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# Chapter 3 : Forces and their components

3.1 Orthogonal components of forces
3.2 Determination of resultant of forces
3.3 Resultant of forces, a sample case
3.4 Three-dimensional aspects
3.5 Summary,Re-statement of Equilibrium,
and Free-Body-Diagram
3.6 Self-Test, computer programs and problems

## 3.1 Orthogonal components of forces NextSec

The determination of the resultant of three or more forces using strictly the Parallelogram Law in the form of Equations 2.5 is somewhat tedious and in the long run almost useless. We need better tools !!! Figure 3.1a 3 forces on a particle
Let's have a look at Figure 3.1a. Three forces , F1, F2, and F3 are shown acting on a particle A . Also shown is an orthogonal coordinate system whose axes I labelled x and y. The location of its origin and the alignment of its axes with the borders of the figure are arbitrary choices of mine. Our task is now to develop a more efficient way to determine the magnitude and direction of the resultant of the three forces shown.

The trick we will be employing is the following. We interpret each of the three forces in Figure 3.1a as the resultant of two forces, one aligned Figure 3.1b Force F1 and its components
with the x- the other with the y-axis as shown in Figure 3.1b for the force F1.
We call these two new forces the x- and y-component of the force F1.
Their values can be easily calculated if the magnitude (the absolute value) of F1 and its orientation (the angle α) are known. Note that the angle α is measured between the positive x-axis and the force in counterclockwise direction.

Also, depending on the value of the angle α one or both of the components might have a negative value, indicating that the component is pointing in the direction of the minus x-axis for example.

## 3.2 Determination of resultant of forces PrecSec   NextSec

In Figure 3.1a we now replace the force F1 by its x- and y-component and repeat this step for the two other forces involved. The result is that we have replaced the original three forces by six new forces, of which three are aligned with the x-axis and three with the y-axis of our coordinate system.

The final step is then to add the three force components in the x-direction (no sweat here, that would be just adding/subtracting numbers) to get the x-component of the resultant. The y-component of the resultant is obtained in similar fashion.
Formally we write this as : Once we have these components we can determine the magnitude of the resultant and the angle β between the resultant and the x-axis : Equation 3.2d has always two solutions for the angle β. If for given Rx and R your calculator gives β=80° for example then β=110° is a solution as well. But which value is correct, 80 or 110° ?
The answer to this question can be found by looking at the signs (+/-) of the components Rx and Ry which inform you in which quadrant of the unit circle your resultant R lies.

## 3.3 Resultant of forces, a sample case PrecSec   NextSec

The Equations 3.1a through 3.2d together form a set of equations which will follow you throughout this course and beyond (together with the statement of equilibrium of particles ).

When applying these equations it is extremely important to know about the sign (plus/minus) conventions which go along with the cos() and sin() function used in the Equations 3.2a and 3.2b. Of course in Statics we don't make up our own rules but follow strictly the rules of trigonometry. Here is a short sample case I would recommend you read carefully. To some of you it might seem silly to harp on sign conventions. However, in practical engineering applications not observing the correct sign amounts often to the difference between a well designed structure and a failing structure with possible loss of human life and/or millions of dollars.

So, here is the problem as depicted in Figure 3.3a. Originally you know only the magnitudes ( 400 N , 350 N, 600 N, and 100 N ) and the orientations (angles 50, 70, 30, and 15 degrees) of the four forces. Our task is to find the resultant of these four forces, that is to find that single force which has the same action on particle A as the four given forces.
If the labels ( F1, F2, F3, and F4) are not given, you must label them. I furthermore entered already an x-y coordinate system. If it is not given, you must make a choice. I aligned the x-axis with the line a-a. Figure 3.3a Find resultant of four forces

After these preliminary steps the real work begins. Here are my calculations for the x-components of the four forces and then the x-component of the resultant : Three points of interest :

1. The angle used as argument of the cosine function is always determined by going on an arc from the positive x-axis in counter-clockwise direction towards the force of which you want to determine the x-component.
2. Two of the forces have negative x-components ( cos(110) is negative as is cos(210) ). A negative value of the x-component means that the x- component is pointing in the direction of the minus x-axis.
3. The plus/minus sign of the obtained x-components has to be entered when calculating the x-component of the resultant. Here Rx comes out to be negative itself, meaning that the combined action of the forces is to pull to the left in minus x-direction (forgetting at the moment about what happens in the y-direction).
Please, do determine the value of the y-component of the resultant yourself. My results are displayed in Figure 3.3b. Figure 3.3b Resultant of four forces

The action of the four original forces becomes now clear. They will pull the body A to the left and upwards. Also, please check out whether I got the magnitude and angle for the resultant right.

