Question:
Three forces of magnitudes $P, Q$, and $R$ act along the three sides of an equilateral $\triangle ABC$. If their resultant passes through the centroid and is parallel to the side $BC$, prove that $\frac{1}{2}P=Q=R$.
My book's attempt:
[This is my book's figure, but it is unclear, so I drew a clearer picture.]
Let resultant=$R'$
$AB, BE$, and $CF$ are medians & $G$ is the centroid.
Let $AD=BE=CF=x$
Note that the medians are perpendicular to the opposite side.
Taking moment about point $B$,
$$Q\cdot BE=R'\cdot GD\tag{1}$$
$$Qx=R'\frac{x}{3}$$
$$Q=\frac{R'}{3}\tag{4}$$
Taking moment about point $A$,
$$P\cdot AD=R'\cdot AG\tag{2}$$
$$Px=R'\frac{2}{3}x$$
$$\frac{1}{2}P=\frac{R'}{3}\tag{5}$$
Taking moment about point $C$,
$$R\cdot CF=R'\cdot GD\tag{3}$$
$$Rx=R'\frac{x}{3}$$
$$R=\frac{R'}{3}\tag{6}$$
From $(4), (5)$ and $(6)$,
$$\frac{1}{2}P=Q=R\ \text{(Proved)}$$
My comments:
(See my book's figure for this) Why is the vector in the middle acting in two directions? Shouldn't $R'$ have one direction? In my figure, it is acting from $G$ to $S$ and from $G$ to $T$. Why is it acting in two directions? Were they not sure which is the right direction, so they just drew it in both directions? Moreover, how did they formulate $(1)$? What was the reasoning/intuition behind $(1)$? If I can understand $(1)$, I'll understand $(2)$ and $(3)$ as well. I think $(1)$ is wrong because, in $(1)$, the author is assuming that $P$ and $R$ have no moment about $A$. Only $Q$ has moment about $A$ according to the author, which I think is wrong.
My questions:
- Why is the vector(s) in the middle acting in two directions?
- What is the reasoning/intuition behind $(1)$?