Video Transcript
In this video, we will learn how to
investigate the equilibrium of a rigid body under the action of two or more coplanar
couples. We will begin by recalling the
definition of coplanar couples and equilibrium. We will also remind ourselves how
we resolve vertically, horizontally and take moments about a point.
A system is said to be in
equilibrium under the action of two or more forces if it remains stationary. When a system is in equilibrium,
there is no resultant force and therefore no acceleration. This means that when we resolve
horizontally, the sum of forces in the π₯-direction is equal to zero. When resolving horizontally, we
usually take the positive direction to be to the right. In the same way, when we resolve
vertically, the sum of the forces in the π¦-direction is also equal to zero. This time, we usually take the
upwards direction to be positive.
We also know that when a system is
in equilibrium and we take moments about a point, then the sum of the moments will
equal zero. In this situation, we normally take
the counterclockwise or anticlockwise moments to be positive. We know that the moment of a force
is equal to the magnitude of the force multiplied by the perpendicular distance
between its line of action and the axis of rotation. This gives us the equation that the
moment of a force is equal to the force multiplied by the perpendicular distance,
where the force is measured in newtons and the distance in either meters or
centimeters. This means that our units for
moments are either newton meters or newton centimeters.
For the purposes of this video, our
system will contain coplanar couples. When all forces are acting in the
same plane, theyβre called coplanar. A couple refers to two forces that
are equal in magnitude, opposite in direction, and do not share a line of
action. For example, the vertical forces at
π΄ and π΅ both have magnitude πΉ newtons. The force at π΄ is acting
vertically upwards, whereas the force at π΅ is acting vertically downwards. This means that they form a
coplanar couple.
A coplanar couple does not have to
be acting vertically or horizontally. For example, the two forces shown
have magnitude π newtons. Theyβre acting in opposite
directions as the forces are parallel. We know this as the forces make an
angle of π with the horizontal. The forces do not share a line of
action; therefore, they are a coplanar couple. When dealing with forces of this
type, we will need to find their vertical and horizontal components. If the force πΉ is acting at an
angle of π degrees to the horizontal, we can use our knowledge of right-angled
trigonometry to calculate the horizontal and vertical components.
The horizontal component will be
equal to πΉ cos π and the vertical component equal to πΉ sin π. This is down to the fact that the
trig ratios tell us that the sine of angle π is equal to the opposite over the
hypotenuse and the cosine of angle π is equal to the adjacent over the
hypotenuse. We will now look at some questions
where we need to calculate unknown forces, angles, or distances.
π΄π΅ is a rod having a length of 90
centimeters and a negligible weight. It is suspended horizontally by a
pin at its midpoint. Two forces each of magnitude 7.5
newtons are acting at its ends, as shown in the figure. It is also pulled by a string whose
tension is 25 newtons in a direction making an angle of 30 degrees with the rod from
point πΆ. If a force πΉ is acting on the rod
at point π· so that the rod is in a horizontal equilibrium position, find the
magnitude of πΉ, its direction π, and the length πΆπ·.
As the rod is in equilibrium, we
know that the sum of the forces in the π₯- or horizontal direction equals zero. Likewise, the sum of the forces in
the π¦- or vertical direction also equals zero. It will also be true that the sum
of the moments about any point in the system will equal zero. We will take the positive
directions to be to the right, vertically upward, and counterclockwise. We are told that the rod is of
length 90 centimeters. If we let the point π be the
center or midpoint of the rod, we know that π΄π is equal to ππ΅, which is equal to
45 centimeters as this is a half of 90.
We will now clear some space so we
can calculate the three unknowns required. The 25-newton force and πΉ-newton
force are not acting horizontally or vertically. This means that we need to find the
horizontal and vertical components before we resolve in these directions. We can do this using our knowledge
of right-angled trigonometry, where the sine of angle π is equal to the opposite
over the hypotenuse and the cosine of angle π is equal to the adjacent over the
hypotenuse.
If we consider the force at point
πΆ, we will let the horizontal component be equal to π₯ and the vertical component
equal to π¦. This means that the sin of 30
degrees will be equal to π¦ over 25 and the cos of 30 degrees is equal to π₯ over
25. We can multiply both sides of these
equations by 25. This means that π¦ is equal to 25
multiplied by the sin of 30 degrees, and π₯ is equal to 25 multiplied by the cos of
30 degrees. The sin of 30 degrees is equal to
one-half, and the cos of 30 degrees is equal to root three over two. This means that we have a vertical
component equal to 25 over two or 12.5 newtons and a horizontal component equal to
25 root three over two newtons.
We can repeat this for the
πΉ-newton force at point π·. This time, we have a horizontal
component acting to the left equal to πΉ cos π and a vertical component acting
downward equal to πΉ sin π. We can now resolve in both the
horizontal and vertical directions. The forces acting in a horizontal
direction are 25 root three over two and πΉ cos π. As the sum of the forces is equal
to zero and the πΉ cos π is moving to the left, we have 25 root three over two
minus πΉ cos π is equal to zero. 25 over two is equal to 12.5, so
this simplifies to 12.5 root three minus πΉ cos π is equal to zero. We can then add πΉ cos π to both
sides, giving us an expression for this equal to 12.5 root three. We will call this equation one.
