Video Transcript
In this lesson, we’ll learn how to
solve problems that involve the equilibrium of a body under the effect of parallel
coplanar forces. Let’s begin by recalling some of
the key facts and terms that we use when working with forces. Informally, we consider that when a
body is in equilibrium, it’s kind of balanced. And so, formally, we say that for a
body to be in equilibrium, the sum of the forces acting on that body must be equal
to zero.
Now, since force can have direction
as well as magnitude, we might also say that the vector sum of the forces is
zero. Alternatively, if we think about a
force acting in two directions, we can define a direction for the force in terms of
the 𝑥- and 𝑦-direction. And we can say that the sum of the
forces in each direction is zero. The sum of 𝐹 sub 𝑥 is zero, and
the sum of 𝐹 sub 𝑦 is zero.
Similarly, for a body to be in
equilibrium, we say that the sum of the moments must be equal to zero, where the
moment is the turning effect of the force. The formula we use to calculate a
moment is 𝐹𝑑. Now, 𝐹 here is the force that acts
at a point, whereas 𝑑 is the perpendicular distance from the line of action of the
force to the point about which the object will turn. When thinking about the equilibrium
of a rigid body, such as a rod, under a system of forces, we often need to combine
both of these ideas to help us find missing information. Let’s see what that might look
like.
A uniform rod having a weight of 35
newtons is resting horizontally on two supports 𝐴 and 𝐵 at its ends, where the
distance between the supports is 48 centimeters. If a weight of magnitude 24 newtons
is suspended at a point that is 38 centimeters away from 𝐴, determine the reactions
of the two supports 𝑅 sub 𝐴 and 𝑅 sub 𝐵.
When presented with a question like
this, before doing anything, we begin by drawing a free body diagram. This is a very simple diagram that
highlights the key forces that are acting on our body. Here is our 48-centimeter rod
resting on supports 𝐴 and 𝐵 at its ends.
Next, we’re told that the rod is
uniform. And this means that the downwards
force of its weight, that’s 35 newtons here, must act at a point exactly halfway
along the rod itself, so 24 centimeters then from either end. We then have another downwards
force. There is a weight of magnitude 24
newtons suspended at a point 38 centimeters away from 𝐴.
So this is everything that we’ve
been given in the question, but we’re not quite finished with our diagram. Newton’s third law of motion, which
is often informally stated as every action has an equal and opposite reaction, tells
us that since the rod is exerting a downwards force on both supports 𝐴 and 𝐵,
there must be a reaction force of the supports on the rod itself. These reaction forces are normal,
that’s perpendicular, to the rod. And we’re going to call them 𝑅 sub
𝐴 and 𝑅 sub 𝐵. And so now we have all of our
forces, what do we do next?
Well, we’re told the rod is
resting, so we’re going to make an assumption. And that assumption is that the rod
itself is in equilibrium. For a body to be in equilibrium,
there are two important criteria. The first is that the sum of all
the forces acting on that body must be equal to zero. And the second is that the sum of
all the moments must also be equal to zero, where the moment, which is essentially
the turning effect of the force, is calculated by multiplying 𝐹 by 𝑑. 𝐹 here is the force that acts at a
point, and 𝑑 is the perpendicular distance from the line of action of this force to
the point about which the object will turn.
So let’s begin by forming an
equation involving the sum of the forces. In this case, forces are only
acting in the vertical direction. So we’re going to define a positive
direction here. We’re going to define upwards to be
positive so that 𝑅 sub 𝐴 and 𝑅 sub 𝐵 are acting in the positive direction. And then 35 and 24 are acting in
the negative direction. We can therefore say that the sum
of the forces acting in this direction must be 𝑅 sub 𝐴 plus 𝑅 sub 𝐵 minus 35
minus 24. And of course we know that the sum
of these forces must be equal to zero for the body to be in equilibrium. Negative 35 minus 24 is negative
59. So we’re going to add 59 to both
sides of this equation. And we find that 𝑅 sub 𝐴 plus 𝑅
sub 𝐵 is equal to 59.
And there’s not a lot else we can
do with the sum of the forces. And so we move on to the second
piece of information. That is, the sum of the moments of
our forces must also be equal to zero. And so we choose a point about
which to take moments. Now, it really doesn’t matter the
point about which we choose to take the moments as long as we’re very careful to
define a direction. In this case, we’re going to take
moments about one of the endpoints of our rod.
Let’s take moments about 𝐴 since
we’ve defined all of the distances from this point. And we’re going to define the
counterclockwise direction to be positive here. And of course we define a moment as
force times the perpendicular distance from the line of action of the force to the
point about which the object will turn.
It’s worth noting that we generally
calculate moments in terms of newton meters. And so our forces are in newtons
and our distance is in meters. However, in this case, we’re
working with newtons and centimeters. And it’s absolutely fine to work in
newton centimeters as long as we’re consistent. So let’s begin by calculating the
moment of this 35-newton force. It’s trying to turn the body in a
clockwise direction, so the moment is going to be negative. And it’s acting at a point 24
centimeters away from 𝐴. So the moment is going to be
negative 35 times 24.
