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,