There are four forces acting in a
vertical direction. From left to right, these are
negative 7.5, positive 12.5, negative πΉ sin π, and positive 7.5. We know that these must also sum to
equal zero. The two 7.5-newton forces are an
example of a coplanar couple. They have the same magnitude but
act in the opposite direction. This means that when resolving
vertically, these forces will cancel as negative 7.5 plus 7.5 is equal to zero. Adding πΉ sin π to both sides of
our equation gives us 12.5 is equal to πΉ sin π. We will call this equation two.
We now have a pair of simultaneous
equations with two unknowns, πΉ and π. Dividing equation two by equation
one, we get 12.5 over 12.5 root three is equal to πΉ sin π over πΉ cos π. We can divide the numerator and
denominator of the left-hand side by 12.5, giving us one over root three. We can divide the numerator and
denominator of the right-hand side by πΉ. This leaves us with sin π over cos
π, which we know is equal to tan π. We can then take the inverse
tangent of both sides of this equation such that π is equal to the inverse tan of
one over root three. This gives us a value of π equal
to 30 degrees. We have calculated the direction of
the force πΉ.
Next, we need to substitute π
equals 30 degrees into equation one or equation two. If we substitute into equation two,
we have 12.5 is equal to πΉ sin 30. The sin of 30 degrees is equal to
one-half. So, 12.5 is equal to πΉ multiplied
by one-half. We can then divide both sides of
this equation by one-half or multiply them both by two to give us a value of πΉ
equal to 25 newtons. The magnitude of force πΉ is 25
newtons.
We will now clear some space and
use moments to calculate the length of πΆπ·. Before doing this, however, it is
worth noting that the force πΉ and the 25-newton force are a coplanar couple. This is due to the fact theyβre of
equal magnitude and act in the opposite direction. The two forces are parallel to one
another. This means that the system in this
question consists of two pairs of coplanar couples. We will take moments about the
midpoint π. But before doing this, we will tidy
up our diagram so we have the relevant forces and distances.
As we have two pairs of coplanar
couples, the distance πΆπ must be equal to the distance ππ·. We will call this distance π₯
centimeters. We know that the moment of a force
is equal to its magnitude multiplied by the perpendicular distance. The force at π΄ is acting in a
counterclockwise direction. Therefore, the moment is equal to
7.5 multiplied by 45. The force at πΆ is acting in a
clockwise direction; therefore, this is equal to negative 12.5π₯. The vertical force at π· has the
same magnitude. This is also acting in the
clockwise direction and is equal to negative 12.5π₯. Finally, the moment of the force at
π΅ is acting in a counterclockwise direction and once again is equal to 7.5
multiplied by 45. We know that the sum of these four
values equals zero.
This simplifies to 337.5 minus
12.5π₯ minus 12.5π₯ plus 337.5 is equal to zero. Collecting like terms, we have 675
minus 25π₯ is equal to zero. Adding 25π₯ to both sides, we have
25π₯ is equal to 675. And then dividing both sides by 25
gives us π₯ is equal to 27. This means that the lengths of πΆπ
and ππ· are equal to 27 centimeters. 27 plus 27 is equal to 54. Therefore, the length πΆπ· is 54
centimeters. We now have our three answers. The magnitude of πΉ is 25 newtons,
the direction π is 30 degrees, and πΆπ· is 54 centimeters.
In our next question, we will need
to find the value of a force that keeps the rod in equilibrium.
π΄π΅ is a rod having a length of 50
centimeters and a negligible weight. Two coplanar pairs of forces are
acting on the rod as shown in the figure. The first couple consists of two
forces acting perpendicularly to the rod, each of magnitude two kilogram-weight. And the second couple consists of
two forces, each of magnitude πΉ. Determine the value of πΉ that
makes the rod in equilibrium.
We recall that a coplanar couple is
two forces that are equal in magnitude and opposite in direction. In this question, we have two such
couples. In order to calculate πΉ, weβll
need to take moments about a point on the rod. We know that when the rod is in
equilibrium, the sum of our moments is equal to zero. We know that a moment is equal to
the force multiplied by the perpendicular distance. This means that we will firstly
need to calculate the vertical components of the forces πΉ.
Using our knowledge of right-angle
trigonometry, we know that the sine of angle π is equal to the opposite over the
hypotenuse. This means that, in this question,
the sin of 45 degrees is equal to π¦ over πΉ. The vertical components are
therefore equal to πΉ multiplied by sin of 45 degrees. The sin of 45 degrees is equal to
root two over two. We will now take moments about
point π΅, taking the counterclockwise direction to be positive.
The first force of magnitude two
newtons will have a moment equal to two multiplied by 10 as it is 10 centimeters
from π΅. Next, we have root two over two πΉ
multiplied by 30. Our final force is acting in a
clockwise direction, giving us negative two multiplied by 50. The sum of these forces equals
zero. Our equation simplifies to 20 plus
15 root two πΉ minus 100 equals zero. 20 minus 100 is equal to negative
80, and we can add this to both sides. Dividing both sides by 15 root two
gives us πΉ is equal to 16 over three root two. We can then rationalize the
denominator, giving us a value of πΉ equal to eight root two over three
kilogram-weight.
We will now summarize the key
points from this video. We found out in this video that a
system is in equilibrium if it remains stationary. As there is no resultant force, the
sum of the forces in the horizontal or π₯-direction equal zero. The sum of the forces in the
vertical direction equal zero, and the sum of the moments about any point equal
zero. We can use our knowledge of
right-angle trigonometry to calculate the horizontal and vertical components of a
force. Finally, we saw that a coplanar
couple consists of two forces of equal magnitude acting in opposite directions.