# Question regarding direction of torque

1. Apr 6, 2014

### Permanence

1. The problem statement, all variables and given/known data
Replace the force and couple system shown by an equivalent resultant force and couple moment acting at point O.

The solution being:

I understand the force summation, but I'm a bit confused about the direction for the moment summation.

3. The attempt at a solution

I don't know how they determined which components had positive or negative moments. The x components on the two forces are in the same direction, but the moment of force is positive for one and negative for the other. I thought it would have been:
(3kN)sin(30)(0.2m) - (3kN)cos(30)(0.1) - (3/5)(5 kN)(0.1m) - (4/5)(5 kN)(0.5m) - (4kN)(0.2m)

Thank you.

2. Apr 6, 2014

### SteamKing

Staff Emeritus
In the problem statement, positive moments are taken counterclockwise w.r.t. point O.

The moments of the horizontal components of the two force vectors have the opposite sign because they are applied on different sides of the x-axis. It does take a certain amount of visualization to see how forces will create moments. After all, the magnitude of a moment is the product of the force and the distance which this force acts from the axis of rotation,

M = F * d.

The direction of the force is not the sole determinant of the direction of the moment: the sign of the moment arm d also plays a role.

3. Apr 6, 2014

### Permanence

Thank you for the prompt response SteamKing. I had a feeling that distance had some role, but I was under the erroneous impression that it either relied on distance or the force. I'll reattempt the problem now.

Edit:
It took me awhile to get it. I had always treated the distance to be a scalar without direction. I've gotten the hang of it by applying by negating what I know would would take place in the first quadrant.

Last edited: Apr 6, 2014