Video Transcript
In this video, we will learn how to
find the sum of moments of a group of forces acting on a body about a point in two
dimensions.
We know that a nonzero net force
acting on a rigid body produces a linear acceleration of the body in the direction
that the net force acts in. This results in the translation of
the center of mass of the body in that direction. However, a force acting on a body
may also result in the angular acceleration of the body about a point, producing
rotation of the body. The magnitude of this angular
acceleration of the body due to the action of the force is proportional to the
moment of the force about the point.
We will begin by considering what
this looks like in practice. In the diagram drawn, we have a
thin rod suspended vertically at a point 𝑃. A force vector 𝐅 acts on the road
horizontally. Since a moment is the turning
effect of a force about a point, the moment of the force here acts to rotate the rod
about 𝑃. We can go one stage further and say
that the rod rotates in a clockwise direction due to the moment of vector 𝐅 about
𝑃.
We will see later in this video
that moment can act in both a clockwise and counterclockwise direction about a
point. Let’s now consider what happens if
the line of action of force 𝐅 changes so that the line passes through point 𝑃 as
shown in the diagram. This time, the rod does not rotate
about 𝑃 due to the moment of 𝐅. And we can conclude that for 𝐅 to
have a nonzero moment about 𝑃, there must be a nonzero distance between 𝑃 and the
line of action of 𝐅.
When the line of action of the
force 𝐅 and the line connecting a point 𝑃 and a point where the force 𝐅 acts are
perpendicular, then the magnitude of the moment of the force about point 𝑃 is the
product of the magnitude of the force 𝐅 and the distance 𝑑 between the point and
the line of action of the force. This can be written as 𝑚 is equal
to 𝐅 multiplied by 𝑑. If the force is measured in newtons
and the distance in meters, then the unit of the moment of a force is in
newton-meter. We will now look at an example
where we need to find the moment of a force about a point.
If a force of magnitude 498 newtons
is eight centimeters away from a point 𝐴, find the norm of the moment of the force
about point 𝐴, giving your answer in newton-meters.
In this question, we are told that
there is a force of magnitude 498 newtons acting at point eight centimeters from a
point 𝐴. Whilst the question does not
specify that the eight-centimeter distance and the line of action of the force are
perpendicular, we can assume this if nothing indicates otherwise. We can therefore add the line of
action of our force to the diagram. We are asked to find the norm of
the moment of the force, which is its magnitude. And this means it doesn’t matter if
the force is acting in a clockwise or counterclockwise direction.
We know that the moment can be
calculated using the formula 𝑚 is equal to 𝐅 multiplied by 𝑑, where 𝐅 is the
magnitude of the force in newtons and 𝑑 is the perpendicular distance from the line
of action of the force to the point at which we’re taking moments. In order for the moment to be in
newton-meters, we need this distance to be in meters. And since there are 100 centimeters
in a meter, the perpendicular distance is equal to 0.08 meters. We can therefore calculate the
moment by multiplying 498 by 0.08. This is equal to 39.84. And we can therefore conclude that
the norm of the moment of the force is 39.84 newton-meters.
We mentioned earlier that a moment
of a force can produce rotation of a body either clockwise or counterclockwise. We will briefly consider this
before moving on to our next example.
Let’s consider a thin rod suspended
vertically at point 𝑃 once again. This time, there are two forces
acting on it, 𝐅 sub one and 𝐅 sub two. The force 𝐅 sub one produces a
clockwise moment 𝑚 sub one about point 𝑃. And if we let the perpendicular
distance between the line of action of this force and point 𝑃 be 𝑑 sub one, then
𝑚 sub one is equal to 𝐅 sub one multiplied by 𝑑 sub one. The force 𝐅 sub two, on the other
hand, produces a counterclockwise moment 𝑚 sub two about point 𝑃, where 𝑚 sub two
is equal to 𝐅 sub two multiplied by 𝑑 sub two.
The sum of the moments 𝑚 sub one
and 𝑚 sub two is Σ𝑚, where Σ𝑚 is equal to 𝑚 sub two minus 𝑚 sub one. This can also be written 𝑚
net. We subtract 𝑚 sub one from 𝑚 sub
two as counterclockwise moments are usually taken as positive. We will now look at an example
where we need to calculate the net moment due to multiple forces.
