# ANALYTICAL METHODS FOR ENGINEERS

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1 UNIT 1: Unit code: QCF Level: 4 Credit value: 15 ANALYTICAL METHODS FOR ENGINEERS A/601/1401 OUTCOME - TRIGONOMETRIC METHODS TUTORIAL 1 SINUSOIDAL FUNCTION 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. 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. D.J.Dunn 1

4 INVERSE FUNCTIONS Some people find it useful to use the inverse functions which are as follows. cosec (θ) = sin -1 (θ) sec (θ) = cos -1 (θ) cot (θ) = tan -1 (θ) 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 a b It follows that sina sinb If we did the same for another perpendicular to side b or a we could show that a sina b sinb c sinc WORKED EXAMPLE No. Find the length of the two unknown side in the triangle shown. SOLUTION a = 50 mm A = 30 o B = 45 o C = 180 o - 30 o - 45 o = 105 o a b c 50 b c sina sinb sinc sin(30) sin(45) sin(105) 50 sin(45) 50 sin(105) b mm c mm sin (30) sin (30) 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 110 o. 4/sin 110 = L 1 /sin 50 = L /sin 0 L 1 = 3.61 m and L = m Next draw the triangle of forces as shown. F 1 /sin 40 o = 300/sin 70 o F 1 = 05. N F /sin 70 o = 300/sin 70 o F 1 = 300 N D.J.Dunn 4

5 COSINE RULE Consider the diagram. Using Pythagoras we have: h = a (b x) and h = c x a (b + x bx) = c x a b - x + bx = c x a = b + x bx + c x a = b + c bx substitute x = c cos(a) a = b + c bc cos(a) b c a cos(a) bc If we repeated the process with h drawn normal to the other sides we could show that : c a b a b c cos(b) cos(c) ca ab 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 a b c cos(c) ab o cos(60 )()(60)(70) c c = 4300 c = mm o c cos(60 ) ()(60(70) 70 c b c a cos(a) bc ()(70)(65.574) A = 5.4 o B = = 16.5 o 0.61 WORKED EXAMPLE No. 5 Find the resultant of the two forces shown. SOLUTION The addition of the two force is done as shown. o R cos(135 ) ()(100(100) o cos(135 )()(100)(100) = 0000 R R = R = N R D.J.Dunn 5

6 SELF ASSESSMENT EXERCISE No. 1. Find the resultant of the two forces shown. (Answer 38.8 N). Vector A has polar coordinates 1 60 o and vector B has polar coordinates 5 0 o Find the resultant in polar form. ( o ) 3. The diagram shows a weight suspended from two ropes. Calculate the angles of the ropes to the horizontal support. (Answers 49 o and 59 o ) 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, 3.59 T and.93 T) D.J.Dunn 6

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. 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 θ = x R sin(ωt). 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 Velocity Acceleration x = R sin(t) v = dx/dt = R cos(t) a = dv/dt = - R sin(t) = - 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 = - R sin(t + ) = - x The plots show the displacement, velocity and acceleration for = 0 on the left and a negative on the right. D.J.Dunn 7

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 = x o sin (ωt) The velocity of the mass is v = ω x o sin(ωt) The acceleration of the mass is a = -ω x o cos(ωt) It can be shown for the frictionless case that ω = (k/m) 1/ 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 f 1 π k m CENTRIFUGAL FORCE When a mass rotates at radius R the centrifugal force is given by: CF = mω R 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: mω R sin(θ) = mω R sin(ωt) D.J.Dunn 8

9 WORKED EXAMPLE No. 6 The displacement of a body performing simple harmonic motion is described by the following equation x = A sin (t + ) where A is the amplitude, is the natural frequency and is the phase angle. Given A = 0 mm, = 50 rad/s and = /8 radian, calculate the following. i. The frequency. ii. The periodic time. iii. The displacement, velocity and acceleration when t = T/4. Sketch the graphs of x, v and a and confirm your answers. SOLUTION First deduce the frequency. f = / = 50/ = 7.96 Hz. Next deduce the periodic time. T = 1/f = 0.16 s Next deduce the time t. t = T/4 = s Next write out the equation for displacement and solve x at t = s π x 0 sin (50 x 0.314) 8 π x 0sin sin mm 8 Next write down the equations for v and a a - 0ω x 0sin ωt v 0ω cos φ ωt φ 0 x 50 x cos mm/s φ - 0 x 50 sin(1.963) mm/s sin ωt The plots of x, v and a confirm these answers. D.J.Dunn 9

10 WORKED EXAMPLE No. 7 A spring of stiffness 0 kn/m supports a mass of 4 kg. The mass is pulled down 8 mm and released to produce linear oscillations. Calculate the frequency and periodic time. Sketch the graphs of displacement, velocity and acceleration. Calculate the displacement, velocity and acceleration 0.05 s after being released. SOLUTION k 0000 ω rad/s M 4 ω f 11.5 Hz π 1 T s f The oscillation starts at the bottom of the cycle so x o = -8 mm. The resulting graph of x against time will be a negative cosine curve with an amplitude of 8 mm. The equations describing the motion are as follows. x = xo cost When t = 0.05 seconds x = -8 cos(70.71 x 0.05) x = mm. (Note angles are in radian) This is confirmed by the graph. If we differentiate once we get the equation for velocity. v = -x o sin t v = -x o sin t = (-8)sin(70.71 x 0.05) v = -17 mm/s This is confirmed by the graph. Differentiate again to get the acceleration. a = - x o cost and since x = x o cost a = - x a = x = mm/s This is confirmed by the graph. D.J.Dunn 10

