1. Three forces acting on a fixed point are represented by the sides of a triangle of dimensions 7.2 cm, 9.6 cm and 11.0 cm. Determine the angles between the lines of action and the three forces.
2. A vertical aerialAB, 9.60 m high, stands on ground which is inclined 12◦to the horizontal. A stay connects the top of the aerialAto a pointCon the ground 10.0 m downhill fromB, the foot of the aerial. Determine (a) the length of the stay, and (b) the angle the stay makes with the ground.
3. A reciprocating engine mechanism is shown in Fig. 26.20. The crankABis 12.0 cm long and the con- necting rodBCis 32.0 cm long. For the position shown
A 40° B
C Fig. 26.20
determine the length ofAC and the angle between the crank and the connecting rod.
4. From Fig. 26.20, determine how farCmoves, correct to the nearest millimetre when angleCABchanges from 40◦to 160◦,Bmoving in an anticlockwise direction.
5. A surveyor, standing W 25◦S of a tower measures the angle of elevation of the top of the tower as 46◦30. From a position E 23◦S from the tower the elevation of the top is 37◦15. Determine the height of the tower if the distance between the two observations is 75 m.
6. Calculate, correct to 3 significant figures, the co-ord- inatesxandyto locate the hole centre atP shown in Fig. 26.21.
P
x
116° 140°
100 mm y
Fig. 26.21
7. An idler gear, 30 mm in diameter, has to be fitted between a 70 mm diameter driving gear and a 90 mm diameter driven gear as shown in Fig. 26.22. Determine the value of angleθbetween the centre lines.
90 mm dia
30 mm 99.78 mmu dia
70 mm dia
Fig. 26.22
8. 16 holes are equally spaced opn a pitch circle of 70 mm diameter. Determine the length of the chord joining the centres of two adjacent holes.
206 Basic Engineering Mathematics
Assignment 12
This assignment covers the material in Chapters 25 and 26. The marks for each question are shown in brackets at the end of each question.
1. Plot a graph of y=3x2+5 from x=1 to x=4.
Estimate, correct to 2 decimal places, using 6 intervals, the area enclosed by the curve, the ordinatesx=1 and x=4, and thex-axis by (a) the trapezoidal rule, (b) the mid-ordinate rule, and (c) Simpson’s rule. (12) 2. A circular cooling tower is 20 m high. The inside diam- eter of the tower at different heights is given in the following table:
Height (m) 0 5.0 10.0 15.0 20.0
Diameter (m) 16.0 13.3 10.7 8.6 8.0 Determine the area corresponding to each diameter and hence estimate the capacity of the tower in cubic
metres. (7)
3. A vehicle starts from rest and its velocity is measured every second for 6 seconds, with the following results:
Timet(s) 0 1 2 3 4 5 6
Velocityv(m/s) 0 1.2 2.4 3.7 5.2 6.0 9.2 Using Simpson’s rule, calculate (a) the distance trav- elled in 6 s (i.e. the area under thev/tgraph) and (b) the average speed over this period. (6)
4. A triangular plot of landABCis shown in Fig. A12.1.
Solve the triangle and determine its area. (10)
B
A
C 15.4 m 15 m 71°
Fig. A12.1
5. A car is travelling 20 m above sea level. It then travels 500 m up a steady slope of 17◦. Determine, correct to the nearest metre, how high the car is now above see
level. (3)
6. Figure A12.2 shows a roof truss PQR with rafter PQ=3 m. Calculate the length of (a) the roof risePP, (b) rafterPR, and (c) the roof spanQR. Find also (d) the cross-sectional area of the roof truss. (12)
Q R
P
P′ 3 m
40° 32°
Fig. A12.2
27
Vectors
27.1 Introduction
Some physical quantities are entirely defined by a numerical value and are called scalar quantities or scalars. Examples of scalars include time, mass, temperature, energy and volume.
Other physical quantities are defined by both a numerical value and a direction in space and these are calledvector quantities orvectors. Examples of vectors include force, velocity, moment and displacement.
