00:01
In this problem we are given the following force configuration in three -dimensional space.
00:08
Let us use the exact figure for sake of precision because this three -dimensional figure is kind of intricate.
00:18
Okay now we have this f1 here, f2 here and with that we are going to find the magnitude of the resultant force and also coordinate direction angles of the result.
00:33
Resultant force, alpha, beta and gamma.
00:37
Okay, now we need to express these forces, f1 and f2, as vectors, so we can add them up in the vector form.
00:45
And to do that, we need the unit vectors in the direction from a to b and from a to c.
00:53
And to do that, we need the coordinates of these points, a, b and c.
00:59
Okay, let's write down the coordinates of these points.
01:05
A is given by 0, 4, 0 feet, b is 2 .0 minus 6 feet, and c is minus 2 .5 .0.
01:26
Okay, for y, for z, we have the following.
01:30
So this grade is 5 to 12, and we have 2 .5 here, which is half of 5.
01:37
So this must be half of 12 which is 6.
01:41
So we have minus 2 .506 in use of feet.
01:47
Now let us write down the vector ab, which is b minus a, which is equal to 2 minus 4 minus 6 feet.
01:59
The norm of this vector is square root of x component square plus y component squared plus that component squared.
02:10
So we have root 56 feet.
02:13
Then the unit vector in this direction is the vector itself divided by the normal of the vector.
02:22
So this will give us 0 .267261 minus 0 .5 .345222 .2 and minus 0 .8.
02:43
Now let's do it for ac.
02:47
We have c minus a, so minus 2 .5, minus 4, 6 in unisof feet.
02:59
And the norm of ac is square root of 58 .25 feet...