Question

With two forces acting at a point, the maximum effect is obtained when their resultant is $4N$. If they act at right angles, then their resultant is $3N$. Then the forces are

• A. $F_1=(2+\frac{1}{2}\sqrt{3})Nand{F}_2=(2-\frac{1}{2}\sqrt{3})N$
• B. $F_1=(2+\sqrt{3})Nand{F}_2=(2-\sqrt{3})N$
• C. $F_1=(2+\frac{1}{2}\sqrt{2})Nand{F}_2=(2-\frac{1}{2}\sqrt{2})N$
• D. $F_1=(2+\sqrt{2})Nand{F}_2=(2-\sqrt{2})N$

Answer 1

The correct option is C

$F_1=(2+\frac{1}{2}\sqrt{2})Nand{F}_2=(2-\frac{1}{2}\sqrt{2})N$

Explanation of correct answer: option (c) is correct.

Step 1: Given,

1. For case 1, Resultant Force is $4N$. When the force acts at a point the $\alpha =0°$.
2. For case 2, Resultant force is $3N$ and $\alpha =90°$.

Formula used: $R=\sqrt{{F_1}^2+{F_2}^2+2F_1{F}_2\cos \alpha }$

Step 2:

Resultant Force: When two forces, P and Q, act on an object simultaneously, their combined effect, known as the resultant force, is the same as the action of a single force.

$\overrightarrow{R}=\overrightarrow{P}+\overrightarrow{Q}$, it is the sum of the forces taken in vectors.

If there is an angle between the forces: Let $\alpha$ be the angle between forces $\overrightarrow{F_1}$ and $\overrightarrow{F_2}$, ${\theta }_1$ the angle between $\overrightarrow{R}$ and $\overrightarrow{F_1}$, and ${\theta }_2$ the angle between $\overrightarrow{R}$and $\overrightarrow{F_2}$. Then the resultant force is $R=\sqrt{{F_1}^2+{F_2}^2+2F_1{F}_2\cos \alpha }$

Case 1: Resultant Force is $4N$.

When the force acts at a point the $\alpha =0°$.

So, the resultant force becomes $R=\sqrt{{F_1}^2+{F_2}^2+2F_1{F}_2\cos 0°}$ $\Rightarrow$ $\sqrt{{({F_1}^2+{F_2}^2)}^2}\Rightarrow (F_1+F_2)$

$(F_1+F_2)=4N\dots \left(i\right)$

Case 2: Resultant force is $3N$ and $\alpha =90°$.

So, the resultant force is $R=\sqrt{{F_1}^2+{F_2}^2+2F_1{F}_2\cos 90°}=\sqrt{{F_1}^2+{F_2}^2}$

$$\begin{gathered} \sqrt{{F_1}^2+{F_2}^2}=3N \\ \Rightarrow {F_1}^2+{F_2}^2=9\dots \left(ii\right) \end{gathered}$$

Step 3: Solving both equations:

From equation $\left(i\right)$

Squaring both sides we get

$$\begin{gathered} {(F_1+F_2)}^2=4^2 \\ \Rightarrow {F_1}^2+{F_2}^2+2F_1{F}_2=16 \end{gathered}$$

Substituting equation $\left(i\right)$ in the above equation we get,

$$\begin{gathered} 9+2F_1{F}_2=16 \\ \Rightarrow 2F_1{F}_2=7\dots \left(iii\right) \end{gathered}$$

We, know that ${(F_1-F_2)}^2={F_1}^2+{F_2}^2-2F_1{F}_2$

From equation $\left(ii\right)and\left(iii\right)$,

$$\begin{gathered} (F_1-F_2)=9-7 \\ \Rightarrow (F_1-F_2)=2\dots \left(iv\right) \end{gathered}$$

Solving $\left(i\right)and\left(iv\right)$ we get $F_1=(2+\frac{1}{2}\sqrt{2})Nand{F}_2=(2-\frac{1}{2}\sqrt{2})N$.

Hence, the answer is (c) .