1. Determine the resultant of the four forces acting through the point A (Fig. 5).
pounds acting upward through A and a point 0.45 feet below C.
2. Determine the resultant of the three forces acting at the point F (Fig. 5).
( 175 pounds acting upward through F and a Ans*P • point 0 57 feet to left of C.
13. Graphical Resolution of Force into Two Concurrent Components. This is performed by applying the triangle law inversely. Thus, if it is required to resolve the 100-pound force of Fig. 5 into two components, we draw first Fig. 10 (a) to show the line of action of the force, and then AB, Fig. 10 (b), to represent the magnitude and direction. Then draw from A and B any two lines which intersect, mark their intersection C, and place arrow heads on AC and CB, pointing from A to C and from C to B. Also draw two lines in the space diagram parallel to AC and CB and so that they intersect on the line of action of the 100-pound force, a.
The test of the correctness of a solution like this is to take the two components as found, and find their resultant; if the resultant thus found agrees in magnitude, direction, and sense with the given force (originally resolved), the solution is correct.
Notice that the solution above given is not definite, for the lines drawn from A and B were drawn at random. A force may therefore be resolved into two components in many ways. If, however, the components have to satisfy conditions, there may be but one solution. In the most important case of resolution, the lines of action of the components are given; this case is definite, there being but one solution, as is shown in the following example.
Example. It is required to resolve the 100-pound force (Fig. 5) into two components acting in the lines AE and AB.
Using the space diagram of Fig. 10, draw a line AB in Fig. 10 (c) to represent the magnitude and direction of the 100-pound force, and then a line from A parallel to the line of action of either of the components, and a line from B parallel to the other, thus loc'ating D (or D'). Then AD and DB (or AD' and D'B) repre sent the magnitudes and directions of the required components.
1. Resolve the 160-pound force of Fig. 5 into components which act in Al' and AE.
S The first component equals 2381 pounds, and its sense Ans. is is from A to F; the second component equals 119k ( pounds, and its sense is from E to A.
2. Resolve the 50-pound force of Fig. 5 into two compo nents, acting in FA and FB.
The first component equals 37.3 pounds, and its sense Ans. is from A to F; the second component equals 47.0 pounds, and its sense is from B to F.
14. Algebraic Resolution of a Force Into Two Components. If the angle between the lines of action of the two components is not 90 degrees, the algebraic method is not simple and the graphical method is usually preferable. When the angle is 00 degrees, the algebraic method is usually the shorter, and this is the only case herein explained.
Let F (Fig. 11) be the force to be resolved into two compo nents acting in the lines OX and OY. If AB is drawn to repre sent the magnitude and direction of F, and lines be drawn from A and B parallel to OX and OY, thus locating C, then AC and BC with arrowheads as shown represent the magnitudes and directions of the required components.
Now if F' and F" represent the components acting in OX and OY, and x and y denote the angles between F and F', and F and F" respectively, then AC and BC represent F' and F", and the angles BAC and ABC equal x and y respectively. From the right triangle ABC it follows that and, If a force is resolved into two components whose lines of action are at right angles to each other, each is called a rectangular component of that force. Thus F' and F" are rectangular components of F.
The foregoing equations show that the rectangular component of a force along any line equals the product of the force and the cosine of the angle between, the force and the line. They show also that the rectangular component of a force along its own line of action equals the force, and its rectangular com ponent at right angles to the line of action equals zero.
Examples. 1. A force of 120 pounds makes an angle of 22 degrees with the horizontal. What is the value of its compo nent along the horizontal ?* Since cos 22° = 0.927, the value of the component equals 120 X 0.927 = 111.24 pounds.
2. What is the value of the component of the 90•pound force of Fig. 5 along the vertical ? First we must find the value of the angle which the 90-pound force of Fig. 5 makes with the vertical.
* When nothing is stated herein as to whether a component is rectan gular or not, then rectangular component is meant.
1. Compute the horizontal and vertical components of a force of 80 pounds whose angle with the horizontal is 60 degrees 40 pounds.
Ans.
69.28 pounds.
2. Compute the horizontal and vertical components of the 100.pound force in Fig. 5. What are their senses ? { 89.44 pounds to the right.
Ans.
44.72 pounds upwards.
3. Compute the component of the 70-pound force in Fig. 5 along the line EA. What is the sense of the component ? Ans. 31. 29 pounds ; E to A.