Revision Test 6 : Quadratics, logarithms and exponentials
17.5 Practical problems involving straight line graphs
142 Basic Engineering Mathematics
horizontal lineBDmeets the vertical axis indicates the equivalent Fahrenheit temperature.
Hence,55◦C is equivalent to 131◦F.
This process of finding an equivalent value in between the given information in the above table is calledinterpolation.
(b) To find the Celsius temperature at 167◦F, a horizontal line EF is constructed as shown in Figure 17.17. The point where the vertical lineFG cuts the horizontal axis indicates the equivalent Celsius temperature.
Hence, 167◦F is equivalent to 75◦C.
(c) If the graph is assumed to be linear even outside of the given data, the graph may be extended at both ends (shown by broken lines in Figure 17.17).
From Figure 17.17,0◦C corresponds to 32◦F.
(d) 230◦F is seen to correspond to 110◦C.
The process of finding equivalent values outside of the given range is calledextrapolation.
Problem 13. In an experiment on Charles’s law, the value of the volume of gas,Vm3, was measured for various temperaturesT◦C. The results are shown below.
Vm3 25.0 25.8 26.6 27.4 28.2 29.0 T◦C 60 65 70 75 80 85 Plot a graph of volume (vertical) against temperature (horizontal) and from it find (a) the temperature when the volume is 28.6 m3and (b) the volume when the temperature is 67◦C
If a graph is plotted with both the scales starting at zero then the result is as shown in Figure 17.18. All of the points lie in the top right-hand corner of the graph, making interpolation difficult. A more accurate graph is obtained if the temperature axis starts at 55◦C and the volume axis starts at 24.5 m3. The axes correspond-ing to these values are shown by the broken lines in Figure 17.18 and are calledfalse axes, since the origin is not now at zero. A magnified version of this relevant part of the graph is shown in Figure 17.19. From the graph,
(a) When the volume is 28.6 m3, the equivalent tem-perature is82.5◦C.
(b) When the temperature is 67◦C, the equivalent volume is26.1 m3.
30 25 20 15 Volume (m3)
10 5
0 20 40 60 80 100
Temperature (8C) y
x
Figure 17.18
29 28.6
28
27 Volume (m3)
26
25
55 60 65 67 70 75 80 82.5 85 26.1
Temperature (⬚C) y
x
Figure 17.19
Problem 14. In an experiment demonstrating Hooke’s law, the strain in an aluminium wire was measured for various stresses. The results were:
Stress (N/mm2) 4.9 8.7 15.0 Strain 0.00007 0.00013 0.00021 Stress (N/mm2) 18.4 24.2 27.3 Strain 0.00027 0.00034 0.00039
Straight line graphs 143
Plot a graph of stress (vertically) against strain (horizontally). Find (a) Young’s modulus of elasticity for aluminium, which is given by the gradient of the graph, (b) the value of the strain at a stress of 20 N/mm2and (c) the value of the stress when the strain is 0.00020
The co-ordinates (0.00007, 4.9), (0.00013, 8.7), and so on, are plotted as shown in Figure 17.20. The graph produced is the best straight line which can be drawn corresponding to these points. (With experimen-tal results it is unlikely that all the points will lie exactly on a straight line.) The graph, and each of its axes, are labelled. Since the straight line passes through the ori-gin, stress is directly proportional to strain for the given range of values.
Stress (N/mm2)
0.00005 0 4 8 12 14 16 20 24 28
0.00015 0.00025
Strain 0.0002850.00035 y
x A
C B
Figure 17.20
(a) The gradient of the straight lineACis given by AB
BC = 28−7
0.00040−0.00010= 21 0.00030
= 21
3×10−4 = 7
10−4=7×104
=70 000 N/mm2
Thus, Young’s modulus of elasticity for alu-minium is 70 000 N/mm2. Since 1 m2=106mm2, 70 000 N/mm2is equivalent to 70 000×106N/m2, i.e.70×109N/m2(or pascals).
From Figure 17.20,
(b) The value of the strain at a stress of 20 N/mm2is 0.000285
(c) The value of the stress when the strain is 0.00020 is14 N/mm2.
