# Why is torque equivalent to $xF_y-yF_x$?

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.

• You are giving a perfect example of why I never liked the Feynman lectures. There is absolutely nothing enlightening about the choice of a coordinate system and the approach fails to teach the student to think in parallel and orthogonal components. – CuriousOne Aug 2 '16 at 20:43

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 $xF_y - yF_x$.

• Right, so is there a good way to maybe visualize why it equals that. Why the x-comp would be paired with the force y-comp and the y with Fx, and why there is a minus sign in there. Maybe I'm not getting your answer. – BoddTaxter Aug 2 '16 at 20:16
• Visualise by making a sketch of the force F and its components acting through some point P with position vector r. $F_x$ is paired with $y$ because for moments the force and distance must be perpendicular. – sammy gerbil Aug 2 '16 at 20:30

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 $\tau=r\times F$

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

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

• Welcome to Physics.SE Moulik! – heather Aug 3 '16 at 16:21