Question:
Two forces of magnitude $4P$ and $3P$ acting at a point O have a resultant of magnitude $5P$. If any transversal cuts the lines of action of the forces at the points R, S, and T respectively, show that $\frac{4}{OR}+\frac{3}{OS}=\frac{5}{OT}$.
My book's attempt:
Let AB be the transversal. OX is drawn perpendicular to AB. Taking resolved parts of the forces $4\vec{P}$, $3\vec{P}$, and $5\vec{P}$ we get,
$$4P\cos (XOR)+3P\cos (XOS)=5P\cos (XOT)$$
$$4\frac{OX}{OR}+3\frac{OX}{OS}=5\frac{OX}{OT}$$
$$\frac{4}{OR}+\frac{3}{OS}=\frac{5}{OT}\text{(showed)}$$
My question:
- The theorem of resolved parts: "The algebraic sum of the magnitude of the resolved parts of two forces acting at a point in any direction is equal to the resolved part of their resultant in the same direction." This theorem was applied by my book for solving this problem. After the drawing of the transversal $AB$ all three vectors are cut and new vectors $\vec{OS}$, $\vec{OT}$, and $\vec{OR}$, which are different than $4\vec{P}$, $3\vec{P}$, and $5\vec{P}$, are formed. Now, my question is, is $\vec{OT}$ the resultant of $\vec{OS}$ and $\vec{OR}$? Because if $\vec{OT}$ isn't the resultant of $\vec{OS}$ and $\vec{OR}$, the theorem doesn't apply and the book has done the math wrongly. So, to reiterate, is $\vec{OT}$ the resultant of $\vec{OS}$ and $\vec{OR}$?
The theorem applies to the original vectors, $\vec{3P}$,$\vec{4P}$, and $\vec{5P}$, as seen in the first equation. $\vec{OT}$ is not the resultant of $\vec{OS}$ and $\vec{OR}$. You are using $OS,OT,OR$ only to calculate the cosines. Otherwise, the theorem applied to these vectors would mean $OB+OB=OB$.