ME102: Subunit 1.2.5: Resultant of Systems
Mehrdad Negahban (1999)
The idea of finding the resultant of forces is to simplify some problems by expressing
two or more forces acting on a point or body as a single force. We consider three cases
here.
a. For the case of forces acting on a point, all we need to do is find the vector sum
of the forces.
Examples:
Here Fa and Fb are equivalent in magnitude but are perpendicular to one
another. The resultant or vector sum of Fa and Fb is given by Fr, which has a
magnitude of sin(45)Fa and acts at a 45-degree angle between Fa and Fb.
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The situation is a little more complicated here. Fa acts horizontally and Fb is
smaller in magnitude than Fa and acts at an angle alpha inclined from Fa. A
simple procedure here is to just calculate the x and y components of Fb (Fbx =
cos(a)Fb, and Fby= sin(a) Fb). Then Frx = Fa + cos(a)Fb, and Fry = sin(a)Fb.
The angle at which Fr acts can by calculated from b= atan(Frx/Fry).
For example, if Fa= 10 N, Fb = 5 N, and a= 75 degrees, then Frx = 10N +
cos(75) 5N = 11.3 N, Fry = sin(75) 5N = 4.83 N, Fr = 11.4 N, and b=23 degrees.
b. For the case of forces acting on a larger object, we must find an equivalent force
and its line of action, so that the resultant force produces the same force and
torque as the multiple forces that it replaces. Although this job may be easily
seen for some simple objects like beams, it may be much more complex for
several forces acting on a large, irregularly shaped object. In these cases, it may
be useful to rely upon a general formula rather than try to analyze each individual
situation (you may wish to refer to the Wikipedia article Net Force for a concise
review), but first let us consider a few simple examples.
Examples:
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Here we have two forces of equal magnitude and direction acting at each end of
a beam in the downward direction. The resultant force is just the total force
acting on the beam Fr=Fa+Fb in the downward direction. Visualize a stick with
two forces of equal magnitude and direction acting at both ends; in this case
there is no tendency for the stick to spin. If we replace the two forces at the ends
with a single force, then we must place it at the center of the stick in order to get
the same tendency not to spin. So, Fr is located at the center of the beam in the
above figure.
In the next example, we let Fa and Fb be unequal in magnitude. The situation is
more complicated to deal with here and we must do some calculations.
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Clearly the total force on the beam is Fr= Fa+Fb, but it is a little trickier to
determine where to place Fr on the beam. Let us calculate the moment, or
torque, about x=0. In this case Fa produces no moment, but the moment from
Fb is just L times Fb or LFb. Fr must be located to produce an equivalent
moment, so RFr = LFb. Hence R = LFb/(Fa+Fb).
You can imagine many more complex situations involving more forces acting in
many different directions. When analyzing a problem, sometimes it makes the
most sense to do the simple resultant calculations first. That is, you can perform
resultant calculations in an iterative sort of way to simplify a problem.
c. Sometimes we would like to express distributed or continuous forces acting on
an object by a single force. This topic is summarized in more detail in subunit
1.2.6.
Example:
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Here we have a beam that is loaded by a block weighing 10 N extending from x=L1 to
x=L2. The main idea we wish to consider here is how to handle forces that may be
applied continuously. In this case, the block exerts a small force at every point on the
beam between L1 and L2, such that the sum, or integral, is 10 N. Here we do not need
to resort to integration to figure out where to place the resultant force. In order for it to
produce the same resultant torque as the block, it should be placed at (L1+L2)/2, or
exactly in the middle of the block.
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