00:03
Let's first express every force and the couple in vector form.
00:11
The force is applied along the line g h and we have to find the unit vector of the vector rgh, which is the vector from point g to point h.
00:25
So we can write rgh is equal to rh minus.
00:37
Rg which is equivalent to 40 60 and minus 120 now the magnitude of this vector is rgh is equal to square 2nd2 square plus 60 square plus minus 120 square is equivalent to 140 then the real factor of this rgh can can be written as ratio of vector rgh to the magnitude of rgh.
01:39
So this is equivalent to 40 upon 140, 60 upon 140, and minus 120 upon 140.
01:58
The force applied at point g is fg is equal to 20 n, sorry 20 i plus 30 j minus 60 k because f is equal to 70 into r.
02:48
Now the moment applied along the line af is r.
03:00
Af is equal to rf minus ra, which is equivalent to 160 minus 120, 0.
03:20
We will follow the same steps.
03:23
So the magnitude of the vector, raf is equal to square root of 160 square plus minus 120 square, gives us the answer 200.
03:39
Then the unit vector of this are af is again the ratio of vector r to magnitude r which gives us the answer 4 by 5 minus 3 by 5 to 0.
04:03
And the moment along the line af in the vector form is m .a .f is equal to 10...