Why is torque equivalent to $xF_y-yF_x $?

Question

In Feynman's lecture on rotations in space, http://www.feynmanlectures.caltech.edu/I_20.html, he introduced the cross product by building upon a definition of torque he had derived in a previous lecture,

$$\tau=xF_y-yF_x$$

He explained that radial distance, $\mathbf{r} = \sqrt{x^2+y^2}$, and force, $\mathbf{F}=\sqrt{F_x^2+F_y^2}$, are just two vectors and that any two vectors can be combined similarly to get a third resultant vector. Rather than try to represent the result on the same plane as the original vectors it makes organizational sense to represent the result in the third dimension, perpendicular to the original vectors. This is the cross product $\mathbf{c = a \times b}$:

$$c_x = a_y b_z - a_z b_y$$ $$c_y = a_z b_x - a_x b_z$$ $$c_z = a_x b_y - a_y b_x$$

Glorious illumination! I had been trying for years to figure out why arbitrarily smooshing vector components together resulted in a vector sticking out of the blackboard somehow equivalent to simple torque. I now understand that the cross product is just a pseudo vector to represent two things that interact orthogonally.

But I still don't feel 100% about about Feynman's original definition for torque. I followed the geometrical proof but I'm hoping there is a straightforward, intuitive, way to understand why $$xF_y-yF_x = \mathbf{rF_{tangential}}$$ If anybody could help me fill in this last hole it would be greatly appreciated.

Answer 1

Suppose the force $F$ with components $(F_x, F_y)$ acts through a point $P(x, y)$ on a rigid body which is pivoted so that it can rotate about the z axis. Make a sketch.

The torque due to $F$ is the same as the moment of $F$, because the pivot provides and equal and opposite force. (Torque = one force in pair x distance between them.)

The moment of $F$ is the same as the moment of its components. The perpendicular distance of the $F_y$ component from the z axis is $x$, so the anticlockwise moment of this force is $+xF_y$. The perpendicular distance of the $F_x$ component from the z axis is $y$, so the anticlockwise moment of this force is $-yF_x$. Adding the moments we get a total torque of $xF_y - yF_x$.

Answer 2

The vector or cross product of 2 vectors $ \vec C=\vec A \times \vec B$ is equal to $AB sin\theta \hat n $, where $\theta$ is the angle between $\vec A$ and $\vec B$ and $\hat n $ is unit vector representing direction(which is perpendicular to both $\vec A$ and $\vec B$ ). If we resolve them in rectangular components, we get $\vec C=(A_x\hat i\ +A_y\hat j+A_z\hat k)\times(B_x\hat i+B_y\hat j+B_Z\hat k) $. On expanding this, $\vec C=(A_x. B_x)(\hat i \times \hat i)+(A_x . B_y)(\hat i\times \hat j)+(A_x. B_z)(\hat i \times \hat k) + (A_y. B_x)(\hat j \times \hat i)+(A_y. B_y)(\hat j\times \hat j)+(A_y. B_z)(\hat j \times \hat k)+(A_z. B_x)(\hat k \times \hat i)+(A_z. B_y)(\hat k\times \hat j)+(A_z. B_z)(\hat k \times \hat k)---(1)$

where $\hat i, \hat j $ and $ \hat k$ are unit vectors in $X,Y$ and $Z$ axis respectively.But, $\hat i\times \hat i=\hat j\times \hat j=\hat k\times \hat k=1\times 1 .sin(\pi /2)=0$ and $\hat i\times \hat j =-(\hat j\times \hat i) =\hat k, \hat j\times \hat k =-(\hat k\times \hat j) =\hat i$ and $\hat k\times \hat i =-(\hat i\times \hat k) =\hat j$ So, $(1)$ becomes $\vec C =(A_x.B_y)\hat k-(A_x.B_Z)\hat j-(Ay.B_x)\hat k+(A_y.B_z)\hat i+(A_z.B_x)\hat j - (A_z.B_y)\hat i ---(2)$.

If only $X$ and $Y$ axis are taken,$(2)$ becomes $((A_x.B_y)-(A_y.B_x))\hat k$.

Now, Torque $\tau=r\times F$

$=(x\hat i+y \hat j)\times (F_x\hat i + F_y\hat j)$

$= (x.F_y-y.F_x)\hat k$