00:02
So the maximum horizontal force that can be applied to this block without it moving is when it's equal to the static frictional force.
00:16
So f is equal to, you know, f s max, which is mu s, fn, you know that fn is simply equal to m g, if i know the fn is simply equal to m g, if i know the force, acting vertically and so this is equal to 0 .5 times m we know that and this case is 11 kg times 11 times 9 .1 and this gives you 56 mutants now what now what if this force is acting at a 60 degrees angle, right? 60 degrees, what's going to be our, what's going to be the magnitude of the force keeps the clock from moving? well, let me just raise the board really quickly.
01:39
So resolving horizontally, we have the absolute magnitude of the cause, it must be equal to the static fictional force, right, which was mu s, fn, where fn is now, because this force is being applied as an angle, fn is mg minus f sine theta.
02:18
So this becomes mu s, m g minus f, sine theta.
02:31
We can rearrange this equation for, you know, just f, and get mu s over ng, the other side.
02:46
We're going to have mu s, fine theta, cos theta, and now we can just plug everything in.
03:00
6 .5 times 11 times 9181 .5 .2.
03:11
9 of 60 plus cos of 60.
03:17
And when you mark that out, you get 59 mutants.
03:24
So that's the force that must be applied at 60 degrees to keep the block from just on the wedge of moving...