Irodov 1.235: Cross Product First, Arm for Free – Amit & Deepali Sinha Foundation

Problem Irodov 1.235

A force $\vect F = A\vect i + B\vect j$ acts at a point $P$ of a body, and $P$ has the position vector $\vect r = a\vect i + b\vect j$ measured from the origin $O$ . Here $a$ , $b$ , $A$ and $B$ are given constants, and $\vect i$ and $\vect j$ are the unit vectors of the $x$ and $y$ axes. Find the moment $\vect N$ of the force about the point $O$ , and the arm $l$ of the force about that same point, the arm being the perpendicular distance from $O$ to the force’s line of action.

The point of application P carries the position vector r from O, its components dashed along the axes as a and b. The force F acts at P, its own components dashed as A and B. — The point of application $P$ carries the position vector $\vect r$ from $O$ , its components dashed along the axes as $a$ and $b$ . The force $\vect F$ acts at $P$ , its own components dashed as $A$ and $B$ .

Given

$\vect r = a\vect i + b\vect j$
$\vect F = A\vect i + B\vect j$
$a$ , $b$ , $A$ , $B$ constant

Find

\vect N

about

O

; the arm

l

about

O

Solution

Step 1Expand the moment as a cross product

The moment of a force about a point is the cross product of the position vector of the point of application with the force. Expanding it term by term keeps every piece of the bookkeeping visible.

\vect N = \vect r\times\vect F = \bigl(a\vect i + b\vect j\bigr)\times\bigl(A\vect i + B\vect j\bigr)

\begin{align*}\vect N &= aA\,(\vect i\times\vect i) + aB\,(\vect i\times\vect j) + bA\,(\vect j\times\vect i) + bB\,(\vect j\times\vect j)\end{align*}

The four products of the unit vectors are fixed by the right handed frame,

\vect i\times\vect i = \vect j\times\vect j = \vect 0, \qquad \vect i\times\vect j = \vect k, \qquad \vect j\times\vect i = -\vect k ,

so only the two mixed terms survive.

\vect N = (aB - bA)\,\vect k

(1)

Step 2What the arm is, and how it reaches the moment

The arm is a geometric quantity: the perpendicular distance from $O$ to the line along which the force acts. Its link to the moment (1) is the definition of the cross product’s magnitude, in which the sine of the angle $\theta$ between $\vect r$ and $\vect F$ picks out exactly the part of $r$ standing perpendicular to the line of action.

The arm about O. The segment OP has length r and ends where F acts; the line of action, produced as the gray dash, carries the perpendicular from O, whose length is the arm l, right angle marked. The angle between r produced and F is θ. — The arm about $O$ . The segment $OP$ has length $r$ and ends where $\vect F$ acts; the line of action, produced as the gray dash, carries the perpendicular from $O$ , whose length is the arm $l$ , right angle marked. The angle between $\vect r$ produced and $\vect F$ is $\theta$ .

|\vect N| = rF\sin\theta = F\,(r\sin\theta)

|\vect N| = F\,l

(2)

Step 3Divide the moment by the force

Both factors on the right of (2) are now known: the magnitude of the moment is the modulus of the scalar in (1), and the magnitude of the force follows from its two components.

\begin{align*}|\vect N| &= |aB - bA| \\ F &= \sqrt{A^2 + B^2}\end{align*}

Dividing the first by the second gives the arm.

l = \frac{|\vect N|}{F} = \frac{|aB - bA|}{\sqrt{A^2 + B^2}}

(3)

Check

Dimensions first: $aB$ and $bA$ are each a length times a force, so their difference divided by a force is a length, as an arm must be. Now the case that matters: if the force points along the position vector, its components are proportional to those of $\vect r$ , so $A/a = B/b$ and therefore $aB - bA = 0$ . Both boxed results vanish together, which is right, because the line of action then passes straight through $O$ and there is no arm and no twist about that point.

Remark

The combination $aB - bA$ is the determinant of the two vectors, that is, the signed area of the parallelogram they span. Read that way the arm is just an area divided by a base, which is why dividing the moment by the force recovers a length at all. There is a second thing hidden in it: sliding the point of application along the line of action, which replaces $\vect r$ by $\vect r + \lambda\vect F$ , changes the moment by $\lambda\,\vect F\times\vect F$ , and that is zero. So the moment never depended on where on the line the force was applied, only on the line itself, and the arm is precisely the piece of information about that line which survives.