## Examples For Practice

1. Resolve the 160-pound force of Fig. 5 into components which act in AF and AE.

 Ans. The first component equals 238½ pounds, and its sense is from A to F; the second component equals 119½ pounds, and its sense is from E to A.

2. Resolve the 50-pound force of Fig. 5 ½into two components, acting in FA and FB.

 Ans. Ihe first component equals 37.3 pounds, and its sense 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 90 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 components acting in the lines OX and OY. If AB is drawn to represent 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, F' = F cos x, and F" = F cos y.

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-pone?U 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 component 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 Ficr. 5 makes with the vertical. Fig. 11.

* When nothing is stated herein as to whether a component is rectangular or not, then rectangular component is meant.

Since tan EAG = EG/AG =½, angle EAG = 26° 34'.

Hence the value of the desired component equals

90 X cos 26° 34' = 90 X 0.8944 = 80.50 pounds.

## Examples For Practice

1. Compute the horizontal and vertical components of a force of 80 pounds whose angle with the horizontal is 60 degrees

 Ans. 40 pounds. 69.28 pounds.

2. Compute the horizontal and vertical components of the 100-pound force in Fig. 5. What are their senses?

 Ans. 89.44 pounds to the right. 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.