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This is module 24 of Applications in Engineering Mechanics.

We're going to continue where we, pick left off last time with this

concept of Coulomb friction to actually

solve some real world engineering problems.

So this is the first problem I want to look at.

what I have here is a bar that can rotate, bar

B2 and then I have a body on that bar B1.

And I want to find the relationship between

this angle theta and the coefficient of static friction between the bodies B1

and B2 that will prevent the body B1 from slipping or sliding down the plane.

So if you slide over here here's the situation.

As I increase the angle theta, at some point the coefficient of friction

is not going to be great enough to keep the body from sliding

down the plane and I want to find that boundary.

I want to find that condition where fric, sliding will not occur.

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And so my question to you is, how might you, what,

what are you going to do to start to analyze this problem?

So think about that, come on back.

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My subs, my suggestion is to go ahead and draw a, a full body diagram

of B1 since that's the body that

we want to prevent form sliding down the plane.

And so, do that, come on back and let's see how you did.

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you'll notice that I have my, my weight force acting straight down.

I've got a normal force acting tangent

to this surface, perpendicular to this surface.

And then I have a friction force

opposing the pending motion, or the impending motion.

And so, I've actually rotated my X and Y.

frame so that it's along the, the

incline.

And what I want you to do now is to use the equation of the

equilibrium for some of the forces in the y direction to solve for the normal force.

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Okay, and what you should have done is found that the normal

force Is just equal to mg times the cosine of the angle.

I'm going to use the equilibrium equation of sum of the forces

in the x direction to come up with the next equation.

And so in this case, I've chosen up and to the right positive.

So I've got f minus mg sine of theta equals 0.

I want the condition here just before slipping a curve.

And so if I look at my graph of the

slipping versus non slip condition that we talked about last

module, I'm at the point where I'm going to get

the very max friction available just before it starts sliding.

And so f has to be f max or mu sub s times N minus mg

sine theta. But we know that m

from the equation where you found up here, is equal to mg cosine theta.

So we get U sub s times mg cosine

theta equals mg sin theta.

And I can cancel the mgs now from both sides and

I find out that the condition from USIV S is that USIV S is sin

theta over cosine theta which is the same as tangent of theta.

And so that's the intending condition just before the, the,

the body would slip, body B1 would slip and so what

we would have to have to keep it from slipping

is mu sub s would have to be greater than or

equal to that value.

So we've got U sub s is greater

than or equal to tangent theta to prevent slipping.

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In this case changed just slightly.

I've got a crate. It's on a fixed plane of 30 degrees.

But I have the ability to have an applied force here P.

I know now that the crate weighs 200 pounds, and I'm given that

the static coefficient of friction between the crate and the plane is 0.4.

And so,

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what I want you to first determine is, if that force P were

not present okay, if it was gone, would the crate slip down the plane?

So go ahead and do that on your own, and

then come on back after you've, you've completed that analysis.

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So to do that, that is exactly like the, the problem we just looked at previous

to this where we were raising the bar to see what, what condition was

just before, slipping would occur if there was no force P here, and so all

you have to do is check the condition we came up with to prevent slipping.

Is u sub s greater than or equal

to tangent theta, well you put your values in.

U sub s is equal

to 0.4, theta is 30 degrees, so 0.4 has to be greater than or equal to 0.577.

And so you see that no, we don't have enough friction, and so unless we put some

sort of forces on this crate, it is going to slide down, down the, the incline.

And so my next question is, what is the smallest

force P that we can use to prevent the crate

from sliding or slipping down the plane.

And again I'd like you to go ahead and try to do this analysis on your own.

once you've given it a good shot come on back and we'll do it together.

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So again, what I'm going to do, what I did is, I've drawn my free body diagram.

The only difference in what we had before is, that we

now have this force P acting up and to the right.

you can again sum forces in the y direction.

I'd ask you to do that. And when you do You'll find that N equals

173 pounds. I'll go ahead together with you and

we'll sum forces in the x direction and so

we've got P plus F minus 200 times the sine

of theta, which is 30 degrees, equals 0.

So F we know is going to be on the verge of slipping.

So, we want the smallest force p that will prevent slipping.

So, that's going to be the maxify your friction I can have which is U sub

s times N, and I know that U sub s is

equal to 0.4. I found N to be 173 pounds.

200 times the sine of 30, well, sine of 30 is 0.5, so this term over here is

100. So I've got minus 100 equals 0.

And you find out that the smallest value of P that'll prevent slipping is

30.8 pounds. And so that's our answer.

If I have anything less than 30.8 pounds, the crate will slide down the plane.

As I get, to, to the point 30.8 and

a little bit higher, that's enough to prevent sliding

down the plane.

So my next question to you in a worksheet for you to do on your own is, okay, what

is the largest force now, P, that can be applied

before the crane, crate will start sliding up the hill?

And so, we know if we have zero force P, Crate slides down as we increase P.

Once P gets to 30.8 pounds, the crate will not slide.

We can increase it,

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we can increase it more and more until there's some point

where P is going to get so large that The friction is

not going to be able to prevent it from sliding up the

plane and that's what I'd like you to do in this worksheet.

I've got the solution in your module handouts and we'll see you next time.