27.2 Vector addition
A vector may be represented by a straight line, the length of line being directly proportional to the magnitude of the quantity and the direction of the line being in the same direction as the line of action of the quantity. An arrow is used to denote the sense of the vector, that is, for a horizontal vector, say, whether it acts from left to right or vice-versa. The arrow is positioned at the end of the vector and this position is called the ‘nose’ of the vector. Figure 27.1 shows a velocity of 20 m/s at an angle of 45◦ to the horizontal and may be depicted byoa=20 m/s at 45◦to the horizontal.
20 m/s 45° o
a
Fig. 27.1
To distinguish between vector and scalar quantities, various ways are used. These include:
(i) bold print,
(ii) two capital letters with an arrow above them to denote the sense of direction, e.g.−→
AB, whereAis the starting point and Bthe end point of the vector,
(iii) a line over the top of letters, e.g.ABora,¯ (iv) letters with an arrow above, e.g.a, A,
(v) underlined letters, e.g.a,
(vi) xi+jy, whereiandjare axes at right-angles to each other;
for example, 3i+4jmeans 3 units in theidirection and 4 units in thejdirection, as shown in Fig. 27.2.
j
i A(3, 4) 4
3 2 1
1
O 2 3
Fig. 27.2
(vii) a column matrix a
b
; for example, the vectorOAshown in Fig. 27.2 could be represented by
3 4
Thus, in Fig. 27.2,OA≡−→
OA≡OA≡3i+4j≡ 3
4
208 Basic Engineering Mathematics
The one adopted in this text is to denote vector quantities in bold print. Thus,oarepresents a vector quantity, butoais the magnitude of the vectoroa. Also, positive angles are measured in an anticlockwise direction from a horizontal, right facing line and negative angles in a clockwise direction from this line – as with graphical work. Thus 90◦is a line vertically upwards and
−90◦is a line vertically downwards.
The resultant of adding two vectors together, sayV1 at an angleθ1andV2at angle (−θ2), as shown in Fig. 27.3(a), can be obtained by drawing oa to represent V1 and then draw- ing ar to represent V2. The resultant of V1+V2 is given by or. This is shown in Fig. 27.3(b), the vector equation being oa+ar=or. This is called the‘nose-to-tail’ methodof vector addition.
V1 u1
u2
V2 (a)
r a
o (b)
O R V1
u1
u2
V2 (c) Fig. 27.3
Alternatively, by drawing lines parallel toV1andV2from the noses ofV2andV1, respectively, and letting the point of inter- section of these parallel lines beR, givesORas the magnitude and direction of the resultant of addingV1andV2, as shown in Fig. 27.3(c). This is called the‘parallelogram’ methodof vector addition.
Problem 1. A force of 4 N is inclined at an angle of 45◦ to a second force of 7 N, both forces acting at a point. Find the magnitude of the resultant of these two forces and the direction of the resultant with respect to the 7 N force by both the ‘triangle’ and the ‘parallelogram’
methods.
The forces are shown in Fig. 27.4(a). Although the 7 N force is shown as a horizontal line, it could have been drawn in any direction.
Using the‘nose-to-tail’ method, a line 7 units long is drawn horizontally to give vectoroa in Fig. 27.4(b). To the nose of this vectoraris drawn 4 units long at an angle of 45◦ tooa.
The resultant of vector addition is orand by measurement is 10.2 units long and at an angle of 16◦to the 7 N force.
Figure 27.4(c) uses the ‘parallelogram’ methodin which lines are drawn parallel to the 7 N and 4 N forces from the noses of the 4 N and 7 N forces, respectively. These intersect at R. VectorORgives the magnitude and direction of the resul- tant of vector addition and as obtained by the ‘nose-to-tail’
method is 10.2 units long at an angle of 16◦ to the 7 N force.