𝐴𝐵 is a rod of length 114
centimeters and negligible weight. Forces of magnitudes 83 newtons,
225 newtons, 163 newtons, and 136 newtons are acting on the rod as shown in the
following figure. 𝐶 and 𝐷 are the points of
trisection of 𝐴𝐵, and point 𝑂 is the midpoint of the rod. Find the algebraic sum of the
moments of these forces about point 𝑂.
We are asked to find the sum of the
moments of the four forces about point 𝑂. In order to do this, we firstly
need to calculate the distances between each point on the rod. We are told that 𝐴𝐵 has length
114 centimeters. 𝐶 and 𝐷 are the points of
trisection of 𝐴𝐵. And this means that 𝐵𝐷 is equal
to 𝐷𝐶, which is equal to 𝐶𝐴. All three of these will be equal to
114 centimeters divided by three. This is equal to 38
centimeters. The lengths 𝐵𝐷, 𝐷𝐶, and 𝐶𝐴
are all equal to 38 centimeters.
We are told that 𝑂 is the midpoint
of the rod, which means that 𝐵𝑂 is equal to 𝑂𝐴. These lengths are equal to 114
divided by two, which is equal to 57. The lengths 𝐵𝑂 and 𝑂𝐴 are equal
to 57 centimeters. As the lines of action of all four
forces are acting perpendicular to the rod, we can calculate the moment of each
force using the equation 𝑚 is equal to 𝐅 multiplied by 𝐷, where 𝐅 is the
magnitude of the force measured in newtons and 𝐷 is the perpendicular distance to
the point at which we are taking moments, in this question, measured in
centimeters. This means that the units for each
moment will be newton-centimeters.
We can take moments about any point
on the rod. However, in this question, we are
asked to find the moment of the forces about point 𝑂. We can find the sum of the moments
by firstly finding the moment of each force at points 𝐴, 𝐵, 𝐶, and 𝐷. We are told that moments acting in
the counterclockwise direction will be positive. From the diagram, it is clear that
this is true for the 83- and 163-newton forces at points 𝐴 and 𝐷,
respectively. As these produce counterclockwise
moments about 𝑂, the moment will be positive. The forces acting at point 𝐵 and
𝐶 produce clockwise moments about 𝑂. This means that the moments of the
136- and 225-newton forces will be negative.
Let’s begin by calculating the
moment of force 𝐴. This is equal to 83 newtons
multiplied by a perpendicular distance of 57 centimeters, which is equal to 4731
newton-centimeters. As already mentioned, the moment of
the force acting at 𝐵 will be negative. This is equal to negative 136
multiplied by 57, which is equal to negative 7752 newton-centimeters. The moment of the force acting at
point 𝐶 is also negative. It is equal to negative 225
multiplied by 19, as the distance 𝑂𝐶 is 19 centimeters. The moment here is equal to
negative 4275 newton-centimeters.
Finally, we have the moment of the
force acting at 𝐷 which is equal to 163 multiplied by 19. This is equal to 3097
newton-centimeters. The sum of the moments is therefore
equal to 4731 minus 7752 minus 4275 plus 3097. This is equal to negative 4199. The algebraic sum of the moments of
the four forces about point 𝑂 is negative 4199 newton-centimeters.
In the examples we have seen so
far, the line of action of the forces have always been perpendicular to the
rods. Let’s now consider what happens
when this is not the case. We once again have a thin rod
suspended vertically at a point 𝑃. This time, however, the line of
action of force 𝐅 makes an angle of 𝜃 degrees with the vertical. We know that if 𝜃 is zero, the
line of force 𝐅 must pass through 𝑃. And this would produce zero moment
about 𝑃.
If 𝜃 is equal to 90 degrees, which
means the line of action is perpendicular to the rod, then 𝐅 produces its maximum
moment about 𝑃. This means that the calculation of
the moment of a force must therefore include the angle at which the force acts. Using our knowledge of right angle
trigonometry, the component of the force acting perpendicular to the rod is
therefore equal to 𝐅 multiplied by sin 𝜃. And we can therefore calculate the
moment of a force about a point using the formula 𝑚 is equal to 𝐅 multiplied by 𝑑
multiplied by sin 𝜃, where 𝐅 is the magnitude of the force and 𝜃 is the angle
between the direction of the force and the direction of the line intersecting 𝑃 and
the point where the force acts.
