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

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 $O A=a, O B=b, O C=c$

Then, the vector representing the force $P$ along $O A$
$=P($ unit vector along $O A)=P \frac{O A}{O A}=\frac{P}{O A} a$.
Similarly, the vector representing the force $Q$ along $O B$
$=\frac{Q}{O B} b$
and the vector representing the force $R$ along $O C=\frac{R}{O C} c$
Since $R$ is the resultant of $P$ and $Q$,
we have $( P / O A) a+( Q / O B) b=( R / O C) c$
or, $(P / O A) a+(Q / O B) b-(R / O C) a=0\,\,\,\, (1)$.
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., $(P / O A)+(Q / O B)-(R / O C)=0$
Hence, $\frac{P}{O A}+\frac{Q}{O B}=\frac{R}{O C}$.