00:02
We have a beam that is cannolievered from the left side, has an implied load at an angle on the right, and then has a distributed load that is uniform over part of it, and then drops down as linearly to zero to at the end.
00:20
And this is, this height is 8 kilnibbean per meter.
00:25
And we'll ask to look at a cut in 3 meters from the canlevered end.
00:30
So we can, let's see here, we can have a, replace this distributed load with a point load at the center of the region, and then this one with a point load at two -thirds of the way in.
00:51
So in the x direction we have, if we look at the entire thing, the x direction we have minus f1x, minus f2 cosine of theta, and theta in this case is this angle here which is 60 degrees.
01:06
Then we have f1y, minus f2 sign of data, minus f3 minus f4 has to be.
01:15
And then we have, for moments, we have minus m0, minus three, this blank here, times f3, minus seven, which is this distance here, times f4, and then minus 9 .3 times f2, the y component of f2, which, which is just f2 sine theta because this kind of acts off the end a little bit here...