The components of a series of vectors can be used to deter-mine the net effect of the vectors. As mentioned, this method is best suited when more than two vectors need to be com-bined. This method involves resolving each individual vector into horizontal and vertical components. It is standard to use the algebraic sign convention for the components as described previously.
All horizontal components may then be added into a single vector component. This component represents the net horizontal effect of the series of vectors. It is worth noting that the component magnitudes can be simply added together because they all lay in the same direction. These components are treated as scalar quantities. A positive or negative sign is used to denote the sense of the component.
This concept can be summarized in the following equation:
(3.11) Rh = Ah + Bh + Ch + Dh + Á
( + 7)
Similarly, all vertical components may be added together into a single vector component. This component represents the net vertical effect of the series of vectors:
(3.12) The two net components may then be added vectorally into a resultant. Trigonometric relationships can be used to produce the following equations:
(3.13)
(3.14)
This resultant is the combined effect of the entire series of vectors. This procedure can be conducted most efficiently when the computations are arranged in a table, as demon-strated in the following example problem.
ux = tan-1a Rv Rhb R = 3Rh2 + Rv2
Rv = Av + Bv + Cv + Dv + Á
EXAMPLE PROBLEM 3.9
Three forces act on a hook as shown in Figure 3.17. Using the analytical component method, determine the net effect of these forces.
SOLUTION: 1. Use x-axis, Angle Method to Determine Resultant Components
The horizontal and vertical components of each force are determined by trigonometry and shown in Figure 3.18. Also shown are the vectors rearranged in a tip-to-tail fashion. The components are organized in Table 3.1.
54 CHAPTER THREE
C = 50 lb Bv
Bh Ch
B
A
C R
Cv A = 30 lb
B = 20 lb
60° 45°
FIGURE 3.18 Components of vectors in Example Problem 3.9.
Rh= 19.14 lb
Rv= 57.44 lb R
θx
FIGURE 3.19 Resultant vector for Example Problem 3.9.
TABLE 3.1 Vector Components for Example Problem 3.9.
Vector
Reference Angle ux
h component (lb) Fh= F cos ux
v component (lb) Fv= F sin ux
A 0° Ah= (30)cos 0°= + 30lb Av= (30)sin 0°= 0
B 45° Bh= (20)cos 45°= +14.14 lb Bv= (20)sin 45°= +14.14 lb
C 120° Ch= (50)cos 120° = -25 lb Cv= (50)sin 120° = +43.30 lb
Rh= 19.14 Rv = 57.44
Notice in Figure 3.18 that adding the magnitudes of the horizontal components is tracking the total
“distance” navigated by the vectors in the horizontal direction. The same holds true for adding the magni-tudes of the vertical components. This is the logic behind the component method of combining vectors. For this problem, adding the individual horizontal and vertical components gives the components of the resul-tant as follows:
and
2. Combine the Resultant Components
The resultant is the vector sum of two perpendicular vectors, as shown in Figure 3.19.
Rv = 57.44 lb.
Rh = 19.14 lb.
C = 50 lb
B = 20 lb
A = 30 lb
60° 45°
FIGURE 3.17 Forces for Example Problem 3.9.
The magnitude of the resultant can be found from equation (2.13):
= 3(19.14lb)2 + (57.44lb)2 = 60.54lb R = 3Rh2 + Rv2
Vectors 55 The angle of the resultant can be found:
Thus the resultant of the three forces can be formally stated as R= 60.54 lb. 71.6°
ux= tan-1aRv
Rhb = tan-1a57.44 lb
19.14 lbb = 71.6°
3.12 VECTOR SUBTRACTION
In certain cases, the difference between vector quantities is desired. In these situations, the vectors need to be sub-tracted. The symbol denotes vector subtraction, which differentiates it from algebraic subtraction [Ref. 5].
