Question

The resultant of two concurrent forces $ n\overrightarrow {OP} $ and $ m\overrightarrow {OQ} $ is $ \left( {m + n} \right)\overrightarrow {OR} $ . Then $ R $ divides $ PQ $ in the ratio
A. $ m:n $
B. $ n:m $
C. $ 1:n $
D. $ m:1 $

Answer 1

Hint: Two forces are said to be concurrent if the forces originate from the same point or end at the same point. We are given two forces and the resultant force. We have to find the ratio that the end point of the resultant force divides the line meeting the end points of the original two forces. We can convert the problem into mathematical terms and use section formula to find the ratio.

Complete step by step solution:
We are given two concurrent forces $ n\overrightarrow {OP} $ and $ m\overrightarrow {OQ} $ . We are also given the resultant force of the two forces as $ \left( {m + n} \right)\overrightarrow {OR} $ .
The resultant of two forces can be written as vector addition of the forces. Thus, we can write,
 $ \left( {m + n} \right)\overrightarrow {OR} = n\overrightarrow {OP} + m\overrightarrow {OQ} \;\;\;\;...\left( 1 \right) $
 $ \left( {m + n} \right)\overrightarrow {OR} $ can also be written as $ \left( {m + n} \right)\left( {Position\;vector\;of\;R - Position\;vector\;of\;O} \right) $ . Since $ O $ is the origin, $ Position\;vector\;of\;O = 0 $ .
Thus, $ \left( {m + n} \right)\overrightarrow {OR} = \left( {m + n} \right)\left( {Position\;vector\;of\;R} \right) $
Similarly we can write $ n\overrightarrow {OP} $ and $ m\overrightarrow {OQ} $ as,
 $
  n\overrightarrow {OP} = n\left( {Position\;vector\;of\;P} \right) \\
  m\overrightarrow {OQ} = m\left( {Position\;vector\;of\;Q} \right) \\
  $
We can simplify equation $ \left( 1 \right) $ as,
 $
  \left( {m + n} \right)\overrightarrow {OR} = n\overrightarrow {OP} + m\overrightarrow {OQ} \\
   \Rightarrow \left( {m + n} \right)\left( {Position\;vector\;of\;R} \right) = n \times \left( {Position\;vector\;of\;P} \right) + m \times \left( {Position\;vector\;of\;Q} \right) \\
   \Rightarrow Position\;vector\;of\;R = \dfrac{{\left\{ {n \times \left( {Position\;vector\;of\;P} \right)} \right\} + \left\{ {m \times \left( {Position\;vector\;of\;Q} \right)} \right\}}}{{\left( {m + n} \right)}}\;\;\;\;\;\;\;\;\;...\left( 2 \right) \\
  $
From the section formula for vectors we have,
If point $ P $ divides the line segment joining the points $ A $ and $ B $ , and $ P $ divides $ AB $ in the ratio $ m:n $ , then the position vector of point $ P $ is given as,
 $ Position\;vector\;of\;P = \dfrac{{\left\{ {n \times \left( {Position\;vector\;of\;A} \right)} \right\} + \left\{ {m \times \left( {Position\;vector\;of\;B} \right)} \right\}}}{{\left( {m + n} \right)}} $
If we compare equation $ \left( 2 \right) $ with the section formula, then we can infer that the point $ R $ divides $ PQ $ in the ratio $ m:n $ .
Hence, option (A) is correct.
So, the correct answer is “Option A”.

Note: The position vector of the origin is taken as zero as it is the reference point. Vector addition is different from simple arithmetic addition as vector addition also takes the direction into account. Note that $ \overrightarrow {OP} ,\overrightarrow {OQ} \;and\;\overrightarrow {OR} $ may not be unit vectors and thus may be having some magnitude. We did not require the value of the magnitudes of the forces.