Irodov 1.236: Slide Both Forces to Their Crossing – Amit & Deepali Sinha Foundation

Problem Irodov 1.236

A force $\vect F_1 = A\vect j$ acts at the point whose position vector from the origin $O$ is $\vect r_1 = a\vect i$ , and a force $\vect F_2 = B\vect i$ acts at the point whose position vector is $\vect r_2 = b\vect j$ . Here $\vect i$ and $\vect j$ are the unit vectors of the $x$ and $y$ axes and $a$ , $b$ , $A$ , $B$ are given constants. Find the arm $l$ of the resultant of these two forces about the point $O$ , the arm being the perpendicular distance from $O$ to the line along which that resultant acts.

The two forces as the statement gives them: F_1 along j, applied at the point a i of the x axis, and F_2 along i, applied at the point b j of the y axis. — The two forces as the statement gives them: $\vect F_1$ along $\vect j$ , applied at the point $a\vect i$ of the $x$ axis, and $\vect F_2$ along $\vect i$ , applied at the point $b\vect j$ of the $y$ axis.

Given

$\vect F_1 = A\vect j$ at $\vect r_1 = a\vect i$
$\vect F_2 = B\vect i$ at $\vect r_2 = b\vect j$
$a$ , $b$ , $A$ , $B$ constant

Find

l

, the arm of the resultant about

O

Solution

Step 1Add the two forces

The resultant is the vector sum, and here the two forces lie along the two axes, so the sum needs no resolving at all: one force contributes the whole $x$ component and the other the whole $y$ component.

\vect F = \vect F_1 + \vect F_2 = A\vect j + B\vect i = B\vect i + A\vect j

Its magnitude is the root of the sum of the squares of those two orthogonal components.

F = \sqrt{A^2+B^2}

(1)

Step 2Find one point the resultant passes through

Knowing the resultant’s magnitude and direction is not yet knowing its line, and it is the line that the arm is about. The way in is that a force may be slid anywhere along its own line of action without altering its moment about any point, since shifting the point of application by any multiple of the force itself adds nothing.

(\vect r + \lambda\vect F)\times\vect F = \vect r\times\vect F + \lambda\,(\vect F\times\vect F) = \vect r\times\vect F

So slide each force to where the two lines cross. The line of $\vect F_1$ is the vertical through the point $a\vect i$ , the line of $\vect F_2$ is the horizontal through the point $b\vect j$ , and those two meet at the single point $S$ with position vector

\vect r_S = a\vect i + b\vect j .

The crossing this step finds: the line of F_2, produced as the gray dash, meets the line of F_1 at the single point S, whose position vector is a i + b j. — The crossing this step finds: the line of $\vect F_2$ , produced as the gray dash, meets the line of $\vect F_1$ at the single point $S$ , whose position vector is $a\vect i + b\vect j$ .

Both forces slid along their lines to the common point S, a distance r_S from O, and added there, dashed, into the solid resultant F. The perpendicular from O onto its line of action is the arm l. — Both forces slid along their lines to the common point $S$ , a distance $r_S$ from $O$ , and added there, dashed, into the solid resultant $\vect F$ . The perpendicular from $O$ onto its line of action is the arm $l$ .

Step 3The moment of the resultant about

O

With the resultant located at $S$ , its moment about the origin is one cross product, expanded term by term as before.

\begin{align*}\vect N &= \vect r_S\times\vect F = \bigl(a\vect i + b\vect j\bigr)\times\bigl(B\vect i + A\vect j\bigr) \\ &= aB\,(\vect i\times\vect i) + aA\,(\vect i\times\vect j) + bB\,(\vect j\times\vect i) + bA\,(\vect j\times\vect j)\end{align*}

The products of a unit vector with itself vanish, and the two mixed ones differ only in sign,

\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 the surviving pair enters with opposite signs.

\vect N = (aA - bB)\,\vect k

(2)

Step 4Turn the moment into a distance

The arm is defined by the moment the resultant produces about $O$ : the magnitude of that moment is the force times the perpendicular distance to its line. Dividing (2) by the magnitude in (1) isolates the distance.

|\vect N| = F\,l

l = \frac{|aA - bB|}{\sqrt{A^2+B^2}}

(3)

Check

The sliding step can be tested by not using it. Taken about $O$ at the points where they were originally applied, the first force’s moment is $aA\vect k$ and the second’s is $-bB\vect k$ , and those two add to exactly the moment found at $S$ in (2). Dimensionally $aA$ and $bB$ are each a force times a length, so the boxed ratio is a length. And the arm vanishes when the two products are equal: the moments about $O$ then cancel, which means the resultant’s line runs through $O$ itself.

Remark

Notice which quantities pair up. The moment mixes $a$ with $A$ and $b$ with $B$ , that is, each force with its own distance from $O$ , and the two enter with opposite signs. With all four constants positive, as the figure draws them, that opposition is the statement that one force twists the origin counterclockwise and the other clockwise; the opposition of the signs itself is a fact about the algebra and survives whatever signs the constants carry, which is why the boxed answer takes the modulus. The arm is what is left after that near cancellation, so it can be made as small as you please with neither force being small, and at $aA = bB$ the pair still has a perfectly good resultant, which simply happens to point through the origin. The arm measures the line’s offset, never the strength of what acts along it.