Problem 15. The following values of resistance Rohms and corresponding voltageV volts are obtained from a test on a filament lamp.
Rohms 30 48.5 73 107 128
V volts 16 29 52 76 94
Choose suitable scales and plot a graph withR representing the vertical axis andV the horizontal axis. Determine (a) the gradient of the graph, (b) the Raxis intercept value, (c) the equation of the graph, (d) the value of resistance when the voltage is 60 V and (e) the value of the voltage when the resistance is 40 ohms. (f) If the graph were to continue in the same manner, what value of resistance would be obtained at 110 V?
The co-ordinates (16, 30), (29, 48.5), and so on are shown plotted in Figure 17.21, where the best straight line is drawn through the points.
147 140
120
8580
60 100
Resistance R (ohms)
40
10 20
24
0 20 40 60 80 100 110 120
Voltage V (volts) y
C B
A
x
Figure 17.21
(a) The slope or gradient of the straight line AC is given by
AB
BC =135−10 100−0 =125
100=1.25
144 Basic Engineering Mathematics
(Note that the vertical lineAB and the horizon-tal lineBC may be constructed anywhere along the length of the straight line. However, calcula-tions are made easier if the horizontal lineBC is carefully chosen; in this case, 100.)
(b) TheR-axis intercept is atR=10 ohms(by extra-polation).
(c) The equation of a straight line isy=mx+c, when yis plotted on the vertical axis andxon the hori-zontal axis.m represents the gradient and cthe y-axis intercept. In this case, Rcorresponds toy, V corresponds tox,m=1.25 andc=10. Hence, the equation of the graph isR=(1.25 V+10) . From Figure 17.21,
(d) When the voltage is 60 V, the resistance is85. (e) When the resistance is 40 ohms, the voltage is24V.
(f ) By extrapolation, when the voltage is 110 V, the resistance is147.
Problem 16. Experimental tests to determine the breaking stressσof rolled copper at various temperaturestgave the following results.
Stressσ(N/cm2) 8.46 8.04 7.78 Temperaturet(◦C) 70 200 280 Stressσ(N/cm2) 7.37 7.08 6.63 Temperaturet(◦C) 410 500 640 Show that the values obey the lawσ=at+b, whereaandbare constants, and determine approximate values foraandb. Use the law to determine the stress at 250◦C and the temperature when the stress is 7.54 N/cm2.
The co-ordinates (70, 8.46), (200, 8.04), and so on, are plotted as shown in Figure 17.22. Since the graph is a straight line then the values obey the lawσ=at+b, and the gradient of the straight line is
a=AB
BC =8.36−6.76
100−600 = 1.60
−500=−0.0032 Vertical axis intercept,b=8.68
Hence, the law of the graph isσ=0.0032t+8.68 When the temperature is 250◦C, stressσ is given by
σ= −0.0032(250)+8.68=7.88 N/cm2
Temperature t (8C) Stress (N/cm2)
8.68
8.36 8.50
8.00
7.50
7.00
6.50
0 100 200 300 400 500 600 700 6.76
y
x C
B A
Figure 17.22
Rearrangingσ= −0.0032t+8.68 gives 0.0032t=8.68−σ, i.e. t=8.68−σ
0.0032 Hence, when the stress,σ = 7.54 N/cm2,
temperature,t=8.68−7.54
0.0032 =356.3◦C
Now try the following Practice Exercise Practice Exercise 69 Practical problems involving straight line graphs (answers on page 347)
1. The resistanceR ohms of a copper winding is measured at various temperaturest◦C and the results are as follows:
R(ohms) 112 120 126 131 134
t◦C 20 36 48 58 64
Plot a graph ofR(vertically) againstt (hori-zontally) and find from it (a) the temperature when the resistance is 122 and (b) the resistance when the temperature is 52◦C.
2. The speed of a motor varies with armature voltage as shown by the following experi-mental results.
Straight line graphs 145
n(rev/min) 285 517 615 V (volts) 60 95 110 n(rev/min) 750 917 1050 V (volts) 130 155 175 Plot a graph of speed (horizontally) against voltage (vertically) and draw the best straight line through the points. Find from the graph (a) the speed at a voltage of 145 V and (b) the voltage at a speed of 400 rev/min.