It is worth noting that in our
previous examples, 𝜃 was equal to 90 degrees. And we know that sin of 90 degrees
is equal to one. This allowed us to use the formula
𝑚 is equal to 𝐅𝑑. We can use this simplified version
of the formula when the line of action of our force is perpendicular to the rod. We will now look at one final
example where the angles at which the forces act must be considered.
Three forces, measured in newtons,
are acting along the sides of an equilateral triangle 𝐴𝐵𝐶 as shown in the
figure. Given that the triangle has a side
length of seven centimeters, determine the algebraic sum of the moments of the
forces about the midpoint of 𝐴𝐵 rounded to two decimal places.
In this question, we are asked to
find the sum of the moments of the three forces 300, 100, and 150 newtons. Whilst we could take moments about
any point, in this question, we are asked to take them about the midpoint of 𝐴𝐵,
labeled 𝑃 on the diagram. Since 𝑃 is the midpoint of 𝐴𝐵,
we know that 𝐴𝑃 is equal in length to 𝑃𝐵. And since the side length of the
equilateral is seven centimeters, then both 𝐴𝑃 and 𝑃𝐵 are equal to 3.5
centimeters. The line of action of the
300-newton force is along 𝐴𝐵. And it therefore passes through
point 𝑃. This means that it produces zero
moment about 𝑃 and can therefore be ignored.
We have two other forces of
magnitude 100 and 150 newtons that we need to consider. The 100-newton force acts at point
𝐵 and the 150-newton force acts at point 𝐶. Recalling that we can calculate the
moment of a force using the formula 𝐅 multiplied by 𝑑 multiplied by sin 𝜃, where
𝑑 is the distance between the point at which the force acts and the point at which
we are taking moments, then we need to calculate the length 𝑃𝐶. We can do this using our knowledge
of right angle trigonometry and the fact that an equilateral triangle has equal
angles equal to 60 degrees.
The tangent ratio tells us that tan
𝜃 is equal to the opposite over the adjacent. This means that in triangle 𝑃𝐵𝐶,
𝑃𝐶 over 3.5 is equal to the tangent of 60 degrees. 60 degrees is one of our special
angles. And the tangent of 60 degrees is
root three. Multiplying through by 3.5, we have
𝑃𝐶 is equal to 3.5 root three centimeters. We can now sketch a diagram showing
the magnitude of the forces acting, the angles at which they act, and the distances
from 𝑃. Convention dictates that moments
acting in a counterclockwise direction are positive. In this question, both the
150-newton force and the 100-newton force act clockwise. This means that the moment of both
forces will be negative.
The moment of the 150-newton force
is therefore equal to negative 150 multiplied by 3.5 root three multiplied by sin of
30 degrees. We know that sin of 30 degrees is
one-half. The moment of this force is
therefore equal to negative 525 root three over two newton-centimeters. Repeating this for the 100-newton
force, we have a moment equal to negative 100 multiplied by 3.5 multiplied by the
sin of 60 degrees. As the sin of 60 degrees is root
three over two, this is equal to negative 175 root three newton-centimeters. The sum of the moments about 𝑃 is
therefore equal to the sum of these two moments. This can be written Σ𝑚 or 𝑚 net
and is equal to negative 757.7722 and so on. Rounding to two decimal places as
required, the sum of the moments about the midpoint of 𝐴𝐵 is negative 757.77
newton-centimeters.
We will now finish this video by
summarizing the key points. The moment of a force 𝐅 about a
point 𝑃 is the distance 𝑑 from 𝑃 to the point where the force acts multiplied by
the component of the force perpendicular to the direction of the line intersecting
𝑃 and the point where the force acts. This can be written as 𝑚 is equal
to 𝐅 multiplied by 𝑑 multiplied by sin 𝜃, where 𝐅 is the magnitude of the force
and 𝜃 is the angle between the direction of the force and direction of the line
intersecting 𝑃 and the point where the force acts. The net moment due to a set of
moments about a point is the sum of the clockwise and counterclockwise moments about
the point, where the counterclockwise moments are positive.
