2 solutions for X in this equation.  There will always be 2 solutions to a quadratic equation, however, usually only one solution will satisfy the conditions of the problem.  After the 2 solutions have been calculated, then they must be evaluated in the context of the problem to determine which solution is the appropriate one for the situation.   The 2 solutions should be substituted back into the original equation to check the work.  The checks will not be shown here, but should be done.

Example 2-4 - Non-Linear Equation With 1 Variable Involving Natural Logs Solve for X in the equation: Take the inverse ln of each side: Solve for X in the equation: X = 270.4

Equations With Several Variables:  In statics, as in most technical courses, problems arise which involve several variables.  If a system of equations has N variables, then there must be N independent equations in those variables to be able to find a solution.  There are several methods which can be used to solve such a system of equations.  The examples below show 3 ways to solve the same system of equations.  The equations used are:

One method used to solve simultaneous equations is to solve one of the equations for one of the variables, then substituting that into one of the other equations, thereby eliminating that variable from the second equation.  That process continues until there is only one variable remaining in an equation.  The lone variable is solved for, then back substituted into one of the other equations, until all of the solutions have been found.

The same system of equations can be solved by using another method of eliminating variables in the equations.  In order to take full advantage of this second method, it is important to understand the concept of equivalent systems of equations.  Two  systems of equations are equivalent if they have precisely the same solution set. There are three operations that can be performed on any system of equations which will produce equivalent systems:

1.) Interchange two equations.
2.) Multiply an equation by a nonzero constant.
3.) Add a multiple of an equation to another equation.

The first of these is important in other methods of solving a system of equations, but will not be used here.  The method that will be used in the next example involves eliminating variables by repeatedly using the second and third operations above.  

The same system of equations can be solved by using a third method known as Cramer's Rule.  This method is efficient for systems of equations involving three variables, but is very tedious for larger systems.  However, this method can be programmed fairly easily, so it lends itself well to computer solutions of larger systems.  Cramer's rule involves finding the determinant of matrices.  Therefore, a brief review will be given for finding the determinant of matrices of order 2 and order 3.  Refer to any linear algebra text for other orders.

A matrix is an array of numbers arranged in row and column format.  For applying Cramer's rule, the matrices will be square, meaning that the number of rows and the number of columns is the same.  A matix is represented by the array shown inside a set of brackets:

The determinant of a matrix is represented by the same array shown inside a set of vertical line:

so we can write:

The determinant of a matrix is a number, which is evaluated by manipulating the numbers in the array. 

Determinant of a matrix of order 2:
To evaluate the determinant of a matrix of order 2, use:

Example 2-6 - Determinant of a Matrix of Order 2 Find the determinant of the following matrix: A = |A| = (2)(2) - (4)(1) |A| = 0

Determinant of a matrix of order  3:
To evaluate the determinant of a matrix of order 3, it is necessary to define two new terms, minors and cofactors for a square matrix.  The minor M_ij the element a_ij is the determinant of the matrix which is left if row i and column j are deleted.  The cofactor C_ij is then given by the expression  C_ij = (-1)^i+j M_ij.

Or for:  

Cofactors will always be either +1 or -1 times the minor, and can be determined from the following matrix:

This shows that the cofactor for a₁₁ = C₁₁ = (+1)M₁₁.

Having defined these two terms, it is now possible to define a procedure for evaluating the determinant of a matrix of order 3:

|A| = a₁₁ (C₁₁) + a₁₂ (C₁₂) + a₁₃ (C₁₃) 

Example 2-7 - Determinant of a Matrix of Order 3 Find the determinant of the following matrix: A = det A = |A| = a11 (C11) + a12 (C12) + a13 (C13)

With this background, it is now possible to discuss Cramer's Rule.  This will be done by way of an example using a matrix of order 3.  Refer to a linear algebra text for applications of Cramer's Rule to other size matrices.

Example 2-8 - Cramer's Rule for a Systems of Equations Write down all of the equations: -X + 2Y - 3Z = 1   (1) 2X + Z = 0   (2) 3X - 4Y + 4Z = 2   (3) Set up a "coefficient matrix".  This is a matrix with the coefficients of X in column 1, coefficients of Y in column 2, and coefficients of Z in column 3:       A = Calculate the determinant of the coefficent matrix using the method shown in Example 2-7: |A| = 10 Set up 3 more matrices by replacing individual columns with the values of the constants on the right side of the equations: Calculate the determinants of these matrices using the method shown in Example 2-7: |A1| = 8 |A2| = -15 |A3| = -16 Apply Cramer's Rule: So, the final solutions are: X = .8 Y = -1.5 Z = -1.6

Trigonometry:  Trigonometry is used extensively in statics.  Trigonometry is the study of relationships in triangles. These relationships include functions of angles such as sine and cosine, and relationships involving the lengths of the sides, such as the Pythagorean theorem.

Basic Trigonometric Functions:  The three basic trig functions used extensively in statics are sine, cosine, and tangent.  Consider the triangle below:

Also, Pythagorean theorem: C2 = A2 + B2

It is important to recognize that these relationships only hold for right triangles.  In statics, right triangles appear quite regularly, so these relationships will be used often, however, there is also a need for relationships between angles and sides on triangles which do not have a right angle.  Those relationships are called the Law of Sines and the Law of Cosines. Consider the triangle below:

Example 2-9 - Trig. Functions For the triangle on the left, find a and b. This is a right triangle, and both a & b can be determined using basic trig. functions:                                                   a = 4.284 M Now that a is known, the tangent function can be used again to find b: 4.284 tan 45 = 3 + b sin = 6(sin 20)/2.6 b = 1.284

Example 2-10 - Law of Cosines & Law of Sines For the triangle on the left, find the length of the unknown side and the angle . Since two sides and the angle between them are given, this is an example of a problem suited to the Law of Cosines.  Call the side opposite the given angle C, and apply the Law of Cosines:                                                   C2 = A2 + B2 + 2AB cos c C2 = 62 + 42 + 2(6)(4) cos 20 C2 = 6.9 C = 2.6 ft. The Law of Sines can now me used to find the angle . 6/sin = 2.6/sin 20 sin = 6(sin 20)/2.6 sin = 0.789 = 52.1o

Geometry:  There are several geometric relationships which will be needed in this course.

Intersecting lines create equal angles as shown below.
	a = c ; b = d
Angles a and b below are supplementary angles. Supplementary angles total 180o.
	a + b = 180o
When a straight line intersects two parallel lines, it produces equal angles as shown below.
	a = d = e = h b = c = f = g
The sum of the angles in a triangle is 180o.
	a + b + c = 180o
Circle relationships:

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