Question

Forces $P,Q$ act at $O$ and have a resultant $R$. If any transversal cuts their lines of action at $A,B,C$, respectively, then

• A. $\frac{P}{OA}+\frac{Q}{OB}+\frac{R}{OC}=0$
• B. $\frac{P}{OA}+\frac{Q}{OB}+\frac{R}{OC}=1$
• C. $\frac{P}{OA}+\frac{Q}{OB}-\frac{R}{OC}=0$
• D. $\frac{P}{OA}+\frac{Q}{OB}-\frac{R}{OC}=1$.

Answer 1

Answer: (C)

Let $O$ be taken as the origin of reference.
Let $a,b,c$ be the position vectors of $A,B,C$ respectively
so that $OA=a,OB=b,OC=c$

Then, the vector representing the force $P$ along $OA$
$=P($ unit vector along $OA)=P\frac{OA}{OA}=\frac{P}{OA}a$.
Similarly, the vector representing the force $Q$ along $OB$
$=\frac{Q}{OB}b$
and the vector representing the force $R$ along $OC=\frac{R}{OC}c$
Since $R$ is the resultant of $P$ and $Q$,
we have $(P/OA)a+(Q/OB)b=(R/OC)c$
or, $(P/OA)a+(Q/OB)b-(R/OC)a=0(1)$.
Since, $a,b,c$ are the position vectors of three collinear points,
therefore, we have the algebraic sum of the coefficients of $a,b,c$ in (i) $=0$
i.e., $(P/OA)+(Q/OB)-(R/OC)=0$
Hence, $\frac{P}{OA}+\frac{Q}{OB}=\frac{R}{OC}$.