# Sign of distance/arm when finding moment/torque

I know that $$M=Fd$$, but when I have a negative moment, does the sign of the moment matter when I'm calculating for distance? If yes, then what would it mean when distance is negative?

You need to write it as an exterior product: $$\vec{M}=\vec{d}\times \vec{F}$$ Now let's assume $$\vec{d}$$ and $$\vec{F}$$ lie in $$xy$$ plane. Then $$\vec{M}$$ will be along the $$z-$$axis, either $$\hat{z}$$, or $$-\hat{z}$$. It depends on the relative situation of $$\vec{d}$$ and $$\vec{F}$$.

In elementary physics, sometimes, a moment in the direction of $$𝑧̂$$ is taken as a positive moment: $$M>0$$, while if it is in the direction of $$-𝑧̂$$, it is taken to be negative: $$M<0$$.

So sign of $$\vec{M}$$ is important because it determines the direction of $$\vec{M}$$.

does the sign of the moment matter when I'm calculating for distance?

No. See the figures below.

The distance $$d$$ is simply the magnitude of the position vector $$\vec r$$ between the point $$o$$ about which the moment is calculated and the point of application of the force. The moment about $$o$$ is given by the following vector cross product.

$$\vec M_{o}=\vec r x \vec F=dF\sin\theta$$

The direction of the force in the figure to the left causes a clockwise (negative) moment about $$o$$ in the z direction whereas the direction of the force vector in the figure to the right causes a counterclockwise (positive) moment about $$o$$.

The direction of the moment can also be shown by taking the right hand and pointing the index finger in the direction of $$\vec r$$ and the middle finger in the direction of $$\vec F$$. The thumb then points in the direction of $$\vec M_o$$.

Hope this helps.

The torque is:

$$\vec M=\vec d\times\vec F$$

assume

$$\vec d=d\,\left(\pm\vec e_x\right)\quad, \vec F=F\,\left(\pm\vec e_y\right)\quad, d>0~,F>0$$

you obtain

case I $$\vec d=d\,\vec e_x~,\vec F=F\,\vec e_y\\ \vec M=d\,F\,\vec e_z$$

case II $$\vec d=d\,\left(-\vec e_x\right)~,\vec F=F\,\vec e_y\\ \vec M=d\,F\,\left(-\vec e_z\right)$$

case III $$\vec d=d\,\vec e_x~,\vec F=F\,\left(-\vec e_y\right)\\ \vec M=d\,F\,\left(-\vec e_z\right)$$

case IV $$\vec d=d\,\left(-\vec e_x\right)~,\vec F=F\,\left(-\vec e_y\right)\\ \vec M=d\,F\,\vec e_z$$

thus the magnitude of the torque is always $$~d\,F~$$ and the direction is $$~+\vec e_z~$$ or $$~-\vec e_z$$