In each one of the following figures there's a pole of length $1.2
\text{m}$ and there's a force $\vec F = 150 \text{N}$ acting on it.
Determine the torque that is created by the force relative to the axis
of rotation $O$.
I had no problem with calculating the magnitude of the torque, but I did have a problem determining its direction in fig. $D$. I used the rule of the right hand and I got that its direction is "outside the monitor" while according to the answers it has the same direction as in all the other figures before (i.e. "inside the monitor"). I'm also wondering whether the vectors directions are important in cross product. For example – is the direction of the displacement vector is important? If it is, then how can I apply the rule of the right hand here when the thumb and the index finger aren't starting from the same position.
Best Answer
The torque is defined as the cross product between the force vector and the displacement vector from the center of the object to the point where the force is applied: $\vec{\tau}=\vec{r}\times\vec{F}$.
The direction of the vectors is important when calculating cross products. Notice if the direction is reversed in the displacement vector:
$$\vec{r}\times\vec{F}=(r_yF_z-r_zF_y)\hat{i}+(r_zF_x-r_xF_z)\hat{j}+(r_xF_y-r_yF_x)\hat{k}$$
$$-\vec{r}\times\vec{F}=(-r_yF_z+r_zF_y)\hat{i}+(-r_zF_x+r_xF_z)\hat{j}+(-r_xF_y+r_yF_x)\hat{k}$$
$$-\vec{r}\times\vec{F}=-(\vec{r}\times\vec{F})$$
So as you can see, reversing the direction of the displacement vector will reverse the direction of the torque. A similar proof will demonstrate that the same will happen if the force vector is reversed.
Another way to look at the right hand rule (because I too have a bit of trouble using the finger rules) is to instead think of it as, put the bases of the two vectors together, curl your fingers in a direction that goes from the first vector ($\vec{r}$ in this case) to the second vector ($\vec{F}$). The direction that your thumb points in is the direction of the torque.