Let’s talk about equilibrium of a particle. In simple terms, equilibrium is used to describe
an object that’s not moving. So for example, a flower pot hung on the ceiling, or a picture
frame hung on a wall, or something like that. The possibilities are endless. In this video,
we are going to look at just coplanar forces, so a system of forces that lie in the x-y plane.
Let’s say we want to figure out the tension or a force in a cable. Then we need to write equations
of equilibrium. One for the x-axis forces, and one for the y-axis forces. All we’re saying is that
all x-axis forces added together must equal zero for the object to stay at rest. The same goes for
y-axis forces. When we do that, we need to make sure we break each force into x-y and components.
Then all we do is sum up all the x components and equal it to zero, and all the y components
and equal it to zero. Now, let’s discuss springs. When we have a spring like this, and we stretch
it or compress it, the spring tries to go back to equilibrium. We can figure out that force
by using hook’s law. K is the stiffness of the spring, and s is the elongation or compression.
Next, we need to discuss cables. A cable can only support a tension or a pulling force. Remember,
you can’t push on a rope. This force will always act in the direction of the cable.
If we have a cable going through a pully, the tension throughout the whole cable
will be the same. Lastly, objects have a weight, and that’s simply the mass of the object times the
acceleration due to gravity, which is 9.81 m/s^2 or 32.2 ft/s^2. Weight is always straight down,
even if an object is on an inclined plane. Now let’s move onto some examples
to see how we can solve for unknown forces. Let’s take a look at this problem, where we
have a cylinder in equilibrium. We need to figure out the tension developed in the wires CA and CB.
The first step is to pick our point where we will write out equations of equilibrium.
For that, we will pick point C. Now, we draw our coordinate system,
so we have the x axis and the y axis. So now, we can draw our forces. We have a force in cable CA,
a force in cable CB, and we have the weight of the object downwards. The weight is the
mass of the object times the acceleration due to gravity, so that would be 98.1 N.
Now we can draw the components of each force. Let’s focus on force CA. We have the x-component,
which can be found using cosine and the y-component, which can be found using sine.
Now for force CB. If this is unfamiliar to you, please check the description for vector
addition of coplanar forces where I explain how to break forces into components step by step.
It is time for us to write our equations of equilibrium. First for x axis forces.
We will assume right is positive. So we are just looking for any force
that lies on the x-axis. We see 2 forces. We have the x components of forces CA
and CB. Notice that the x component of force CA is facing to the left, so it’ll be negative.
Next, we look at y-axis forces. We will assume up to be positive. There are 3 forces to consider.
We have the y components of force CA and CB, and we have the weight which is straight down.
Since the weight is downwards, it’ll be negative. We now have 2 equations with 2 unknowns.
You can solve them anyway you like. So these are the tensions in the cables
to hold the 10 kg cylinder in equilibrium. Let’s take a look at this problem where we have
multiple points to consider. In this question, we are told that each cable can support a
maximum tension of 500 N, and we need to find the largest mass of the pipe that can be supported.
So straight away, we know that the tension of cord HA would be equal to the weight of the pipe. Now,
leaving that to the side, we can look at point A and draw a free body diagram.
We have the force of cable HA downwards, force of cable AE to the right,
and we have the x and y components of force BA. So let’s write our equations of equilibrium,
first for x-axis forces and we will assume right to be positive.
Next, we look at y-axis forces. We will assume up to be positive. Instead of writing FAH,
we can write the weight since they are equal. Let’s isolate this equation for FAB.
Now we can use this value and plug it in to our previous equation to isolate that one for F_AE.
You will see at the end, why we do this. Okay, let’s set those equations to the side
and draw a free body diagram for point B. So we have force FBA that now points downwards,
we have force BD, and we have force BC. We can also draw the components of force BA like this.
The components of force BD can be found using the ratio triangle given to us.
Now, we will write our equations of equilibrium. First for x axis forces. Next, y axis forces.
Now, if we look at the equations we came up with before, we see that we already wrote force AB
in terms of weight. So we can actually plug that value in for force BA. Let’s isolate for force BD.
Now we can plug this value in to the previous equation to write force BC in terms of weight.
So the reason why we wrote every force in terms of one weight is to compare which
cable would carry the largest force. If we look at the 4 values we have, we see that force BC
would have the largest tension since it’s the biggest number. So all we have to do is
figure out what weight would cause force BD to equal 500 N. Let’s solve for the weight.
Now if we divide the weight by the acceleration due to gravity, we get the mass.
So the maximum mass of the pipe that can be supported is 26.67 kg.
Let’s take a look at a question involving springs. Here, we need to find the stiffness of the spring
to hold the crate in the position shown. Before we draw a free body diagram, let’s quickly calculate
the angles in our diagram. This angle can be found using tan inverse. Now for the other angle.
Okay, now we will draw the free body diagram about point D.
So we have the force in cable DC, the force in the spring DB,
and we have the weight of the crate. Now we can write our equations of equilibrium.
Let’s start with the x-axis forces. So we have the 2 x components of our cable and spring. Next,
we can write an equation for the y-axis forces. We have the 2 y components plus
the weight which is straight down. Now we have 2 equations with 2 unknowns, so let’s solve them.
So we figured out the force in the spring. The question tells us that the unstretched
length of the spring is 2 m. Lets figure out the length of the spring after it stretched.
We can use the Pythagorean theorem for that. This tells us the current length of the spring.
Let’s subtract the initial length from this value to figure out how much it stretched.
So the spring stretched 1.61 m after the crate was hung. Now we can use hooks law
to figure out the stiffness. The force is what we found before, so that’s the force in
spring DB. The x value is the length the spring stretched. Let’s solve and that’s our answer.
Let’s take a look at one last example. This is the type of question that involves geometry
more so than the usual. In this question, we need to find the x value and the tension
in the single cable required for equilibrium. So if we draw a free body diagram of point B,
the forces that goes both ways can be represented the same, since it’s a
single cable. In other words, the tension in the cable is the same throughout the whole length.
Now, there are 2 angles that’s created, and both of these angles would be the same as well.
Lastly, we have the weight of the 100 kg sack. Let’s write an equation of equilibrium
for the y-axis forces. We will pick up to be positive. Writing an equation of
equilibrium for the x-axis forces won’t help us since it’ll just equal zero. So instead,
we need to look at the geometry of the problem. Let’s draw this diagram separately.
Notice how these 2 angles are the same as the angles on our free body diagram.
This is because these are alternate interior angles. Now, the total length of the cable is 5
m. So that means if we add up the 2 hypotenuses of the triangles we have here, they must equal 5 m.
Since both are right angle triangles, we can use cosine to write it out. For the small triangle,
we can say the cosine of the angle is equal to the hypotenuse of the triangle divided by length x.
Let’s isolate it for the hypotenuse. Next the big triangle.
Now both of these hypotenuses added together must equal 5 m.
Let’s simplify. Notice how the 2 x’s cancel each other out, so we can solve for theta.
Now we need to figure out the x-value. To do that, we need one more equation. For that, we will look
at the opposite lengths in our triangles. Let’s label that y. From the perspective of the small
triangle, we can write the tan like this. So remember, it’s opposite over adjacent.
Let’s isolate for y. Next, from the side of the big triangle, we can write tan like
this. Let’s isolate this for y. Now we can plug one equation into the other. Let’s solve for x.
Now we can use the angle we found before in our force equation to figure out the
tension in each cable. Those are our answers.
That should cover the types of problems you will face when it comes to 2D equilibrium.
In the next video, we will go over how to solve 3-dimensional equilibrium problems.
Thanks for watching and best of luck with your studies!