Handle forces f1 and f2 are applied to the electric drill. Replace this force system by an equivalent resultant force and couple moment acting at point o. Express the results in cartesian vector from.

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,

The moment of the resultant force about point O (M_o) is 1.3i +3.3 j -0.45k.

pe = weightxheight = 2000x1000=2,000,000joules