Working from left to right, we now
see we have to deal with this 24-newton force. Once again, this is trying to move
the body in a clockwise direction. So its moment is going to be
negative. Force times distance here is 24
times 38. Then there’s one more moment we
need to consider. And that’s the moment of the
reaction force at 𝐵. This time, this is acting in a
counterclockwise direction, so the moment is going to be positive. It’s 𝑅 sub 𝐵 times the distance
from 𝐴, which is 48. So that’s the sum of our
moments. And of course we know those are
equal to zero. Negative 35 times 24 minus 24 times
38 is negative 1752. So our equation simplifies as
shown.
We can now solve this equation for
𝑅 sub 𝐵 by adding 1752 to both sides. So 48𝑅 sub 𝐵 is 1752. Then, finally, we divide through by
48. 1752 divided by 48 is 36.5. And so we can say that 𝑅 sub 𝐵 is
36.5 newtons. We still need to calculate the
value of 𝑅 sub 𝐴. So we’re going to clear some space
and go back to this equation here. 𝑅 sub 𝐴 plus 𝑅 sub 𝐵 is 59. We can now replace 𝑅 sub 𝐵 with
36.5. And so this equation becomes 𝑅 sub
𝐴 plus 36.5 equals 59. Let’s solve for 𝑅 sub 𝐴 by
subtracting 36.5 from both sides. So 𝑅 sub 𝐴 is 59 minus 36.5,
which is 22.5. So we’ve calculated the reactions
of the two supports. 𝑅 sub 𝐴 is 22.5 newtons, and 𝑅
sub 𝐵 is 36.5 newtons.
So we saw that by considering the
forces and the moments acting on the body simultaneously, we could create a system
of linear equations which we could solve to find missing forces. We can also use this very same
process to find other missing information, such as the length of the rod. Let’s have a look at an example of
this.
In the figure, forces having
magnitudes of 61, 43, 100, and 𝐹 newtons are acting on the light rod, and the rod
is in equilibrium horizontally. Determine the length of the line
segment 𝐷𝐴 and the magnitude of 𝐹.
The key to answering this question
is to spot that the rod is in equilibrium. So what does that mean? Well, it means two things. Firstly, the sum of all the forces
acting on the rod is equal to zero. In this case, we’ll consider the
sum of the forces acting in a vertical direction. Secondly, it also means that the
sum of the moments of our forces is also equal to zero, where the moment is
calculated by multiplying the force acting at a point by the perpendicular distance
from the line of action of this force to the point about which the object will
turn.
So let’s begin by considering the
sum of the forces acting on our diagram. Let’s define the positive direction
to be upwards so that 𝐹 is acting in the positive direction, whilst 61, 43, and 100
are acting in the negative direction. We can therefore say that the sum
of our forces must be 𝐹 minus 61 minus 43 minus 100. And of course we know that this sum
is equal to zero.
Now, in fact, the expression on the
left-hand side simplifies to 𝐹 minus 204. So 𝐹 minus 204 is equal to
zero. We’ll solve for 𝐹 by adding 204 to
both sides. So 𝐹 is equal to 204 or 204
newtons. So we’ve now calculated the value
of 𝐹 and we’ve done everything we can with the sum of our forces. So we move on to the second
criteria. The sum of the moments of our
forces is equal to zero.
Now, we’ve been defined a positive
direction here. That’s the counterclockwise
direction. And we know that a moment is
calculated by multiplying the force by the perpendicular distance of this force from
the point about which the object is trying to rotate. We have a few distances on here,
but there is one distance that we’re missing.
Let’s define the distance between
𝐴 and 𝐷, so the length of the line segment 𝐷𝐴, which coincidently we’re trying
to find, to be 𝑥 centimeters. And then once we have that
information, we pick a point about which to calculate our moments. Now, we can pick any point on the
rod itself. We’re going to pick 𝐷 here. Now, the reason we’re going to pick
𝐷 is because the force 𝐹 is acting at this point. And had we not calculated 𝐹 yet,
we still would’ve been able to calculate the moments about this point since the
moment of 𝐹 would’ve been zero.
It’s also worth noting that we
generally choose to work in newton meters when calculating moments. But actually, our dimensions are in
centimeters. So we’re going to be working in
newton centimeters throughout. And that’s absolutely fine as long
as we’re consistent. So let’s find the moment of our
61-newton force. This force is trying to turn the
object in a counterclockwise direction. So its moment will be positive. It’s 100 centimeters away from 𝐷,
so the moment is 61 times 100.
Moving from left to right, we’re
now going to deal with the force at 𝐶. Once again, this is trying to move
the object in a counterclockwise direction. So its moment is positive. But now it’s 43 times 50. As we said, the force 𝐹 is zero
centimeters away from 𝐷. So we don’t need to worry about
that moment. And instead we move on to the force
at 𝐴.
