Two analytical methods can be used to determine the net effect of vectors. The first method is best suited when the resultant of only two vectors is required. As with the
( + 7)
graphical method, the two vectors to be combined are placed tip-to-tail. The resultant is found by connecting the tail of the first vector to the tip of the second vector.
Thus, the resultant forms the third side of a triangle. In general, this is an oblique triangle, and the laws des-cribed in Section 3.6.2 can be applied. The length of the third side and a reference angle must be determined
FIGURE 3.11 The combined effect of vectors A, B, C, and D for Example Problem 3.6.
EXAMPLE PROBLEM 3.6
Graphically determine the combined effect of force vectors A, B, C, and D, as shown in Figure 3.10.
Q
Vectors 51
A= 46 ft/s2
B= 23 ft/s2
20° 75°
FIGURE 3.12 Vectors for Example Problem 3.7.
R B
R = A +> B
A 75°
20° β θ
FIGURE 3.13 Combined effect of vectors A and B for Example Problem 3.7.
EXAMPLE PROBLEM 3.7
Analytically determine the resultant of two acceleration vectors as shown in Figure 3.12.
SOLUTION: 1. Sketch a Rough Vector Diagram
The vectors are placed tip-to-tail as shown in Figure 3.13. Note that only a rough sketch is required because the resultant is analytically determined.
2. Determine an Internal Angle
The angle between A and the horizontal is 20°. By examining Figure 3.13, the angle between vectors A and B is:
Therefore, the problem of determining the resultant of two vectors is actually a general triangle situation as described in Section 3.6.2 (Case 3).
3. Determine Resultant Magnitude
By following the procedure outlined for a Case 3 problem, the law of cosines is used to find the magnitude of the resultant:
4. Determine Magnitude Direction
The law of sines can be used to find the angle between vectors A and R:
5. Fully Specify Resultant
The angle from the horizontal is . The resultant can be properly written as:
or
R = 53.19 ft./s2 134.5°
R = 53.19 ft./s2 45.5°
20° + 25.5° = 45.5°
= sin-1e (23 ft/s2)
(53.19 ft/s2) sin 95° f = 25.5°
b = sin-1e aB Rb sin u f
= 3(46ft/s2)2 + (23ft/s2)2 - 2(46ft/s2)(23 ft/s2){cos 9°} = 53.19 ft/s2 R = 2A2 + B2 - 2ABcosu
u = 20° + 75° = 95°
through the laws of sines and cosines to fully define the resultant vector. This method can be illustrated through an example problem. To clearly distinguish
quantities, vectors are shown in boldface type (D) while the magnitude of the vector is shown as non-bold, italic type .(D)
a Q
52 CHAPTER THREE
3.10 COMPONENTS OF A VECTOR
The second method for analytically determining the resultant of vectors is best suited for problems where more than two vectors are to be combined. This method involves resolving vectors into perpendicular components.
Resolution of a vector is the reverse of combining vectors. A single vector can be broken into two separate vectors, along convenient directions. The two vector com-ponents have the same effect as the original vector.
In most applications, it is desirable to concentrate on a set of vectors directed vertically and horizontally; therefore, a typical problem involves determining the horizontal and vertical components of a vector. This problem can be solved by using the tip-to-tail approach, but in reverse. To explain the method, a general vector, A, is shown in Figure 3.14.
important. Therefore, it is irrelevant whether the horizontal or vertical vector is drawn first. Figure 3.14c illustrates the components of a general vector in the opposite order.
Notice that the magnitude of the components can be found from determining the sides of the triangles shown in Figure 3.14. These triangles are always right triangles, and the methods described in Section 3.3 can be used. The directions of the components are taken from sketching the vectors as in Figure 3.14b or 3.14c. Standard notation con-sists of defining horizontal vectors directed toward the right as positive. All vertical vectors directed upward are also defined as positive. In this fashion, the direction of the components can be determined from the algebraic sign asso-ciated with the component.
An alternative method to determine the rectangular components of a vector is to identify the vector’s angle with the positive -axis of a conventional Cartesian coordinate system. This angle is designated as . The magnitude of the two components can be computed from the basic trigono-metric relations as
(3.9) (3.10) The importance of this method lies in the fact that the directions of the components are evident from the sign that results from the trigonometric function. That is, a vector that points into the second quadrant of a conventional Cartesian coordinate system has an angle, , between 90°
and 180°. The cosine of such an angle results in a negative value, and the sine results in a positive value. Equations (3.9) and (3.10) imply that the horizontal component is negative (i.e., toward the left in a conventional coordinate system) and the vertical component is positive (i.e., upward in a conventional system).
FIGURE 3.14 Components of a vector.
EXAMPLE PROBLEM 3.8
A force, F, of 3.5 kN is shown in Figure 3.15. Using the analytical triangle method, determine the horizontal and vertical components of this force.
F= 3.5 kN 35°
FIGURE 3.15 Force vector for Example Problem 3.8.
F= 3.5 kN Fh
Fv 35°
FIGURE 3.16 Force components for Example Problem 3.8.
Two vectors can be drawn tip-to-tail along the hori-zontal and vertical that have the net effect of the original. The tail of the horizontal vector is placed at the tail of the original, and the tip of the vertical vector is placed at the tip of the original vector. This vector resolution into a horizontal component, , and the vertical component, , is shown in Figure 3.14b. Recall that the order of vector addition is not
Av Ah
SOLUTION: 1. Sketch the Vector Components
The horizontal vector (component) is drawn from the tail of vector F. A vertical vector (component) is drawn from the horizontal vector to the tip of the original force vector. These two components are shown in Figure 3.16.
Vectors 53 2. Use Triangle Method
Working with the right triangle, an expression for both components can be written using trigonometric func-tions:
Both these expressions can be solved in terms of the magnitude of the desired components:
3. Use x-axis, Angle Method
An alternative solution is obtained by using equations (3.9) and (3.10). The angle from the positive -axis to the vector F is 215°. The components are computed as follows:
= 2.0 kNT
Fv = F sin ux= (3.5 kN) sin 215° = -2.0 kN
= 2.87kN ;
Fh = F cos ux = (3.5 kN) cos 215° = -2.87 kN
x ux
= -2.00 kN
Fv = (3.5kN) sin 35° = 2.00 kNT
= -2.87 kN
Fh = (3.5kN) cos 35° = 2.87kN ; cos 35° = adjacent side
hypotenuse = Fh 3.5 kN sin 35° = opposite side
hypotenuse = Fv 3.5 kN