QCF Level: 4 Credit value: 15 OUTCOME 2 - …

1 1 UNIT 1: ANALYTICAL METHODS FOR ENGINEERS Unit code: A/601/1401 QCF Level: 4 Credit value: 15 OUTCOME 2 - TRIGONOMETRIC METHODS TUTORIAL 1 SINUSOIDAL FUNCTION 2 Be able to analyse and model engineering situations and solve problems using trigonometric methods Sinusoidal functions: review of the trigonometric ratios; Cartesian and polar co-ordinate systems; properties of the circle; radian measure; sinusoidal functions Applications: angular velocity, angular acceleration, centripetal force, frequency, amplitude, phase, the production of complex waveforms using sinusoidal graphical synthesis, AC waveforms and phase shift Trigonometric identities: relationship between trigonometric and hyperbolic identities; double angle and compound angle formulae and the conversion of products to sums and differences; use of trigonometric identities to solve trigonometric equations and simplify trigonometric expressions You should judge your progress by completing the self assessment exercises.

2 Trigonometry has been covered in the NC Maths module and should have been studied prior to this module. This tutorial provides further studies and applications of that work. 2 1. RADIAN In Engineering and Science, we use the radian to measure angle as well as degrees. This is defined as the angle created by placing a line of length 1 radius around the edge of the circle as shown. In mathematical words it is the angle subtended by an arc of length one radius. This angle is called the RADIAN. The circumference of a circle is 2 R. It follows that the number of radians that make a complete circle is RR 2 or 2 . There are 2 radians in one revolution so 360o =2 radian 1 radian = 360/2 = 2. TWO DIMENSIONAL COORDINATE SYSTEMS CARTESIAN In a two dimensional system the vertical direction is usually y (positive up) and the horizontal is direction is x (positive to the right).

3 Other letters may be used to designate an axis and they don t have to be vertical and horizontal. The origin o is where the axis cross at x = 0 and y = 0 A point p on this plane has coordinates x, y and this is usually written as p (x,y) POLAR If a line is drawn from the origin to point p it is a radius R and forms an angle with the x axis. The angle is positive measured from the x axis in a counter clockwise direction. A vector with polar coordinates is denoted R CONVERSION The two systems are clearly linked as we can convert from one to the other using trigonometry and Pythagoras theorem. y = R sin x = R cos y/x = tan R = (x2 + y2) WORKED EXAMPLE No. 1 The x, y coordinates of a point is 4, and 6. Calculate the polar coordinates. SOLUTION R = (42 + 62)1/2 = = tan-1 (6/4) = 3 SELF ASSESSMENT EXERCISE No.

4 1 1. Convert 60o to radian. ( rad) 2. Convert /6 radian into degrees. (30o) 3. The x, y Cartesian coordinates of a vector are 2 and the vector in Polar coordinates. ( 74o) 4. A vector with Cartesian coordinates 10, 20 is added to a vector with coordinates -20,10. What are the polar coordinates of the resulting vector? ( ) 3. REVIEW OF TRIGONOMETRIC RATIOS The ratios of the lengths of the sides of a right angle triangle are always the same for any given angle . These ratios are very important because they allow us to calculate lots of things to do with triangles. In the following the notation above is used with the corners denoted AB and C SINE The ratio ABCBH ypotenuseOpposite is called the sine of the angle A. (note we usually drop the e on sine) Before the use of calculators, the values of the sine of angles were placed in tables but all you have to do is enter the angle into your calculator and press the button shown as sin.

5 For example if you enter 60 into your calculator in degree mode and press sin you get If you enter into your calculator in radian mode and press sin you get Note that sin( ) = -sin (- ) and sin (180 ) = sin( ) COSINE The ratio ABACH ypotenuseAdjacent is called the cosine of the angle A. On your calculator the button is labelled cos. For example enter 60 into your calculator in degree mode and press the cos button. You should obtain If you enter into your calculator in radian mode and press cos you get Note that cos( ) = cos (- ) and cos(180 ) = -cos( ) TANGENT The ratio ACBCA djacentOpposite is called the tangent of the angle A. On your calculator the button is labelled tan. For example enter 60 into your calculator in degree mode and press the tan button and you should obtain If you enter into your calculator in radian mode and press tan you get Note that sin( )/cos( ) = tan( ) and tan(90 ) = 1/tan( ) 4 INVERSE FUNCTIONS Some people find it useful to use the inverse functions which are as follows.