Now, this force is trying to move
the object in a clockwise direction. So its moment is actually going to
be positive. And it’s 100 times the distance
from 𝐷, which is 𝑥. We know of course that the sum of
these moments is zero. So we can form an equation in
𝑥. This equation simplifies to 8250
minus 100𝑥 equals zero. Then we add 100𝑥 to both sides and
finally divide through by 100. So 𝑥 is 82.5. And we see that the length of line
segment 𝐷𝐴 is 82.5 centimeters.
In our final example, we’ll see how
we can use the information about an object in equilibrium and a little bit of
intuition to model problems in which a rod is on the point of rotating.
The length of a rod 𝐴𝐵 is 111
centimeters, and its weight is 95 newtons, which is acting at its midpoint. The rod is resting horizontally on
two supports, where one of them is at end 𝐴 and the other is at a point 𝐶, which
is 30 centimeters away from 𝐵. A weight of 71 newtons is suspended
from the rod at a point that is nine centimeters away from 𝐵. Find the magnitude of weight 𝑤
that should be suspended from end 𝐵 so that the rod is about to rotate, and
determine the value of the pressure 𝑃 exerted on 𝐶 in that situation.
Let’s begin by drawing a free body
diagram showing this scenario. Here is our rod. Now, the downwards force of its
weight acts at its midpoint. So that’s 55.5 centimeters from
either end. Now, we also have this 71-newton
force. Now, that’s acting downwards at a
point nine centimeters away from 𝐵. We want to add a weight 𝑤 at point
𝐵. So we add that to the diagram and
two further forces. Those are the reaction force of the
support on the rod. We’ll call them 𝑅 sub 𝐴 and 𝑅
sub 𝐵, respectively.
Now, the rod is about to rotate, so
it’s essentially in limiting equilibrium. So we can say two things. Firstly, the sum of all the forces
acting on the body is zero. And secondly, the sum of the
moments is also equal to zero. So we’ll begin by considering the
sum of the forces. We’re going to take the direction
in which the reaction forces are acting as being positive.
And so this means that the
95-newton force, the 71-newton, and the weight force 𝑤 must be acting in the
opposite direction. And so we can therefore say that
the sum of the forces, which we know is equal to zero, is 𝑅 sub 𝐴 plus 𝑅 sub 𝐶
minus 95 minus 71 minus 𝑤 equals zero. And if we add 95 and 71 to both
sides of this equation, we get 𝑅 sub 𝐴 plus 𝑅 sub 𝐶 minus 𝑤 equals 166.
There’s not a lot more we can do
with this. And so we move on to the next bit
of information. And this is the sum of the moments
of our forces is also equal to zero. Let’s clear some space and find a
point about which we want to take moments.
Now, since we have this unknown
force 𝑤 at 𝐵, let’s take moments about 𝐵. And of course we define a direction
to be positive. This time, let’s choose
counterclockwise. We begin by thinking about the
reaction force at 𝐴. We know that this is trying to move
the body in a clockwise direction. And so its moment is going to be
negative. It’s negative 𝑅 sub 𝐴 times
111.
The moment of the weight force is
acting in a counterclockwise direction, so it’s positive. It’s 95 times 55.5. We then have negative 𝑅 sub 𝐶
times 30. That’s the moment of the reaction
force at 𝐶. And then we have 71 times nine. That’s the moment of this 71-newton
force. The sum of these moments, which
we’re of course measuring in newton centimeters here, is equal to zero. And this simplifies as shown.
Now, there’s one bit of information
that we haven’t yet used. And that is that the rod is on the
point of rotating. Now, since we’re adding a weight at
𝐵, we know the object is going to be rotating about 𝐶. And so this means that, at this
exact moment, there must be no downwards force of the rod on point 𝐴. And so, similarly, the reaction
force at 𝐴 must also be equal to zero. So we can rewrite our equation by
letting 𝑅 sub 𝐴 be equal to zero. And we’ll solve by adding 30𝑅 sub
𝐶 to both sides. So 30𝑅 sub 𝐶 is 5911.5. And when we divide both sides of
this by 30, we get 𝑅 sub 𝐶 equals 197.05. But how does this help?
We were trying to find the value of
the pressure 𝑃. But of course pressure is the
amount of force acting on an area. So since this force is acting on a
point, we can say that 𝑅 sub 𝐶 must actually be equal to 𝑃. And so 𝑃 must be equal to 197.05
newtons. Let’s clear some space and use this
information in our first equation here. Using 𝑅 sub 𝐴 equals zero and 𝑅
sub 𝐶 as 197.05, our equation is as shown. We add 𝑤 to both sides and then
subtract 166. And so we get that 𝑤 is 31.05 or
31.05 newtons.
We’re now going to recap the key
points from this lesson. In this video, we saw that if a
body is in equilibrium, we can say that the sum of all forces acting on that body
must be equal to zero. Since force can have direction as
well as magnitude, we define a direction for our forces and say that the sum of the
forces in each direction is zero. We can also say that the sum of the
moments must also be equal to zero. We saw that we can consider the
forces and the moments acting on the body to help us create a system of linear
equations, which we can solve to find missing values.