Subtracting vectors is accomplished in a manner similar to adding them. In effect, subtraction adds the negative, or opposite, of the vector to be subtracted. The negative of a vector is equal in magnitude, but opposite in direc-tion. Figure 3.20 illustrates a vector A and its negative,
-7A.
-7
( - 7)
A −>A
FIGURE 3.20 Negative vector.
Whether a graphical or analytical method is used, a vector diagram should be drawn to understand the pro-cedure. Consider a general problem where vector B must be subtracted from A, as shown in Figure 3.21a.
This subtraction can be accomplished by first drawing the negative of vector B, . This is shown in Figure 3.21b. Then, vector can be added to vector A, as shown in Figure 3.21c. This subtraction can be stated mathe-matically as
Notice that this expression is identical to the subtraction of scalar quantities through basic algebraic methods. Also, the outcome of the vector subtraction has been designated J.
The notation R is typically reserved to represent the result of vector addition.
Figure 3.21d shows that the same vector subtraction result by placing the vector B onto vector A, but opposite to the tip-to-tail orientation. This method is usually preferred, after some confidence has been established, because it elimi-nates the need to redraw a negative vector. Generally stated, vectors are added in a tip-to-tail format whereas they are subtracted in a tip-to-tip format. This concept is further explored as the individual solution methods are reviewed in the following example problems.
3.13 GRAPHICAL VECTOR SUBTRACTION
As discussed, vector subtraction closely parallels vector addition. To graphically subtract vectors, they are relocated to scale to form a tip-to-tip vector diagram. The vector to be subtracted must be treated in the manner discussed in Section 3.12.
Again, the process of subtracting vectors can be completed graphically, using either manual drawing techniques or CAD software. Whatever method is used, the underlying concepts are identical. The specifics of the process are shown in the following examples.
FIGURE 3.21 Vector subtraction.
EXAMPLE PROBLEM 3.10
Graphically determine the result of subtracting the velocity vector B from , as shown in Figure 3.22.
FIGURE 3.22 Vectors for Example Problem 3.10.
Q
56 CHAPTER THREE
FIGURE 3.23 J = A - 7 Bfor Example Problem 3.10.
A = 200 lb
B = 226 lb
D = 300 lb
C = 176 lb
0 100
lb Scale:
200 65°
60° 15°
FIGURE 3.24 Vectors for Example Problem 3.11.
EXAMPLE PROBLEM 3.11
Graphically determine the result,J = A -7 B -7 C +7 D, of the force vectors as shown in Figure 3.24.
SOLUTION: 1. Construct the Vector Diagram
To determine the result, the vectors are located in the tip-to-tail form, but vector B points toward vector A. Again, this occurs because B is being subtracted (opposite to addition). The vector diagram is shown in Figure 3.23.
2. Measure the Result
The resultant extends from the tail of A, the origin, to the tail of B. The length vector J is measured as 56.8 in./s.
The direction is also required to fully define the vector J. The angle from the horizontal to vector J is measured as 99°. Therefore, the proper manner of presenting the solution is as follows:
or
J = 56.8 in./s 99°
J = 56.8 in./s 81°Q a
Vectors 57 SOLUTION: 1. Construct the Vector Diagram
To determine the result, , the vectors must be relocated tip-to-tail or tip-to-tip, depending on whether they are added or subtracted. Vector B must be drawn pointing toward vector A because B is being subtracted. A similar approach is taken with vector C. The tail of vector D is then placed on the tail of C because D is to be added to the series of previously assembled vectors. The completed vector diagram is shown in Figure 3.25.
J = A -7 B -7 C +7 D
FIGURE 3.25 Result for Example Problem 3.11.
From viewing the vector polygon in Figure 3.25, it appears that vectors B and C were placed in backward, which occurs with vector subtraction.
2. Measure the Result
The length of vector J is measured as 365 lb. The angle from the horizontal to vector J is measured as 81°.
Therefore, the proper manner of presenting the solution is as follows:
J = 365 lb 81°