Irodov 1.237: Three Forces Collapse to One Line – Amit & Deepali Sinha Foundation

Problem Irodov 1.237

A square plate has the vertices $A$ , $B$ , $C$ , $D$ in order around it, so that $AC$ and $BD$ are its two diagonals. Three forces act on the plate, all in its own plane: at the vertex $A$ a force of magnitude $F$ parallel to the diagonal $BD$ , in the sense from $D$ to $B$ ; at the vertex $B$ a force of magnitude $F$ parallel to the diagonal $AC$ , in the sense from $A$ to $C$ ; and at the vertex $D$ a force of magnitude $F\sqrt{2}$ parallel to the side $AD$ , in the sense from $A$ to $D$ . No force acts at $C$ . Find the modulus and the direction of the resultant of these three forces, and the point at which it is applied, that point being taken on the side $BC$ .

The plate, the diagonal AC horizontal, with the three applied forces: F at A perpendicular to AC, F at B along AC, and sqrt(2) F at D along AD. Nothing acts at C. — The plate, the diagonal $AC$ horizontal, with the three applied forces: $\vect F$ at $A$ perpendicular to $AC$ , $\vect F$ at $B$ along $AC$ , and $\sqrt{2}\,\vect F$ at $D$ along $AD$ . Nothing acts at $C$ .

Given

square plate $ABCD$
$F$ at $A$ , along $D\to B$
$F$ at $B$ , along $A\to C$
$F\sqrt{2}$ at $D$ , along $A\to D$

Find

the modulus and direction of the resultant, and its point of application on

BC

Solution

Step 1A frame the square chooses for itself

The three given directions are two diagonal directions and one side direction, so the diagonals are the natural axes: let $\vect i$ run along $AC$ from $A$ to $C$ , let $\vect j$ run along $BD$ from $D$ to $B$ , and put the origin at the center of the plate, where the diagonals cross. Writing $h$ for the distance from that center to any vertex, half a diagonal, the four vertices and the three forces are

\begin{align*}\vect r_A &= -h\vect i, \quad \vect r_B = h\vect j, \quad \vect r_C = h\vect i, \quad \vect r_D = -h\vect j, \\ \vect F_A &= F\vect j, \quad \vect F_B = F\vect i, \quad \vect F_D = F(\vect i - \vect j) .\end{align*}

The force sqrt(2) F at D resolved along the two diagonals: a component F along AC and a component F against the sense D-> B. — The force $\sqrt{2}\,\vect F$ at $D$ resolved along the two diagonals: a component $F$ along $AC$ and a component $F$ against the sense $D\to B$ .

Step 2The resultant force, by adding the three

The modulus and direction come from the vector sum alone; where the forces act cannot affect it.

\begin{align*}\vect F_{\mathrm{res}} &= \vect F_A + \vect F_B + \vect F_D \\ &= F\vect j + F\vect i + F(\vect i - \vect j) \\ &= 2F\vect i\end{align*}

The two contributions along $\vect j$ were equal and opposite, so nothing at all survives across the diagonal $AC$ .

F_{\mathrm{res}} = 2F

(1)

Parallel to the diagonal

AC

, in the sense from

A

toward

C

Step 3The moment the three forces exert about the center

A resultant is not fully known until its line of action is, and the line is fixed by requiring the single force to twist the plate exactly as the three do. So take the moment of each force about the center of the plate and add.

\begin{align*}\vect N &= \vect r_A\times\vect F_A + \vect r_B\times\vect F_B + \vect r_D\times\vect F_D \\ &= (-h\vect i)\times(F\vect j) + (h\vect j)\times(F\vect i) + (-h\vect j)\times F(\vect i - \vect j) \\ &= -hF\vect k - hF\vect k + hF\vect k\end{align*}

\vect N = -hF\vect k

(2)

Step 4Where that moment puts the line of action

Let the resultant act at a point with position vector $x\vect i + y\vect j$ from the center. Its moment there is

(x\vect i + y\vect j)\times 2F\vect i = -2Fy\,\vect k ,

which is independent of $x$ , as it must be, since sliding the force along its own line changes nothing. Equating this to (2) fixes $y$ .

-2Fy = -hF

y = \frac{h}{2}

(3)

The resultant: one force of modulus 2F parallel to the diagonal AC, whose line of action meets the side BC at its midpoint M. It is drawn at M because that is where the problem asks the point of application to be taken. — The resultant: one force of modulus $2F$ parallel to the diagonal $AC$ , whose line of action meets the side $BC$ at its midpoint $M$ . It is drawn at $M$ because that is where the problem asks the point of application to be taken.

The side $BC$ runs from $\vect r_B$ to $\vect r_C$ , so its points are

\vect r(t) = (1-t)\,h\vect j + t\,h\vect i , \qquad 0\le t\le 1 ,

and the one with the required $y$ is picked out by

(1-t)\,h = \frac{h}{2} \qquad\Longrightarrow\qquad t = \frac{1}{2} .

Applied at the midpoint of the side

BC

Check

Redo the moment about a different point and the same line must come out. About the vertex $B$ , the force at $B$ has no moment, the force at $A$ (whose position relative to $B$ is $-h\vect i - h\vect j$ ) contributes $-hF\vect k$ , and the force at $D$ (at $-2h\vect j$ from $B$ ) contributes $2hF\vect k$ , giving $hF\vect k$ in total. The single force $2F\vect i$ placed at the midpoint of $BC$ , which sits at $(h/2)(\vect i-\vect j)$ from $B$ , gives exactly $hF\vect k$ as well. The magnitude is also reasonable: the three forces total $F+F+F\sqrt{2}\approx3.41F$ in magnitude, and $2F$ is duly less, since one pair cancels across $AC$ .

Remark

The resultant of a plane force system is a sliding vector: what the three forces determine is a line, not a point, and every point of that line does equally well as a point of application. That is why the problem has to say where to take the point, and why the answer is the intersection of the line with the side $BC$ rather than something the physics picks out on its own. The one case where this breaks down is a system whose vector sum vanishes while its moment does not, a couple, which has no line of action at all: no single force anywhere can reproduce it.