3. The following table gives the force F newtons which, when applied to a lifting machine, overcomes a corresponding load of L newtons.
ForceF(newtons) 25 47 64 LoadL(newtons) 50 140 210
ForceF(newtons) 120 149 187 LoadL(newtons) 430 550 700 Choose suitable scales and plot a graph of F(vertically) againstL(horizontally). Draw the best straight line through the points.
Determine from the graph (a) the gradient, (b) theF-axis intercept, (c) the equation of the graph,
(d) the force applied when the load is 310 N, and
(e) the load that a force of 160 N will overcome.
(f ) If the graph were to continue in the same manner, what value of force will be needed to overcome a 800 N load?
4. The following table gives the results of tests carried out to determine the breaking stress σof rolled copper at various temperatures,t.
Stressσ(N/cm2) 8.51 8.07 7.80 Temperaturet(◦C) 75 220 310
Stressσ(N/cm2) 7.47 7.23 6.78 Temperaturet(◦C) 420 500 650 Plot a graph of stress (vertically) against temperature (horizontally). Draw the best straight line through the plotted co-ordinates.
Determine the slope of the graph and the vertical axis intercept.
5. The velocityv of a body after varying time intervalstwas measured as follows:
t(seconds) 2 5 8 v(m/s) 16.9 19.0 21.1
t(seconds) 11 15 18 v(m/s) 23.2 26.0 28.1 Plotvvertically andthorizontally and draw a graph of velocity against time. Determine from the graph (a) the velocity after 10 s, (b) the time at 20 m/s and (c) the equation of the graph.
6. The massmof a steel joist varies with length Las follows:
mass,m(kg) 80 100 120 140 160 length,L(m) 3.00 3.74 4.48 5.23 5.97 Plot a graph of mass (vertically) against length (horizontally). Determine the equa-tion of the graph.
7. The crushing strength of mortar varies with the percentage of water used in its prepara-tion, as shown below.
Crushing strength, 1.67 1.40 1.13 F(tonnes)
% of water used,w% 6 9 12 Crushing strength, 0.86 0.59 0.32 F(tonnes)
% of water used,w% 15 18 21
146 Basic Engineering Mathematics
Plot a graph of F (vertically) against w (horizontally).
(a) Interpolate and determine the crushing strength when 10% water is used.
(b) Assuming the graph continues in the same manner, extrapolate and deter-mine the percentage of water used when the crushing strength is 0.15 tonnes.
(c) What is the equation of the graph?
8. In an experiment demonstrating Hooke’s law, the strain in a copper wire was measured for various stresses. The results were
Stress 10.6×10618.2×10624.0×106 (pascals)
Strain 0.00011 0.00019 0.00025 Stress 30.7×106 39.4×106 (pascals)
Strain 0.00032 0.00041
Plot a graph of stress (vertically) against strain (horizontally). Determine
(a) Young’s modulus of elasticity for cop-per, which is given by the gradient of the graph,
(b) the value of strain at a stress of 21×106Pa,
(c) the value of stress when the strain is 0.00030,
9. An experiment with a set of pulley blocks gave the following results.
Effort,E(newtons) 9.0 11.0 13.6 Load,L(newtons) 15 25 38 Effort,E(newtons) 17.4 20.8 23.6 Load,L(newtons) 57 74 88 Plot a graph of effort (vertically) against load (horizontally). Determine
(a) the gradient,
(b) the vertical axis intercept, (c) the law of the graph,
(d) the effort when the load is 30 N, (e) the load when the effort is 19 N.
10. The variation of pressure pin a vessel with temperatureT is believed to follow a law of the formp=aT+b, whereaandbare con-stants. Verify this law for the results given below and determine the approximate values ofa andb. Hence, determine the pressures at temperatures of 285 K and 310 K and the temperature at a pressure of 250 kPa.
Pressure, p(kPa) 244 247 252 Temperature,T (K) 273 277 282 Pressure, p(kPa) 258 262 267 Temperature,T (K) 289 294 300