Question

The maximum resultant of two forces is$P$and the minimum resultant is$Q$, the two forces are at right angles, the resultant is

• A. $P+Q$
• B. $P–Q$
• C. $\frac{1}{2\left(\sqrt{P^2+Q^2}\right)}$
• D. $\sqrt{\frac{P^2+Q^2}{2}}$

Answer 1

The correct option is D

$\sqrt{\frac{P^2+Q^2}{2}}$

Step 1. Given data:

The maximum resultant of two forces is$P$.

The minimum resultant is$Q$.

Step 2. Formula used:

The magnitude of the resultant of two vectors is given by,

$R=\sqrt{A^2+B^2+2AB\cos \theta }$ (Where $A,B$ are two vectors and$\theta$ is the angle between them)

Step 3. Calculating the resultant at an angle ${90}^{\circ }$,

Let us consider two vectors$A,B$.

According to the given question, for the resultant to be maximum it is required that-,

$\begin{array}{rcl}P& =& \sqrt{A^2+B^2+2AB}(\cos \theta =1,\theta =0^o)\\ P& =& A+B·····\left(1\right)\end{array}$

If it is a minimum, then,

$\begin{array}{rcl}Q& =& \sqrt{A^2+B^2-2AB}(\cos \theta =-1,\theta ={180}^o)\\ Q& =& A-B·····\left(2\right)\end{array}$

When the angle between them is ${90}^{\circ }$ then,

$\begin{array}{rcl}R& =& \sqrt{A^2+B^2}·····\left(3\right)(\cos \theta =0^{°},\theta ={90}^{°})\end{array}$

Add the equation $\left(1\right)$ and $\left(2\right)$ to obtain the value of $A$ in terms of $P$ and $Q$.

$$\begin{gathered} 2A=P+Q \\ A=\frac{P+Q}{2} \end{gathered}$$

Similarly, subtract the equation $\left(1\right)$ and $\left(2\right)$ to obtain the value of $B$ in terms of $P$ and $Q$.

$$\begin{gathered} 2B=P-Q \\ B=\frac{P-Q}{2} \end{gathered}$$

Substitute value of $A$ and $B$ in equation $\left(3\right)$.

$\begin{array}{rcl}R& =& \sqrt{{\left(\frac{P+Q}{2}\right)}^2+{\left(\frac{P-Q}{2}\right)}^2}\\ & =& \sqrt{\frac{P^2+Q^2+2PQ+P^2+Q^2-2PQ}{4}}\\ & =& \sqrt{\frac{2P^2+2Q^2}{4}}\\ & =& \sqrt{\frac{P^2+Q^2}{2}}\end{array}$

Thus, the resultant is $\sqrt{\frac{P^2+Q^2}{2}}$.

Hence, option D is the correct answer.