We have a uniform crate, it's being pushed

to the right by a slowly increasing force, P.

And we want to find the condition for which the crate will

slide before it tips over.

And so, what might you do first to analyze this problem?

Hopefully by now right on the tip of your

tongue is, I want to draw that free body diagram.

That graphical tool for the equations of equilibrium and so draw

the free body diagram of that crate, and then come on back.

Okay, here's a good free body diagram.

We've got our, our weight of our crate

down, mass times its, the acceleration due to gravity.

we've got a normal force acting up and a friction force that opposes motion.

And since we have a force pushing to the right, we know

that the normal force is not necessarily going to act in the center.

And so after we have our free body diagram what do you do next?

What you should say is, you need to go

ahead and come up with those equations of equilibrium.

So I'd like you to write the equations of equilibrium

by summing the forces in the x direction, the y

direction, and do a moment equation about point A, which

is at the center of the crate at the bottom.

And when you've got that complete, come on back

and make sure that you got those answers correct.

'Kay. This is what you should have arrived at.

we find that the applied force P is equal to the friction force N, the

normal force is equal to the weight, and N delta is equal to P h.

And so let's look at those three equations that we came up

with, and now let's go ahead and, and, and solve our problem.

So, it says find the condition for which the crate will slide before it tips over.

And so I'm going to assume that sliding occurs first.

So assume

sliding, or assume that the

crate slips, before it tips.

And if that's the case, we know that friction has to be what in that case?

If we're, if we're on the verge of slipping what is the friction going to be?

Well, what you should say is that the friction has got to be the maximum

friction we can get. So F must equal mu sub S times N.

And so with that I now know therefore, a symbol for therefore

that N delta here equals Ph, but P is

equal to f so I am going to get equal to

fh and we know what f is now, it's mu sub s N times h.

And so I can now solve for delta. Delta equals mu sub s h.

For sliding, delta has to be between the center and the corner

once it gets to the corner then we know that the body will have a tendency to tip.

And so delta has to be less than, in this case, b divided

by two, which is the width of the create.

So delta is less than b over 2 for slipping first.

which we know is equal to mu sub s N, or excuse me, mu sub S h, has

to be less than b over 2. Or B has to be greater than

2 U sub s h, for slipping to occur first.

So, that's the condition that will ensure

it's slipping will, will, conti, happen first.

And so let's just go through a little bit of a thought process here.

If b is greater than 2 mu sub s times N, the crate will slip first.

And so, that means the wider the crate, the more

apt it is to slip before it will tip over.

Now, if b is equal to 2 mu sub s, that means that delta is equal

to b over 2, and we'll write at the corner with my, our normal force.

And so slipping and tipping will occur simultaneously.

And then if indeed B becomes less than 2

times mu sub s N, the crate's going to tip first.

And so that says as the crate gets skinnier, as b

becomes less, you have more tendency for tipping, rather than slipping.

And that should make physical sense to you.

And so, that's a good analysis of this

problem, which deals with Coulomb friction and static equilibrium.

And we'll do another problem with coulomb friction next time.