Figure 3.1a
3 forces on a particle
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Figure 3.1a 3 forces on a particle |
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 |
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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 that component is pointing in the direction of the minus x-axis for example.
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 :
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Once we have these components we can determine the magnitude of the resultant and the angle beta between the resultant and the x-axis :
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Equation 3.2d has always two solutions for the angle beta. If for given
Rx and R your calculator gives beta=30o for example
then beta=180-30=150 is a solution as well. But which value is
correct, 30 or 150o ?
The answer to this question can be found by looking at the signs (+/-)
of the components Rx and Ry which informs you
in which quadrant of the unit circle your resultant R lies.
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 :
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Three points of interest :
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.
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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 :
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For Figure 3.4b the magnitude and the angle αx,
αy, and αz are given. Now the three
components are evaluated according to :
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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 :
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And the magnitude of a force (like the resultant) is calculated according to :
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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
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or in equation form :
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
Apart from the difficulty of realizing which forces are acting on a given body
and what can be said about each of these forces much confusion arises
from some sign conventions which are unfortunately necessary.
Please, look very carefully through
this example.
Click here to do the test.
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 :
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
Zig Herzog, hgn@psu.edu
Last revised: 08/28/09