11 SELF ASSESSMENT EXERCISE No Calculate the frequency and periodic time for the oscillation produced by a mass spring system given that the mass is 0.5 kg and the spring stiffness is 3 N/mm. (1.3 Hz, s).. A mass of 4 kg is suspended from a spring and oscillates up and down at Hz. Determine the stiffness of the spring. (631.6 N/m). The amplitude of the oscillation is 5 mm. Determine the displacement, velocity and acceleration 0.0 s after the mass passes through the mean or rest position in an upwards direction. (1.43 mm, mm/s and mm/s ) 3. From recordings made of a simple harmonic motion it is found that the frequency is Hz and that at a certain point in the motion the velocity is 0.3 m/s and the displacement is 0 mm, both being positive downwards in direction. Determine the amplitude of the motion and the maximum velocity and acceleration. Write down the equations of motion. Note that the data given is at time t = 0. You will have to assume that x = x o cos(t + ) at time t=0 Ans. x= cos(4t - 50 o ) v = sin(4t - 50 o ) a = x D.J.Dunn 11

12 ELECTRICAL EXAMPLES ALTERNATING ELECTRICITY Electricity is generated by rotating a conductor relative to a magnetic field at angular velocity ω rad/s (very simplified case shown). The voltage generated is directly proportional to the angle of rotation. This explains why the voltage in our mains electrical system is sinusoidal. The voltage at any moment in time is given by the equation v = V sin(ωt) where V is the maximum voltage (amplitude) in the cycle and ω the angular velocity or frequency. If we choose to measure the angle from a different starting point then v = V sin(ωt + ) where is the starting angle. RESISTANCE When a sinusoidal voltage is applied across a resistor the current is sinusoidal and in phase with the voltage. V sinωt i R CAPACITANCE When a sinusoidal voltage is applied across a capacitor C the current is given by: dv i C ωcv cosωt dt INDUCTANCE When a sinusoidal voltage is applied across an inductor L the current is cosωt given by: i V ωl If we plot these we see that the current in the capacitor is displaced -90 o from that in the resistor and the current in the inductor is displaced +90 o. This is a similar relationship to that of the displacement, velocity and acceleration in a mechanical system. D.J.Dunn 1

13 SELF ASSESSMENT EXERCISE No Mains electricity has a frequency of 50 Hz. What is the periodic time and angular frequency? (0.0 s and 314 rad/s). An alternating current has a periodic time of s. What is the frequency? (400 Hz) 3. A alternating voltage has a peak to peak amplitude of 300 V and frequency of 50 Hz. What is the amplitude? (150 V) What is the voltage at t = s? (106 V) 4. Determine the following from the graph shown. The amplitude. The offset displacement. The periodic time. The frequency. The angular frequency. (Answers 5,, 1.57 s, Hz, 4 rad/s) 5. A resistor of value 10 Ω, a capacitor of value 40 μf and an inductor of value 10 mh are all connected in parallel to a voltage source as shown. The voltage is 50 sin(000t). Determine an expression for the current drawn from the source. Determine the peak current. Determine the phase of the source current. 3cos 000t i 5 sin000t (5. A, 0.9 radian) D.J.Dunn 13

14 6. COMPLEX WAVEFORMS FUNDAMENTAL FREQUENCY A sinusoidal voltage or current is described by the mathematical formula v = V sin t or i = I sin t The sinusoidal voltage formula is then v = V sin(ft) frequency. In this formula f is the fundamental HARMONICS A harmonic is a multiple of the fundamental frequency. f is the second harmonic. 3f is the third harmonic nf is the n th harmonic. SYNTHESISING COMPLEX WAVES Waveforms with shapes that are not sinusoidal may be synthesised from one common sinusoidal waveform. The proof of this is not given here but the following is mathematically correct. This graph shows the result of adding the first and third harmonic with equal amplitudes. In reality the amplitude of the harmonic is likely to be less than the amplitude of the fundamental. This graph shows the affect of adding the third harmonic with 1/3 of the amplitude. GENERATION OF HARMONICS Harmonics are generated when a sinusoidal signal passes through a non-linear amplifier. An ideal amplifier increases a sinusoidal signal perfectly. D.J.Dunn 14

15 SQUARE WAVES Square waveforms are really d.c. levels that suddenly change from plus to minus. It can be shown that the following formula relates voltage and time. The formula is an infinite series. V V V v Vsin(ω t) sin3(ω t) sin5(ω t) sin7(ω t) TRIANGULAR WAVES It can be shown that the following formula relates voltage and time. The formula is an infinite series. V V V v Vsin(ω t) sin(3ω t π) sin(5ω t) sin(7ω t π) Note that in this series, a phase shift of radians is added to each D.J.Dunn 15

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