Question

Two forces ${F_1}$ and ${F_2}$ are acting at a point having resultant as $F$ . If ${F_2}$ is doubled, $F$ is also doubled. If ${F_2}$is reversed then also $F$ is doubled. Then ${F_1}:{F_2}:F$ is
A) $\sqrt 2 :\sqrt 2 :\sqrt 3 $
B) $\sqrt 3 :\sqrt 3 :\sqrt 2 $
C) $\sqrt 3 :\sqrt 2 :\sqrt 3 $
D) $\sqrt 2 :\sqrt 3 :\sqrt 2 $

Answer 1

Hint: We can use the Triangle law of vector addition to find the resultant and then solve the question by changing the values as given in the question. The reversal of direction can be signified by a minus sign or adding 180 to the initial angle.

Complete Step by step answer: We shall assume that the angle between ${F_1}$ and ${F_2}$ is $\theta $ .
According to the question, applying Triangle law of vector addition on ${F_1},{F_2},F$ we get,
$
  F = \sqrt {F_1^2 + F_2^2 + 2{F_1}{F_2}\cos \theta } \\
   \Rightarrow {F^2} = F_1^2 + F_2^2 + 2{F_1}{F_2}\cos \theta \\
 $
Consider this as equation 1.
Now, if ${F_2}$ is doubled then $F$ is also doubled. We can write that by triangle law of vector addition as
\[
  {(2F)^2} = F_1^2 + {\left( {2{F_2}} \right)^2} + 2{F_1}\left( {2{F_2}} \right)\cos \theta \\
   \Rightarrow 4{F^2} = F_1^2 + 4F_2^2 + 4{F_1}{F_2}\cos \theta \\
 \]
Consider this as equation 2
Now, if ${F_2}$is reversed then also $F$ is doubled. We can write that by triangle law of vector addition as
\[
  {(2F)^2} = F_1^2 + {\left( { - {F_2}} \right)^2} + 2{F_1}\left( { - {F_2}} \right)\cos \theta \\
   \Rightarrow 4{F^2} = F_1^2 + F_2^2 - 2{F_1}{F_2}\cos \theta \\
 \]
Consider this as equation 3
On adding equations 1 & 3, we get
$5{F^2} = 2F_1^2 + 2F_2^2$
Consider this as equation 4
On equating equations 2 & 3, we get
\[F_1^2 + 4F_2^2 + 4{F_1}{F_2}\cos \theta = F_1^2 + F_2^2 - 2{F_1}{F_2}\cos \theta \]
$
   \Rightarrow 3F_2^2 = - 6{F_1}{F_2}\cos \theta \\
   \Rightarrow \cos \theta = \dfrac{{ - {F_2}}}{{2{F_1}}} \\
 $
Now, we can substitute this value in equation 1 to get the following,
$
  {F^2} = F_1^2 + F_2^2 + 2{F_1}{F_2}\left( {\dfrac{{ - {F_2}}}{{2{F_1}}}} \right) \\
   \Rightarrow {F^2} = F_1^2 \\
 $
Substituting this value in equation 4, we get
$
  5{F^2} = 2F_1^2 + 2F_2^2 \\
   \Rightarrow 5F_1^2 = 2F_1^2 + 2F_2^2 \\
   \Rightarrow 3F_1^2 = 2F_2^2 \\
   \Rightarrow {F_1} = \dfrac{{\sqrt 2 }}{{\sqrt 3 }}{F_2} \\
 $
Also, $F = {F_1}$ so, $F = {F_1} = \dfrac{{\sqrt 2 }}{{\sqrt 3 }}{F_2}$
So the ratio ${F_1}:{F_2}:F = \sqrt 2 :\sqrt 3 :\sqrt 2 $ , and the correct answer is (D)

Note: While making equation 3 we can also write the angle as $180 + \theta $ instead of writing ${F_2}$ as $ - {F_2}$ . And because $\cos \left( {180 + \alpha } \right) = - \cos \alpha $ , we would still get the same result. Also, any other method of solving the three equations is good, so you must go with whatever is comfortable with you.