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PROFESSOR: Being able to draw good force diagrams is one of the most
important skills that you need to do well in this course.
Force diagrams, or free body diagrams, are diagrams that
show the body in isolation.
The body, or more generally the system under discussion, must be shown
isolated from its surroundings.
If something is pushing on it, for example, if it's a block being held up
by the floor, we remove the floor, and just draw the block in isolation.
After the body is isolated, then the forces that act on it must be shown.
Here's where you put the environment, that it's around, back in.
If it's a block, resting on the floor, then you put a normal force from the
floor that acts on the block.
Let's make this discussion concrete by picking a block on a floor.
So here's the block.
And we consider now the various sources of interactions.
There's gravity.
So when we have gravity acting on the block, then we have to draw a vector,
an arrow, representing the gravity that starts at the center of mass of
the block and goes down.
And we label that W for the weight.
The other force acting on the block is the contact
interaction with the floor.
Now we're ignoring the horizontal motion here, which might be friction.
And so the only important component of the contact force is the normal force,
the one perpendicular to the surface of both the block and the floor.
So that is acting an upward direction, and it acts on the very
bottom of the block.
And we label it N_BF, the normal force on the block due to the floor.
Whenever you think that you're done with your free body diagram, you
should check two key points.
The first one is, think of Newton's Second Law--
will the sum of all the forces, the vector sum of all forces that you've
drawn, give it the motion that you think it ought to have?
The second one is Newton's Third Law--
have you properly accounted for third law force pairs in your diagram?
Consider the second law, first.
This block is not moving.
Therefore, the sum of all the forces, the vector sum, ought to be zero.
Well if we look at it, we can see that the normal force and the weight have
approximately the same length.
So that looks satisfied.
Now consider Newton's Third Law.
In this case, the two forces shown are equal in magnitude
and opposite in direction.
Does that make them a third law force pair?
Absolutely not.
According to the third law, the forces must be generated by and interaction.
And I think Newton's phrasing exactly was "the mutual interaction between
the two bodies." So they have to be generated by the same interaction, and
they have to be on different bodies.
But one of these forces is a contact force.
The other one is a gravitational force.
So they don't have the same interaction.
And secondly, they act on the same body, so they cannot be
Newton's Third Law pairs.
In the example that we just considered, the block was stationary.
Sometimes those problems are called statics problems.
Now let's change the problem a little bit by imagining that the block is
resting on the floor of an elevator, and that the elevator is
accelerating upwards.
We'll indicate the acceleration direction with an arrow upward.
And we'll label it with the symbol, little a.
We do not draw the acceleration vector acting on the block.
The acceleration is not a force, nor is the mass times the
acceleration of a force.
These should not appear on the force diagram.
Rather, they should appear in the equation F equals ma--
the sum of the forces you get from the force diagram and the acceleration you
get from the motion.
Or the acceleration measures the motion that results from the forces
acting on it.
In order for the block to have an acceleration upwards, the net force
acting on it must also be in the upwards direction.
Now the weight going down is not going to change.
And therefore, the normal force has to get bigger.
So we'll erase the normal force and draw it back in, longer than it was
before, so that the net force on the elevator will be
in the upwards direction.
This satisfies Newton's Second Law.
Now let's turn to the third law.
We can't really talk about the third law, unless we have the two objects
that interact.
So in this case, we're going to need the floor underneath the block.
And so we draw it in, and now the normal force on the block due to the
floor is match, or is a third law pair, with the normal force on the
floor due to the block.
So we'll draw that one in.
And we'll label it N_FB.
N_FB stands for the normal force on the floor due to the block.
Our convention will be the first of these subscripts is the object that is
being acted upon, because in free body diagrams, your draw
the force on an object.
And the second one is the other object that is the third law pair object at
the interactions between those two pairs of objects.
To really drive home this relationship, we're going to note that
these two normal forces act on different bodies, that they have the
same magnitude, that they have opposite directions, and that they
satisfy the third law.
Now checking over our diagram for the floor, we notice that it would be
falling down, unless we put on some additional forces, which we're just
drawing in here, that might be the force of the cables holding the
elevator up.
Or if this were a block simply balanced on a board, not an elevator,
or the elevator had a person up at the top of the building pulling on a rope
on each of the two ends, then this situation might exist.
And it would be important for the two forces to be equal, so that the board
or the floor didn't tilt.
Yeah.
Looking at it, I also see that we've left off the weight of the floor.
And so we should draw that in, as well.
I see I've left off the weight of the board, which probably isn't
negligible.
So I can draw that in and indicate that that's the weight of the board,
or the weight of the floor, depending.
Now, we've pretty much finished with the two force diagrams for the block
and for the floor.
I've drawn them here as bodies with a substantial size.
Because I think that's a good habit to get into.
It's not important when you're just talking about the acceleration of
something, whether the force acts at the top or the bottom.
After all, Newton's law F=ma applies to a point particle or to the center
of mass of a more extended object.
However, when we get to torques and rotational motion, it will be very
important to put the forces on the exact part of the body where they, in
fact, impinge, or where they act, so that the rotational motion of the body
will be represented correctly.
And as I pointed out in this problem, there's really no rotation.
And that was the reason, in fact, that we picked these two tension forces
from the cable holding the board up to be equal.