Let’s talk about frames and machines. We see them all over the place and use them to make our lives
easier. Frames support loads and machines have moving parts that transmit forces or change the
effect of a force. So for example, pliers, where a small force applied at the handles
create a much larger force at the jaws. Usually, frames and machines are structures which contain
pin connected members. So in this object, these are pins, and these are the members. We can figure
out the forces in the pins, and we can figure out the forces in the members. Usually, it is best to
first think of this as a whole and figure out the reactions at the supports. When we do that,
we only consider external forces being applied. These include support reactions and forces applied
manually. After that, we can sperate all the members. At each pin, there will be 2 supports,
one in the x direction and one in the y direction. These are the forces effecting the
pin, but when we draw our members individually, they are now considered external forces. When
we think of the whole object, they are internal forces, so we don’t consider them. The directions
of these forces can be assumed. However, there is something you should keep in mind. Let’s say,
in this member, we drew our x and y reactions at the pin facing down and to the left. On the
other member, at the same exact location, the forces must be opposite to our assumption in
the first member. The magnitude is the same, but the direction flips. So now, the x reaction faces
right and y reaction faces up. After that, you can write equations of equilibrium or moment equations
for each individual member and solve for unknowns. One thing that will make finding these forces much
easier is figuring out two force members. In a two-force member, it only has forces
applied at only two points. The forces must be equal in magnitude but opposite in directions.
Lastly, they must have the same line of action, directed along the line
joining the two points where these forces act. When we see a member with just 2 forces applied,
it makes the math very simple. For example, in the usual case, we will sperate the components
into x and y components at each end, leaving us with 4 unknowns. However,
by realizing a member is a 2-force member, we don’t need to break it into components and we
will only have a single unknown, since the force at the other end would be facing the opposite
direction but will have the same magnitude. All of this will become clear with a few examples.
So without further ado, let’s get started. Let’s take a look at this problem
where we need to figure out the reactions pin C exerts on member ABC. To solve this problem,
we can either focus on member EDC and DBF, or ABC and DBF. I am going to use members ABC and DBF to
solve this problem which means I don’t need to figure out the x and y reactions at support E.
But we will have to figure out the support reaction at A since that’s applied to member ABC.
We can do that by thinking of this whole frame as a single object, without breaking it into pieces.
Note that the 600 N force is applied to the pin, which means it’s not an internal force.
At support E, we would have 2 reactions, one in the x direction and another in the y direction.
At A, which is a roller, we’d have a single reaction in the x direction. If we write a
moment equation about point E, we can eliminate those reactions and figure out the force at A.
Now we will draw the 2 members we will focus on, separately.
First member DBF. At pin D, we’d have 2 reactions. At B, it would be the same.
We also have the 300 N force applied at point F. And I just assumed the directions for these
reactions. Next, we draw member ABC. At A, we have the force we figured out in the previous step.
At B, we would have opposite reactions to our first diagram. So notice in the diagram for
member DBF, we have BX facing right, so in this diagram, we will have it facing left.
The same for BY. We have it facing up, which means here, it’ll be facing down. It’s always opposite.
At pin C, we would have 2 reactions, however, we do not include the 600 N force. To understand why,
let me draw just the pin and show the forces being applied to it. Here, this is the effect of member
ABC on the pin, and this is the effect of member EDC on the pin. Notice how the 600 N force is now
an internal force applied to the pin. When we draw each member separately, we are now considering
CX and CY as external forces and we do not include forces applied to a pin. However, note that when
we thought of the whole body as a single object, then we do not care about CX or CY since those
are now considered internal forces, while the 600 N force applied to the pin is an external force.
Now let’s focus on the member DBF. We can figure out BX if we write a moment equation
about point D. That eliminates forces DY and DX, along with BY since it’s line of action
goes through point D. Let’s solve. We don’t need to figure out BY,
and I will show you why. Let’s look at member ABC. We can figure out CX by writing
an equation of equilibrium for the x axis forces. We only have 3 forces. Let’s solve.
Now, if we write a moment equation about point B, notice how it eliminates the x and y reaction at
pin B. That means we don’t need to find BY to figure out CY. In fact, you can see that there
is only one force that would create a moment about pin B, which is force CY, which means it has to be
zero for this member to stay in equilibrium. And those are our answers.
Let’s take a look at this problem where we have to find the horizontal and vertical components
of the forces at the pins B and C. Usually, it’s good to figure out the reactions at the supports
as a first step, however, what we are looking for are the reactions at pins B and C. Notice
that they are both part of member CB. So let’s just focus on that and draw a few body diagram.
At pin C, we would have 2 reactions, one in the x direction, and one in the y direction. The
directions are assumed. Now you should notice that member AB is a 2 force member. That means
the resultant force at B would be like this. The angle can be found by using the inverse of tan.
We also have the wheel attached, but we don’t really care about the reactions at the pin
of the wheel. What we care about is the tension in the cable. That tension is the
same throughout the cable, and that’s equal to the weight of the cylinder multiplied by
the acceleration due to gravity. Now let’s mark down the distances needed for our calculations.
