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In the figure to the left I sketched ever so roughly a horizontal
beam and the forces acting on it. In the terminology of Engineering Mechanics the shown beam constitutes a rigid body. P=5 kN is a given load and F1 and F2 are unknown support forces which we want to determine. To make matters a little bit easier I assumed that the two support forces are acting in a vertical direction and that the weight of the beam itself can be neglected. Consider the lengths a and b as given, for example a=2.5m and b=1.1m. |
None of the forces has an x-component, hence the sum of the forces in x-direction is zero automatically.
The sum of the forces in y-direction gives us :
This is one equation for two unknowns which cannot be solved, meaning that for example F1=1kN and F2=4kN will satisfy the above equation as well as any other pair of numbers as long as they add up to 5kN.
If it happens to be that a=b then
is the correct answer because of symmetry.
From daily experience we also know that if the
load P slides over to one end of the beam, the support force
at that end will ultimately carry the entire 5 kN.
The conclusions we have to draw is the following :
looking at the sum of the forces alone and requesting this sum to be zero is certainly not wrong but obviously not enough.
The way out : In addition for the center of gravity of the beam not to move we certainly request that the beam does not rotate around its own center of mass, nor around any other point.
So let's investigate rotational aspects a little bit and see whether we can come up with additional equations ( relationships between the known and unknown forces ) which might help us to solve above and then many more problems.
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In Figure 5.2a you see a horizontal weightless beam which at point A is
anchored by a pine/hole connection (acting like a hing). At distance
Lw a constant weight W is applied.
At distance L the force F is applied. The beam is exactly horizontal and the forces W and F are directed exactly vertically. |
We repeat the experiment with different values for the length L and find :
| (5.2a) L F = Lw W |
The product L times F is called the moment of the force F
with respect to point A, it shows the ability of a force to rotate a body
around a certain point. Equilibrium then means that the moment of all the
forces around the same point ( here F and W w.r.t. point A)
have to cancel each other.
Note the length L and force F have to be perpendicular
to each other. If they don't we cannot use equation (5.2a).
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This experiment is an extension of the one from section 5.2, except that
now the force F is not pointing vertically up but forms an angle
with respect to the axis of the beam.
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For different values of L and
we determine experimentally that F which keeps the beam at rest against the
tendency of W to rotate the beam in clockwise direction around point A
and find :
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(5.3a) L F sin(
) = Lw W
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Note that Equation 5.2a is just a special case of Equation 5.3a with sin(90o)=1
Definition :
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M = L F sin() = moment of force F w.r.t point A Point A = point of rotation Point B = point of application (of force F) |
Correspondingly, we would call the product Lw W
the moment
of the force W w.r.t. the point A. In addition to the numerical value
of the moment of forces we have to consider as to how a body (here the beam)
would rotate if the force under consideration would have its way.
In both experiments, the force F
and the force W would rotate the beam around
an axis perpendicular to the viewing plane, force F would like to rotate
the beam in counterclockwise direction (at least for the shown angle)
and force W in clockwise direction.
We conclude :
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A body does not rotate around an axis through a specified point if the sum* of the moments of ALL FORCES around this axis is zero. * counting ccw positive and cw negative |
| In Figure 5.3b we replaced the force F by two orthogonal components, F= being directed along the line AB and F| perpendicular to that. |
From experience we know that the force F= cannot rotate the beam because it acts radially outward.
The moment of the force F| according to Equation (5.2a) would be :
which when incorporating how F| is related to F becomes again :
)The seond way of how to relate Equation 5.3a back 5.2a is shown in Figure 5.3c.
| Here I entered a line through the point B (which is the point of application of the force F) and which is exactly parallel to the force F and basically extends to infinity on either end. The point C on this line is chosen such that the lines AC and BC form a right angle at point C. |
The moment of the force Fy according to Equation (5.2a) would be :
which when including the geometric relationship between L and LAC becomes again :
)Definition :
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A line going through the point of application and along the direction of action of a force is called the line of action .
The line from the point of rotation perpendicular |
Figure 5.3c also suggests the following :
| The moment of force F does not change with changes in the location of the point of application as long as the point of application remains on the original line of action. |
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In Figure 5.4a an arbitrarily shaped body is supported at point A.
A force F is acting on that body at point B. The angle
between the force F
and the location vector
going from point A to point B is assumed to be given as well.
The shown picture is strictly 2-dimensional.
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As before, the moment of the force F with respect to the point A is given by :
)
and according to Figure 5.4a it would try to rotate the body in counter-clockwise direction about point A.
Let me now determine the moments of the force components Fx and Fy and then show that the sum of their moments is identical to the result in Equation 5.4a.
As you can see from Figure 5.4a, Fx has the moment arm ry and tries to rotate the body around point A in clockwise direction. Fy on the otherhand tries to rotate the body in counter-clockwise direction and has the moment arm rx. Because the forces try to turn the body in opposite direction I take the difference of the moments in order to determine their combined moment :
In trying to recover from this Equation 5.4a I express now in Equation 5.4b
the x- and y-components of
and
in terms of their respective magnitudes and the angles
and
.
The result is :
which is exactly equal to the result of Equation 5.4a.
Or in summary :
| The moment of an arbitrary force can be calculated by taking the sum (difference) of the moments of the components of the force. |
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Figure 5.5a Moment vector |
In this coordinate system the force vector
will have certain values for its x- y- and z-components which depend
on the force F itself but also on how the beam is oriented
with respect to the x-y-z coordinate system.
Along the axis of the beam I drew an arrow pointing from the
point of rotation, A, to the point of application, B, of the force F.
