Sol. (i)

The moment arm to the 600-N force is

d = 4 cos 40º + 2 sin 40º = 4.35 m

By M = Fd the moment is clockwise and has the magnitude

M O = 600 (4.35) = 2610 N.m

(ii)

Replace the force by its rectangular components at A

F 1 = 600 cos 40º = 460N, F 2 = 600 sin 40º = 386 N

By Varigonon's theorem, the moment becomes

M O = 460(4) + 386(2) = 2610 N.m

(iii)

By the principle of transmissibility, move the 600-N force along its line of action to point B, which eliminates the moment of the component F 2 . The moment arm of F 1 becomes

d 1 = 4 + 2 tan 40º = 5.68 m

and the moment is M O = 460(5.68)

= 2610N.m

(iv) Moving the force to point C eliminates the moment of the component F 1 . The moment arm of F 2 becomes

d 2 = 2 + 4 cot 40º = 6.77 m

and the moment is M O =386(6.77)

= 2610 N.m.

(v) By the vector expression for a moment, and by using the coordinate system indicated on the figure together with the procedures for evaluating cross products, we have

M 0 = r × F = (2i + 4j) × 600 (i cos 40º – j sin 40º)

= – 2610k N.m

The minus sign indicates that the vector is in the negative z-direction. The magnitude of the vector expression is

M O = 2610 N.m

Helpful Hints:

1. The required geometry here and in similar problems should not cause difficulty if the sketch is carefully drawn.

2. This procedure is frequently the shortest approach.

3. The fact that points B and C are not on the body proper should not cause concern, as the mathematical calculation of the moment of a force does not require that the force be on the body.

4. Alternative choices for the position vector r are r = d 1 j = 5.68 j m and r = d 2 i = 6.77 i m.