The steps to solve a friction problem
aren't really any different than any other kinds of equilibrium problems.
Remember that friction is just a point force. It's just a force.
What I do want to make sure that you understand is when you can assume that motion is impending.
So our steps to solve friction problems: read the problem, draw a free-body diagram,
Assume equilibrium.
So I'm assuming that nothing's moving, or if it's moving it's not going to change
its motion.
Apply your friction opposite motion.
So ask yourself if there were no friction at all,
which way would my object move?
Put friction in the other direction.
Apply your normal or whatever normals you have at the different surfaces
at unknown locations.
You just don't know where they're going to act;
put them in some sort of x location.
List your forces in Cartesian form; write your equations of equilibrium.
I haven't assumed anything yet.
This is the question, the key phrase for you: Is motion impending, and if so, where?
Where motion is impending, you can assume that F is equal mu_s times N but _only_ there.
And what we end up with is a bunch of different cases.
So you might have, "it slips here but not here"
or "it slips here and here."
And you need to be able to consider all of your options.
So this sort of steps will let you get to this
without having assumed anything at this point.
So your equations of equilibrium will be good no matter what happens.
Let's look at a couple examples.
The first one a hundred-pound crate on the floor.
Just a flat surface.
It's four feet wide and six feet tall.
I want to find out what P value I need here at a 3-4-5 triangle to move this where mu_s is 0.5.
That's my free-body diagram with distances and everything.
I'm going to put F opposing motion.
So if there were no friction on this surface, this crate would move this way.
So friction points that way.
The normal force I put at some distance x here.
I don't know where it is exactly.
I'm going to let my equations of equilibrium tell me.
So my equations of equilibrium say, four-fifths of P is F.
That's the x component.
Three-fifths of P is 100, plus 100 equals N.
And four-fifths of P, that's the x component, the moment arm is 6 ...
three-fifths of P times the moment arm of 2, ...
those are in opposite directions when I'm taking the sum of the moments about that point.
So I'll have one of them positive, the other one negative,
and those have to be balanced by N acting at whatever distance x it is.
So if I simplify that equation, I have 18P = 5 * Nx
These, so far I haven't assumed anything.
My unknowns, though, I've got four unknowns and three equations.
That, I can't solve that without actually assuming something.
So we'll go back to the problem: what do I want to know?
I want to know what P I need to have to slide this thing or to tip this thing.
I mean, I have an object here. It could move: it could tip or it could slip.
I need to solve this problem in either case.
So assuming only that it tips.
Now the tipping condition is that N has to act at the very corner of the body.
If it has to act way over here to maintain equilibrium,
then it's not on the object, and that doesn't work.
So my tipping assumption is that x = 2.
If I put that in and solve, I get P = 83.
It could also slip.
My slipping condition is the same one it always is, F = mu_s * N.
I plug that into my equations of equilibrium, and I get P = 100.
Now to try to figure out which of these is the actual answer, which one happens first?
Go back to the problem.
If P were 0, would it move? No, it wouldn't go anywhere.
So as you're increasing from 0,
sooner or later you're going to get to either one of these.
Well, sooner or later you get to this one.
So this crate will tip over before it slips.
The only way to make it slip would be to move P down or to put P at a smaller angle.
You have to change the problem because this problem is going to tip over first.
This is where it's important not to assume this.
If you assume that F is equal to mu_s * N, you're going to get this problem wrong.
The second one I want to look at is the same crate except on a slope.
If you look at this crate, and you put it on the slope,
the friction force acts, has to act tangent to this slope.
It's along the edge.
Normal has to be normal to the surface.
So this is my free-body diagram.
And my equations of equilibrium without assuming anything look like this.
Divide one by the other, you can get F / N = tan theta.
Now the nice thing about this is when I assume that motion is impending,
this is what I want to know.
This does not assume motion is impending; this does.
If I put in F = mu_s * N, I get F / N is mu_s.
That's tan theta.
The block slides when theta is the arctan of mu_s.
Well, that's your phi, phi_s. That's your angle of repose.
The block is going to slide when this angle theta ends up being phi_s, the angle of repose.
And the other way that you can do this is to look at a single resultant.
If you take the resultant of F and N, you have to have,
you only have two forces.
You have R, and you have F, uh W. R and W.
If you only have two forces, they have to be equal and opposite.
And that angle theta has to be the angle phi which would be the angle between N and R.
So the last example says, what happens if I'm pushing on it, if Ihave a force P?
If I have a force P, I want to find the range of P to maintain equilibrium.
I'm going to take an actual number here for mu_s: this is 0.35.
theta is 30 degrees.
What P do I have to have to keep the block still?
So the first thing to check is the angle of repose.
If I take the arctan of phi_s, I get 19.
That means without a P, that block slides down
because theta is bigger than that angle of repose.
Assuming then that P is not 0, that it's greater than 0, I have two choices:
it could move down the slope, or it could move up the slope.
Those are the two things that could happen.
These are my free-body diagrams.
They are the same except for the direction of F.
F is either opposing the motion [of the block] down, which means it [F] is going up,
or it opposing the motion [of the block] up, in which case F is coming down.
Same equations of equilibrium in both cases except for the sign on F.
If the block is sliding down, F and P are in the same direction.
If the block is sliding up, F and P are in opposite directions.
So these are my equations of equilibrium.
After I write my equations of equilibrium, I can say,
at impending motion, F is equal to mu_s*N, and in both cases I can solve.
I get 19.7 in one case and 80.3 in the other, 80.3 pounds.
Those are my boundaries.
As long as P is in between those two, the block stays where it needs to be.
What does that look like when I'm looking at my resultant forces instead?
Here's my resultant between N and F if F is going up the slope
or as F is going down the slope.
So the opposite motions.
The thing here is that phi_max, phi_max is the angle of repose.
That's as big as it can get.
So phi_max, phi_s is the inverse of tan(mu_s).
That's 19.29. We did that a minute ago.
So if you look at this angle here, N makes an angle of 30 degrees.
That's the slope to the vertical.
So R is going to be on this sort of edge of this.
It's not all the way up at W; it's down a bit.
This angle here between R and W is theta - phi_s. That gives me 10.71.
And I know that this angle up here between P and W is already known.
That's 90 minus theta. This is 60 degrees.
So now I have all of my angles, and I have one side.
I can use the law of sines to find out what P and R should be in this case.
And it's the same thing down here except that now my R is in the other direction.
This one, R is pointing up the slope with F, and here it's pointing down the slope with F.
So this phi_s now is the same, but it's on the other side.
So now I have theta + phi_s. It's 49.29.
Still, I know all my angles.
I know that W is 100 pounds, and I can find all this with the law of sines.
So when you're doing these friction problems, the key, the very key is to make sure you've
not assumed anything
when you're writing your free-body diagram and your equations of equilibrium.
Then ask yourself,
where is it going to slip? Where do I know that motion is impending?
You may have to take different cases.
It's going up. It's coming down. Or it slips here and not here.
There may be different cases that you have to take,
but that's problem specific,
and it comes from the answer to the question, where is motion impending?
That's your key.
Thanks.