Answer:
The resultant force is F=6i-j-14k while the resultant Moment is about point O (M_o) is 1.3i +3.3 j -0.45k.
Explanation:
As the complete question is not given, the complete question is attached herewith
The coordinates of the points from the Free-body diagram are given as
0=(0,0,0) m
A=(0.15,0,0.3) m
B=(0,-0.25,0.3) m
the position vector of OA is
roa=(0.15-0)i +(0-0)j +(0.3–0)k
= 0.15i +0j +0.3k
the position vector of OB is
rob =(0-0)i +(-0.25 - 0); +(0.3–0)k
= 0i -0.25j +0.3K
Now
The equivalent resultant force is expressed as,
F = F1+ F2
Substitute 6i - 3j -10k for F1, and 2j -4K for F2.
F =6i -3j -10k +2j - 4k
= 6i - 1j -14k
So the resultant force is F=6i-j-14k.
Resultant couple moment at point O is expressed as,
[tex]M_o=r_{OA}\times F_1+r_{OB}\times F_2\\M_o=\left|\begin{array}{ccc}i&j&k\\0.15&0&0.3\\6&-3&-10\end{array}\right|+\left|\begin{array}{ccc}i&j&k\\0&-0.25&0.3\\0&2&-4\end{array}\right|\\M_o=0.9 i+3.3j-0.45 k+0.4 i+0 j+0k\\M_o=1.3 i+3.3 j-0.45 k[/tex]
The moment of the resultant force about point O (M_o) is 1.3i +3.3 j -0.45k.
