Let’s look at how to find unknown values when a particle is in equilibrium in 3
dimensions. We have the following equations to keep in mind. All we are saying is that
for an object to stay in equilibrium, or in simple terms, at rest, all the components of
the forces effecting it must sum up to zero. Let’s say we have an object hung up like this,
we can assume the directions of the forces. After you solve the problem,
if you get a negative answer, that means its opposite to the direction you chose.
To solve these problems, you must know how to express forces in Cartesian form.
If you don’t remember, or need a refresh, please check the description for force vectors along a
line. Without further ado, let’s get started. Let’s take a look at this problem where we
need to find the force in each cable after a 20kg flowerpot is hung. Let’s draw a free body diagram.
We have a force from A to B, a force from A to C and a force from A to D. We also have the weight
of the pot which is straight down. So we need to express each of these forces in cartesian form.
The weight is easy, it’s straight down along the zed axis so it’ll just have a k component,
which is mass times the acceleration due to gravity. Now for the other cables,
we need 3 position vectors for each of them. So let’s write down the locations of interest.
Now that we have all the points, we can write our position vectors.
Next, we can calculate the magnitude of each position vector.
Now comes the unit vectors. So we are just dividing each component of our position vectors
by the magnitude. We can finally express each force in Cartesian from. All we do is multiply
the magnitude of the force by the unit vector. Now that we have all the forces, we can equate
all the individual components to zero. Let’s start with the x components. So we see that we have 2 x
components. Both of them added together must equal zero. Next the y components.
Lastly, we have the zed components. From this equation,
we can directly solve for force AB. Now, we have 2 equations left with 2 unknowns. Let’s solve them.
Don’t forget to use the value of force AB we just found. Those are our answers.
Let’s take a look at this problem where a plate is hung from cables. Each cable can support a
maximum tension of 15 kN, and we need to find the largest mass of the plate that can be hung.
So first, we need to express the forces in each cable in cartesian form. The process is identical
to what we did before. Let’s start with our locations. Now that we have our locations,
let’s write our position vectors for each cable. I am assuming all the forces start from the plate
and travel towards point A, so the position vectors will be from B to A, C to A, and D
to A. Now that we have the position vectors, it’s time for the magnitudes. Next is the unit vectors.
Now to express each force, all we do is multiply the magnitude of the force by the unit vector.
Don’t forget, we also have the weight of the plate, which will be straight now.
Now we are going to each component to zero. First, x components. Next,
y components. Lastly, we have the zed components. Now we need to make an assumption. Let’s say we
assume cable AB will carry the largest force possible. So we will assume it to have a tension
of 15 kN. If we do that, we need to check to see if the other cables are less than or equal to 15
kN. If they go over, then cable AB can’t be 15 kN. Assuming cable AB is 15 kN, if we solve the
equations, we get the following values. This tells us that cable CA carries no force,
and cable DA also carries the maximum it can. So all the conditions we require are met. Now
all that’s left is to divide the weight by the acceleration due to gravity to get the mass.
But let’s say we assumed cable CA to be 15 kn. Then when you solve them, you will see there no
solutions exist. If we assumed cable DA to be 15 kN, then we will get the following answers.
So the same answer as when we assumed cable BA to be 15 kN.
Let’s take a look at one last example involving springs. In this question,
we need to find the stretch of the two springs in order to hold the 20 kg crate in equilibrium.
So the way to solve it pretty much the same as before, where we need to express each
force in cartesian form. If we look at the 2 springs, spring OA lies on the negative y axis,
and spring OB lies in the negative x-axis. That means spring OA will only have a j component
while spring OB will have an I component. Next, we need to express the force in cable OC.
So the position vector for OC can be written like this.
Now the magnitude. The unit vector comes next.
Finally, we can express the force in cartesian form.
We have one more force, which is the weight of the crate.
That can be expressed in Cartesian form like this, since it’s straight
down and only has a k component. Now we can equal each component to zero.
We now have 3 equations with 3 unknowns, so let’s solve them.
To figure out the stretch in each of the springs, all we need to do is to use hooks law.
K is the stiffness of the spring and s is the elongation or compression. Let’s start with
spring OA. Next, we have spring OB. These values tell us how much the
springs stretched after a 20 kg crate was hung. That should cover the types of problems you will
face. One thing you should be very careful with is writing down the initial locations of the points.
Always double check, because if you make a mistake at the initial step, pretty much everything else
will be wrong. Anyways, thanks so much for watching and best of luck with your studies!