0 2 4 6
4 N
4 N 4 N 45°
45° 45°
O
o O
7 N
7 N 7 N
r R
a
Scale in Newtons
(a)
(b) (c)
Fig. 27.4
Problem 2. Use a graphical method to determine the mag- nitude and direction of the resultant of the three velocities shown in Fig. 27.5.
v2
v1
v3 10°
20° 7 m/s
15 m/s
10 m/s
Fig. 27.5
Often it is easier to use the ‘nose-to-tail’ method when more than two vectors are being added. The order in which the vectors are added is immaterial. In this case the order taken isv1, thenv2, thenv3but just the same result would have been obtained if the order had been, say,v1,v3and finallyv2.v1is drawn 10 units long at an angle of 20◦to the horizontal, shown byoain Fig. 27.6.v2 is added tov1by drawing a line 15 units long vertically upwards froma, shown asab.
Finally,v3is added tov1+v2by drawing a line 7 units long at an angle at 190◦fromb, shown asbr. The resultant of vector addition isorand by measurement is 17.5 units long at an angle of 82◦to the horizontal.
Thusv1+v2+v3=17.5 m/s at 82◦to the horizontal.
Vectors 209
o 2
0 4 6 8 10
10°
82° 20° 12
Scale in m/s
a r
b
Fig. 27.6
27.3 Resolution of vectors
A vector can be resolved into two component parts such that the vector addition of the component parts is equal to the original vector. The two components usually taken are a horizontal com- ponent and a vertical component. For the vector shown asFin Fig. 27.7, the horizontal component isFcosθand the vertical component isFsinθ.
F sin u
F cos u u
F
Fig. 27.7
For the vectors F1 and F2shown in Fig. 27.8, the horizontal component of vector addition is:
H=F1cosθ1+F2cosθ2
and the vertical component of vector addition is:
V =F1sinθ1+F2sinθ2 V
H F1sin u1
F1 F2 F2sin u2
u1
u2
F2 cos u2
F1 cos u1
Fig. 27.8
Having obtainedHandV, the magnitude of the resultant vector Ris given by√
H2+V2and its angle to the horizontal is given bytan−1(V/H)
Problem 3. Resolve the acceleration vector of 17 m/s2at an angle of 120◦to the horizontal into a horizontal and a vertical component.
For a vectorAat angleθto the horizontal, the horizontal com- ponent is given byAcosθand the vertical component byAsinθ. Any convention of signs may be adopted, in this case horizontally from left to right is taken as positive and vertically upwards is taken as positive.
Horizontal component H=17 cos 120◦=−8.50 m/s2, acting from left to right.
Vertical componentV=17 sin 120◦=14.72 m/s2, acting verti- cally upwards.
These component vectors are shown in Fig. 27.9.
⫹V
+H
−V
⫺H
17 m/s2
14.72 m/s2
8.50 m/s2 120°
Fig. 27.9
Problem 4. Calculate the resultant force of the two forces given in Problem 1.
With reference to Fig. 27.4(a):
Horizontal component of force, H=7 cos 0◦+4 cos 45◦
=7+2.828=9.828 N Vertical component of force, V=7 sin 0◦+4 sin 45◦
=0+2.828=2.828 N
210 Basic Engineering Mathematics
The magnitude of the resultant of vector addition
=
H2+V2=
9.8282+2.8282
=√
104.59=10.23 N
The direction of the resultant of vector addition
=tan−1 V
H
=tan−1 2.828
9.828
=16.05◦
Thus, the resultant of the two forces is a single vector of 10.23 N at 16.05◦to the 7 N vector.
Problem 5. Calculate the resultant velocity of the three velocities given in Problem 2.
With reference to Fig. 27.5:
Horizontal component of the velocity,
H =10 cos 20◦+15 cos 90◦+7 cos 190◦
=9.397+0+(−6.894)=2.503 m/s Vertical component of the velocity,
V =10 sin 20◦+15 sin 90◦+7 sin 190◦
=3.420+15+(−1.216)=17.204 m/s Magnitude of the resultant of vector addition
=
H2+V2=
2.5032+17.2042
=√
302.24=17.39 m/s
Direction of the resultant of vector addition
=tan−1 V
H
=tan−1 17.204
2.503
=tan−16.8734=81.72◦
Thus, the resultant of the three velocities is a single vector of 17.39 m/s at 81.72◦to the horizontal
Now try the following exercise