## 3.4 Three-dimensional aspects PrecSec   NextSec

Although this course will mostly deal with 2-dimensional structures and forces ( planar cases ) from time to time we will dive into the three-dimensional space.
The concept of orthogonal components we looked at in the preceeding chapters will be employed again, but the direction of a force in 3-D is harder to describe and therefore the calculations of the component of a 3-D force is a little bit more involved. In the Figures 3.4a and 3.4b I show the same force F embedded in identical x-y-z coordinate systems. Also, the dashed line 0B lies in both Figures in the x-y plane.
The direction of the force F is expressed in different ways though. Figure 3.4a Force in 3-D
For Figure 3.4a : Given is the magnitude of the force and the values of the angle beta and gamma. We then determine the three components of the force by :  Figure 3.4b Force in 3-D
For Figure 3.4b the magnitude and the angle αx, αy, and αz are given. Now the three components are evaluated according to : On the other hand, to determine the components of the resultant of several forces Equation 3.2a to 3.2c have to be modified only slightly : And the magnitude of a force (like the resultant) is calculated according to : ## 3.5 Summary,Re-statement of Equilibrium,       and Free-Body-Diagram PrecSec   NextSec

The concept of orthogonal components of forces greatly facilitates the determination of resultants of multiple forces. Its usefulness will become much more apparent and wide-spread in subsequent chapters.

The choice of the orientation of the coordinate system (x-y or x-y-z) is arbitrary and does not affect the final outcome of your calculations.

In order to determine the components of a force only simple trigonometric functions ( cosine and sine ) are needed, the sign (plus/minus) conventions for angles greater than 90o have to be obeyed.

The value of the component of a force (and later that of a force) can be positive, negative, or zero. If negative, it indicates that the action of that force component is directed in the negative direction of the corresponding coordinate axis.

By the magnitude of a force (or force component) we mean the value of it stripped of its sign. In mathematics we know this also as the absolute value.

In section 2.7 we concluded that in order for a particle to remain at rest the resultant of all forces acting on the particle has to be zero. But if the resultant itself is zero then all its components are zero!!! Hence, the statement of equilibrium of a particle can be re-phrased :

 A particle remains at rest if the sum of the x-components and the sum of the y-components and the sum of the z-components of ALL forces acting on it, are zero. Taking into account ALL of the forces acting on a particular body is not always an easy task. A particularly useful tool is the Free-Body-Diagram

 The Free-Body-Diagram ( F.B.D ) of a body is a rough sketch of a body which includes ALL forces acting on that body from the outside.

## 3.6 Self-Test, computer programs and problems PrecSec

#### Self-Test

The self-test is a multiple-choice test. It allows you to ascertain your knowledge of the definition of terms and your understanding of important results.

#### Two computer programs

Using the concept of x-, y-, and z-components is also a great way to write computer programs which can tackle generalized problems.
Two programs are available for you. The first determines the resultant of up to 10 forces , the second program calculates that force which is necessary to hold a body in equilibrium against up to 10 forces of your choice.

My recommendation to you is to use these programs in three ways :

1. You make up a problem with 2 or 3 forces pointing into various directions and determine yourself the resultant and the equilibrium force. Then you use the programs to check your answers. A sample case is provided with each program.
2. Both programs provide numerical as well as graphical output. Use the graphical output to visualize magnitude and direction of the resultant or equilibrium force changing one of the given forces. For example, choose two given forces of equal magnitude and change the direction of one until the two forces are aligned or until the two forces oppose each other.
3. In general, problems as listed below require a multi-step approach which involves first the analysis and understanding of the problem, then writing out the mathematical equations to be used and finally solving these equations. The computer programs can help you nicely with the last step.

#### Problems

Klick on any of the problem titles below. The problem statement itself will contain links to additional help and to the solution.

Problem 3.6a : Resultant of 2 Forces
Problem 3.6b : Resultant of 3 Forces
Problem 3.6c : Equiblibrium of particle
Problem 3.6d : Crate on inclined surface
Problem 3.6e : Bearing force
Problem 3.6f : Broadcast pole I, 2-D
Problem 3.6g : Broadcast pole II, 2-D
Problem 3.6h : Broadcast pole, 3-D
Problem 3.6i : Anchor of a broadcast pole

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Zig Herzog; hgnherzog@yahoo.com

Last revised: 09/14/13