6 Cosec ( ) = sin-1( ) sec ( ) = cos-1( ) cot ( ) = tan-1( ) 4. SINE AND COSINE RULE The following work enables us to solve triangles other than right angles triangles. SINE RULE Consider the diagram. h = b sin A = a sin B It follows that sinBbsinAa If we did the same for another perpendicular to side b or a we could show that sinCcsinBbsinAa WORKED EXAMPLE No. 2 Find the length of the two unknown side in the triangle shown. SOLUTION a = 50 mm A = 30o B = 45o C = 180 o - 30o - 45o = 105o sinCcsinBbsinAa sin(105)csin(45)bsin(30)50 mm (30)sin sin(45) 50b mm (30)sin sin(105) 50c WORKED EXAMPLE No. 3 A weight of 300 N is suspended on two ropes as shown. Calculate the length of the ropes Draw the vector diagram for the three forces in equilibrium. Calculate the forces in the ropes. SOLUTION The third internal angle is 110o. 4/sin 110 = L1/sin 50 = L2/sin 20 L1 = m and L2 = m Next draw the triangle of forces as shown.

7 F1/sin 40o = 300/sin 70o F1 = N F2/sin 70o = 300/sin 70o F1 = 300 N 5 COSINE RULE Consider the diagram. Using Pythagoras we have: h2 = a2 (b x)2 and h2 = c2 x2 a2 (b2 + x2 2bx) = c2 x2 a2 b2 - x2 + 2bx = c2 x2 a2 = b2 + x2 2bx + c2 x2 a2 = b2 + c2 2bx substitute x = c cos(A) a2 = b2 + c2 2bc cos(A) 2bcacbcos(A)222 If we repeated the process with h drawn normal to the other sides we could show that : 2cabaccos(B)222 2abcbacos(C)222 You can see a pattern for remembering the formulae. This is a useful formula for solving a triangle with three known sides or two known sides and the angle opposite the unknown side. WORKED EXAMPLE No. 4 Find the length of the unknown side in the triangle shown. Find the other internal angles. SOLUTION 2abcbacos(C)222 (2)(60(70)c0706)cos(60222o 222oc07060))(2)(60)(7cos(60 2c85002004 c2 = 4300 c = mm ) )(70)(2( (A)222222 A = B = 180 60 = o WORKED EXAMPLE No.)

8 5 Find the resultant of the two forces shown. SOLUTION The addition of the two force is done as shown. 0)(2)(100(10R010010)cos(135222o 222oR010010100))(2)(100)(cos(135 = 20000 R2 R2 = R = N 6 SELF ASSESSMENT EXERCISE No. 2 1. Find the resultant of the two forces shown. (Answer N) 2. Vector A has polar coordinates 12 60o and vector B has polar coordinates 5 20o Find the resultant in polar form. ( ) 3. The diagram shows a weight suspended from two ropes. Calculate the angles of the ropes to the horizontal support. (Answers 49o and 59o) 4. A weight of 4 Tonne is suspended on two ropes as shown. Calculate the length of the ropes and the forces in them. (Answers m, m, T and T) 7 5. SINUSOIDAL FUNCTIONS In Nature and in Engineering there are many things that oscillate in some form or other and produce a repetitive change of some quantity with respect to time.)

9 Examples are mechanical oscillations and alternating electricity. In many cases a plot of the quantity against time produces a sinusoidal graph and the change is said to be sinusoidal. MECHANICAL EXAMPLES SCOTCH YOKE and ECCENTRIC CAM The Scotch Yoke is a device that produces up and down motion when the wheel is rotated. The displacement of the yoke from the horizontal position is x = R sin = x R sin( t). Plotting x against time or angle will produce a sinusoidal graph. The eccentric cam is really another version of this. In all cases we should remember that velocity is the first derivative of displacement and acceleration is the second derivative. It follows that: Displacement x = R sin( t) Velocity v = dx/dt = R cos( t) Acceleration a = dv/dt = - 2R sin( t) = - 2 x Anything that obeys these equations is said to have SIMPLE HARMONIC MOTION The starting point of the oscillation could be at any angle so in that case the equations become: Displacement x = R sin( t + ) Velocity v = dx/dt = R cos( t + ) Acceleration a = dv/dt = - 2R sin( t + ) = - 2 x The plots show the displacement, velocity and acceleration for = 0 on the left and a negative on the right.

10 8 MASS ON A SPRING A mass on the end of a spring will oscillate up and down and produce identical motion to the Scotch Yoke without the rotation of a wheel. The displacement is x = xo sin ( t) The velocity of the mass is v = xo sin( t) The acceleration of the mass is a = - 2 xo cos( t) It can be shown for the frictionless case that = (k/m)1/2 where k is the spring stiffness in N/m and m the mass in kg. This is called the natural frequency. The natural frequency of oscillation is hence mk2 1f CENTRIFUGAL FORCE When a mass rotates at radius R the centrifugal force is given by: CF = m 2R If this is a machine mounted on a platform as shown that can only move in one direction, the force acting in that direction is the component of the force in that direction. In this example the force exerted on the spring is: m 2R sin( ) = m 2R sin( t) 9 WORKED EXAMPLE No.