In the last video, we had a ten kilogram mass
sitting on top of an inclined plane at a 30 degree angle
And in order to figure out what would happen to this block
we broke down the force of gravity on this block into
the components that are parallel to the surface of the plane
and perpendicular to the surface of the plane
and for a perpendicular component, we got 49 times the square root of 3 N downwards
That's 98 times this quantity over here, downwards
But we said look! We don't see this block of ice accelerating downwards into this wedge
because the wedge is supporting it
So there must be a counteracting force that the wedge is exerting on the block
And that counteracting force is the normal force of the wedge on the block of ice
And that is exactly opposite to the force of gravity in this direction
The normal force of gravity is the normal force of the wedge
And these completely balance each other out in that normal, perpendicular direction
And that is why this block is not accelerating either in that direction
or in this direction over here
But the one component of the force of gravity that did not seem to have any offset
(at least the way we set up the problem in the last video)
is the force/component that is parallel to the surface of the plane
And we figured that out to be 49N, it was
essentially the weight of the block times the sin of this angle
And we said look, if there is no other forces
then it would be accelerated in this direction
And to figure out the rate of acceleration
you take the force in that direction divided by the mass of the block
and you would get 4.9 m/s^2
Now let's say that wasn't happening. Let's say that you were to look at this system
right over here and the block was just stationary
And now, for the sake of argument, let's assume it is not ice on ice
Let's assume that they are both made out of wood
And now all of a sudden we have a situation where the block is stationary
If it is stationary, what is necessarily the case?
Well we already determined that if it is not accelerating in this normal direction
there must be zero net forces on it
But if it is stationary as a whole
then there must be zero net forces in the parallel component too
So there must be some force counteracting this 49 N that wants to take it down the slope
So there must be some force counteracting
the component of gravity that wants to accelerate it down the slope
And the question is what is this force? We're dealing with a situation now
where we're dealing with a stationary block, a block that is not accelerating
So what is that force?
I think you know from experience maybe what is the difference between
a block of wood on top of wood and a block of ice on top of ice
A block of ice on top of ice is much more slippery; there is no friction
between ice and ice, but there is friction between wood and wood
To make it a little bit more tangible, maybe we'd put some sandpaper on the surface over here
And then it becomes a little bit clearer. The force
that is keeping this block from sliding down in this situation is the force of friction
and the force of friction will always act in a direction opposite
to the motion if there was not any friction
or the potential acceleration if there was not any action
So what is the force of friction in this case?
Well, this block is completely stationary. It's not accelerating down the ramp
The force of friction over here is going to be 49 N, upwards, up the ramp
Now I want think about, this is something that can be determined experimentally
as long as you have some way of measuring force, you can do this experimentally
But the interesting question here is
how much do I have to push on this block
until it starts to move down the ramp? How much do I have to push on it?
Let's say you are able to experimentally determine
that if you can apply another 1 N on this
then all of a sudden, I can at least get the box to start accelerating down
Not the rate which it would do naturally, but I can just start to nudge it down
if I give it another push of one N in the parallel direction
So what is the total force--so exactly one N
So the total force at this point that's acting on it in order to just start to budge it
I'll call this the budging force
Remember here isn't a traditional class
F sub B for the budging force
The budging force is in the parallel direction
If I'm applying 1 N in this direction and it already has 49 N due to
the component of gravity in this direction, then my budging force is 50 N
And so an interesting thing that you can determine
based on the materials that are coming in contact with each other is
just how much force you need to just start to overcome friction
In this case, it's the budging force. That's a term that I made up
And the interesting ratio which tends to hold for given materials pretty well
is the ratio between the amount of force just to budge it
and the amount of force between the two objects
between how much force they are exerting on each other
And in this case, the amount of force that is being exerted by this, by the wedge on the block
is the normal force, is 49 sqrt of 3
Maybe I should say the magnitude of the budging force
over the magnitude of the force that is putting these two things in contact
In this case, it is 49 square root of 3 N
Over the magnitude of the normal force, and that is 49 sqrt of 3
We call this the coefficient of static friction
We're gonna use this a little bit more deeply in other problems
but it tends to hold true for different materials so that
in the future if you have a different mass, you have a different incline
but you have the same materials
given the normal force, you can figure out the budging force
You can figure out exactly how much force you need to put
if you know this, which you usually figure out experimentally
So what would be the value in this case? You have 50 N over 49 square root of 3 N
Let's get the calculator out
So I have 50 divided by 40 times the square root of 3
Gives me .72--I'll just round to two significant digits--0.72
This is 0.72
And then you can use this information. This is the coefficient of static friction
We call it the coefficient of static friction because
this deals with the ratio of the force of friction relative to the normal force
I guess the force just to overcome the force of friction, just kind of get right over that
the most friction that can be applied by kind of the abrasiveness of the 2 things
when the object is stationary
I'll do a whole video on how this is different
when an object is stationary to when it's moving
A lot of times, they're very very very close, but for certain materials
you have a very--at least, a noticeably different coefficient
of friction when the object is stationery as opposed to when it is moving
So I'll leave you there. In the next few videos, we'll use coefficient of friction
or calculate coefficient of friction to do some more problems