Notice how we have 3 unknowns, the 2 reactions at C, and force AB. We can directly solve for
force AB if we write a moment equation about point C. That eliminates the reactions at C.
Note that only the y component of force AB creates a moment since the x component’s line
of action goes through point C. So we have the y component times the perpendicular distance,
then we have the force applied at the pully at the top, and lastly the force applied at the pully
on the bottom. Don’t forget to add the radius of the pulley for the perpendicular distance.
Also keep in mind that since we picked clockwise to be positive,
the weight of the cylinder creates a positive moment about point C
while the other forces create counter clockwise moments, meaning they are negative. Let’s solve.
Now, we have 2 unknowns left. To find the y axis force, we can write an equation of equilibrium for
the y axis forces. So we have the y component of force AB, the force pulling straight down
at the pully, and reaction CY. Let’s solve. To find the x reaction, we can do the same,
but with an equation of equilibrium from the x axis forces.
We have the x-component of force AB, the x-axis force at the pully and reaction CX. Let’s solve.
Now since we need the x and y components of the reaction at B,
we need to break force AB into components. We can do that by using the angle we found.
Those are our answers. Let’s take a look at this problem where
we need to find the reactions at the supports. Usually, it’s good to figure out the supports
before thinking about individual pieces, however, this is not the case here. There are 3 supports,
which means it would be difficult and we’d have to solve many equations simultaneously to get an
answer. Instead, let’s think about just member DC and draw it’s separately. At D, we would have
2 reactions, DX and DY. At C, we would have one force straight upwards since it’s a roller. Now
we can easily figure out CY by writing a single moment equation about point D. That eliminates
DX and DY. Note that only the y component of the 7 kN force creates a moment about point D.
Now we can figure out DY by writing an equation of equilibrium for the y axis forces. Let’s solve.
To figure out DX, we write an equation of equilibrium for the x axis forces. Let’s solve.
Now we figured out all the reactions for member DC. Let’s move onto member ABD. At A,
we have a roller, which means we would have a single force upwards.
At B, we have a pin support, so we will have x and y components.
At D, we have the values we found, but note that our reactions are facing opposite to the
direction we chose from our other diagram. We also have the forces that are applied to member ABD.
Now, let’s start with a moment equation about pin B.
This eliminates BX and BY along with DX since it’s line of action goes through point B. Let’s solve.
To figure out BX, all we need is an equation of equilibrium for the x axis
forces. The only other force, excluding BX, is just the x reaction at D. Let’s solve.
Lastly, to find BY, we just need an equation of equilibrium for y
axis forces. Let’s solve. And those are our answers.
Let’s take a look at one last example. In this question, we need to figure out the angle theta
for this object to stay in equilibrium. Before we take apart the members, it’s good to figure out
the force in the spring. To do that, we can use hooks law. We already know the stiffness of the
spring, but we don’t have the length of extension. So that’s what we need to find. Notice how we can
draw a right angle triangle like this. Using sin, we can represent half of the length of the spring.
That means the full length of the spring is 2 times that. Originally, the spring has an
unstretched length of 0.3, so to find how much it extended, we need to subtract this amount.
This gives us the extension. Now we can plug this back into our force equation and simplify.
Next, we will analyze each member separately. First, let’s look at member CB.
At C, we would have 2 reactions since it’s a pin support. We also have the force of the spring
downwards. In the question, we are told that each member has a mass of 20 kg. We can show the weight
of the bar at the center. That’s mass times the acceleration due to gravity. Lastly, at B, we
would have 2 reactions. Now we need the distances. To do so, let’s draw a right-angle triangle like
this. This angle would be the same since these are alternate interior angles. That means the
side opposite to our angle is 2sin theta, and the adjacent length is 2cos theta. Since the
weight is at the center, each piece would be cos theta. Now there isn’t a way to write a moment
equation to eliminate enough unknowns for a direct solution. So let’s move on to the other member.
Starting at B, we have the same reactions as before, but in opposite directions.
At A, we would have a support force at the roller in the x reaction,
and the spring force pulling upwards. Let’s not forget the weight of this member as well.
The distances are the same as before. We can now start with a moment equation
about point C using member CB. This eliminates the reactions at C along with the force of the spring.
Now if we write a moment equation about point A on the other member,
we can eliminate the reaction there, and have another equation with just the forces at B.
We now have 2 equations with 3 unknowns. So we need one more equation. For that, let’s write
an equation of equilibrium for the y axis forces using member AB. We don’t want to use member CB
since we’d have to consider the reactions at C and we will end up with more unknowns.
Now we have 3 equations with 3 unknowns. You can solve them anyway you like. We get an angle
of 23.7 degrees, which is our answer. You may have noticed we got a negative value for Bx,
which means our assumption for its direction is wrong, it faces the other way.
That should cover how to solve frames and machines problems. By isolating for members and thinking
of the object as a whole, you can figure out a multitude of reactions and forces both internal
and external. Thanks for watching and best of luck with your studies!