Question

The resultant of two forces $F_1$​ and $F_2$ is P. If $F_2$​ is reversed, then resultant is Q. Then the value of ($P^2$+$Q^2$) in terms if $F_1$ and $F_2$​
A) $2\left( {F_1^2 + F_2^2} \right)$
B) $F_1^2 + F_2^2$
C) $2F_1^2 + F_2^2$
D) None of these

Answer 1

Hint
in order to solve this question, we should have some knowledge of vector formulas like ${\left| P \right|^2} = {\left| {{F_1}} \right|^2} + {\left| {{F_2}} \right|^2} + 2\left| {{F_1}} \right|\left| {{F_2}} \right|\cos \theta $, Now reverse the sign of $F_2$​ then write one more equation after that adding both the equations, we will get the result.

Complete Step by step solution
As it is given that the resultant of the two forces $F_1$​ and $F_2$ is P then by using the vector formula, we get ${\left| P \right|^2} = {\left| {{F_1}} \right|^2} + {\left| {{F_2}} \right|^2} + 2{F_1}{F_2}\cos \theta $ ……………………. (1)
Where, $F_1$ and $F_2$​ are two forces and P is the resultant forces
$\theta$ is the angle between the two forces.
Now reverse the sign of the $F_2$ i.e. ${F_2} = - {F_2}$, Then the resultant is Q then by using the vector formula, we get ${\left| Q \right|^2} = {\left| {{F_1}} \right|^2} + {\left| { - {F_2}} \right|^2} - 2{F_1}{F_2}\cos \theta $ ……………………… (2)
Now as we have to the value of the $\left( {{P^2} + {Q^2}} \right)$ , then for this adding the equation (1) and equation (2), we get
$ \Rightarrow \left( {{P^2} + {Q^2}} \right) = F_1^2 + F_2^2 + 2{F_1}{F_2}\cos \theta + F_1^2 + F_2^2 - 2{F_1}{F_2}\cos \theta $
$ \Rightarrow \left( {{P^2} + {Q^2}} \right) = 2F_1^2 + 2F_2^2 = 2\left( {F_1^2 + F_2^2} \right)$
Hence, (A) option is correct.

Note
In these types of questions we need to remember the formula of vector law. Firstly, write this after that put the condition for all the forces and angle between them, after that simplify the equation, we will get the required result.