By going from point A to point B you will advance by a certain amount
in the x-direction, say rx, in the y-direction by ry,
and in the z-direction by the amount rz.
These three values make up a vector
:
= (rx,
ry,rz)
Now, the force F will try to rotate the beam exactly as before, namely
around an axis through point A which is exactly perpendicular to the
plane spanned by the vectors
and
.
In Figure 5.5a this axis of rotation is represented by the arrow (vector)
. If you look along this
vector (from its tail towards its head) the tendency of F would be to turn
the beam AB around this axis in clockwise direction as indicated.
What I have so painstakingly gathered here --
from the point of physics -- are all the properties
(direction, sense of rotation, and magnitude r F
sin() ) of
a vector which results from calculating the cross-product of
and
as defined in section 4.7 .
Formally, we wrote
|
We now refer to
as the moment vector.
The values of its components can be calculated according to
Equation
4.7c and 4.7d provided
and
are known.
have physical significance.
As an example I bring below the interpretation of Mx.
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In Figure 5.5b you see the same beam, the same force, and the same
x-y-z coordinate system as in Figure 5.5a. Our
direction of view though is now exactly in the minus x-direction.
Hence, the vectors
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I want to determine now what the moment of the force F around the x-axis is. Clearly, Fx does not contribute to it; this components may have the tendency to rotate the beam AB into the x-axis but not around it. The z-component of F on the other hand tries to rotate the beam in a counter-clockwise direction having the moment arm ry. Fy has as moment arm rz but tries to rotate the beam in clockwise direction. The net-result is exactly :
| Mx = ryFz -rzFy |
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The value of Mx, the x-component of the moment vector,
represents the strength of the force
to rotate a body around an axis parallel to the x-axis
going through the point of rotation.
If Mx is positive the body will rotate in clockwise direction when looking along the x-axis in +x direction. Analog statements are true for My and Mz. |
As an exercise you might want to try your luck on Prob. 5.9d .
If you like to have some additional information about the vector product feel free to take a look at the cross-product of two vectors from the point of you of the physicists at the University of Syracuse. Your web-browser needs to be "java-enabled" to appreciate all of their work. The page you will see has a larger picture and a lot of hidden code attached to it .... be patient while everything downloads. You can modify their big picture by clicking and dragging your left mouse button on various displayed items. Give it a shot .......
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In Figure 5.6a such a case is displayed. A wheel mounted on an axis AB is
acted upon by a force
and we wish to
determine its moment around the axis AB.
If we had a choice in the orientation of the x-y-z coordinate system
we could determine the desired moment by, for example, aligning the
x-axis with the axis AB and then calculate the moment with respect
to point A using the vectors
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Mx is related to the magnitude M of the moment vector by :
where ax is the angle between
and the x-axis.
All seems to be nice and tandy, except if the x-y-z coordinate system is prescribed otherwise and none of its axes are aligned with the line AB.
Well, here is my suggestion :
or in equation form :
(5.6a) |
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Why ?
By definition of the dot-product the value of the dot-product
equals :
where ax is the angle between unit vector along
the line AB and the moment vector
.
But this is exactly equal to the value I had obtained when I did choose a coordinate system aligned with the line AB !!!!!
The expression on the right-hand side of Equation 5.6a is often
referred to as the triple product of vector algebra.
Instead of first evaluating the cross-product and then multiplying the
resulting moment vector with the unit vector you can also evaluate the
determinant in the equation below :
(5.6b) 
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In order for a rigid body not to experience rotation under influence of external forces :
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The sum of the moments around ANY point of ALL the
forces acting on a rigid body has to be ZERO.
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Equation 5.7a summarizes in essence the requirement that the sum of the moments of all forces around the x-axis, the y-axis, and the z-axis has to be zero. A body that does not rotate about any of three mutually orthorgonal axes DOES NOT rotate at all !!!!!
Here are some important points to be kept in mind :
support forces F1 and F2 for the horizontal beam.
Sum of the forces equal to zero :
Sum of the moments around point B equal to zero :
With a=2.5m and b=1.1m we get :
The determination of the moment of forces around certain points or axes
is an essential part of statics.
In 2-dimensional problems where the moment arm and the force vector lie in the plane of the display area we have three alternatives available :
In 3-dimensional cases we have to distinguish between the moment about a point and the moment around an axis.
Moments around a point are characterized by the moment vector which is determined as the cross-product of location vector ( pointing from the point of rotation to the point of application ) and the force vector ( see Equation 4.7b ). The axis of rotation coincides with the direction of the moment vector, the magnitude of the moment vector reflects the strength with which the force wants to rotate the body under investigation.
In 3-dimenions the moment around a specific axis (which may be of interest when a given body is allowed to rotate only about a prescribed axis because of the presence of bearings) is a scalar value to be determined by the triple product of a unit vector along the axis of rotation, a location vector which points from anywhere on the axis of rotation to the point of application of the force), and the force vector (see Equation 5.6a and 5.6b).
Prob. 5.9a :
Evaluate moment , 2-D
Prob. 5.9b :
Evaluate moment , 2-D
Prob. 5.9c :
Evaluate moment , 2-D
Prob. 5.9d :
Evaluate moment about a point in 3-D
Prob. 5.9e :
Evaluate moment about an axis in 3-D
Prob. 5.9f :
Moment of two opposing forces in 3-D
Zig Herzog, hgn@psu.edu Last revised: 12/07/00
pointing along the line AB. This can be done fairly easily if you
know the coordinates of the point A and B for example.
