Torque problem please help?

Question

Calculate the net torque (magnitude and direction) on the beam in Figure P8.3 about the following axes.
http://www.webassign.net/sercp/p8-3.gif...
figure 8.3 is above

(a) an axis through O, perpendicular to the page
(b) an axis through C, perpendicular to the page

http://www.webassign.net/sercp/p8-3.gif

Answer 1

Just plug in the values into the formula. Use the "right hand" angle to the axis to where the force causes the machine to move. (The degrees to use in the formula are the always right hand angle from the machine's perspective because that's the way things are made so everything is universal and you don't strip the threads or gears.).

Answer 2

The link doesn't work

Answer 3

Resolve the forces into perpendicular and horizontal components.

The vertical components of the forces are:

25 sin 30 = 12.5 N up.

10 sin 20 = 3.42 down.

30 sin 45 = 21.21 up.

The horizontal components of the forces are:
25 cos 30
10 cos 20
30 cos 45

Taking moments of these forces about the point O.

The horizontal components pass through O and their moments are zero.
The vertical component 21.21 also passes through O and its moment is also zero.

The sum of the moments of 12.5 N up and 3.42 down about the point O is
12.5 x 2 - 3.42 x 4 = 38.68 Nm and is in the anticlockwise direction. (Since we have taken anticlockwise moment as positive)

Taking moments of these forces about the point C.
The horizontal components pass through C and their moments are zero.
The vertical component 12.5 N up also passes through C and its moment is also zero.

The sum of the moments of 21.21 up and 3.42 down about the point C is(Both are rotating in clockwise direction and hence minus sign)
- 3.42 X 2 - 21.21 x 2 = -49.26 Nm clockwise.