Question

The three forces operating at a point, 1 N, 2 N, and 3 N, are parallel to the sides of an equilateral triangle in order. Their resultant magnitude is

• A. $\sqrt{\frac{3}{2}}N$
• B. $3\mathrm{N}$
• C. $\sqrt{3}\mathrm{N}$
• D. $\frac{3}{2}\mathrm{N}$

Answer 1

The correct option is C

$\sqrt{3}\mathrm{N}$

Step 1: Given Data and the required figure:

Let,

$F_1=1N$

$F_2=2N$

$F_3=3N$

Equilateral triangle

Step 2: Formula used:

The resultant force can be found as-

$F=\sqrt{{F_x}^2+{F_y}^2}$

Where, $F_x,F_y$ are forces in x-direction and y-direction respectively.

Step 3: Calculate the force in x and y directions:

The components of force in x and y direction can be found according to the below shown figure,

First $X$ component of force $F_1=1N$-

$F_{1x}=1N\dots \dots \dots \dots \left(1\right)$

$Y$ component of force $F_1=1N$-

$F_{1y}=0\dots \dots \dots \dots \left(2\right)$

$X$ component of force $F_2=2N$-

$F_{2x}=-2\cos 60°=-1\dots \dots \dots \dots \left(3\right)$

$Y$ component of force $F_2=2N$-

$F_{2Y}=2\sin 60°=\sqrt{3}\dots \dots \dots \dots \left(4\right)$

$X$ component of force $F_3=3N$-

$F_{3x}=-3\sin 30°=-\frac{3}{2}\dots \dots \dots \dots \left(5\right)$

$Y$ component of force $F_3=3N$-

$F_{3y}=-3\cos 30°=-\frac{3\sqrt{3}}{2}\dots \dots \dots \dots \left(6\right)$

Step 4: Calculate the net force in X and Y direction:

The net force in x-direction will be-

From equation (1), (3), and (5) will be-

$\begin{array}{rcl}{F}_X& =& F_{1x}+F_{2x}+F_{3x}\\ & =& 1-1-\frac{3}{2}\\ F_X& =& -\frac{3}{2}N\cdots \cdots \cdots \cdots \left(7\right)\end{array}$

The net force in y-direction will be-

$\begin{array}{rcl}{F}_y& =& F_{1y}+F_{2y}+F_{3y}\\ & =& 0+\sqrt{3}-\frac{3\sqrt{3}}{2}\\ F_y& =& -\frac{\sqrt{3}}{2}\cdots \cdots \cdots \cdots \left(8\right)\end{array}$

Step 5: Calculating the resultant:

The resultant will be-

From equation (7) and (8)-

$F=\sqrt{{F_x}^2+{F_y}^2}$

$$\begin{gathered} \mathrm{F}=\sqrt{{\left(-\frac{3}{2}\right)}^2+{\left(-\frac{\sqrt{3}}{2}\right)}^2} \\ F=\sqrt{\frac{9}{4}+\frac{3}{4}} \\ F=\sqrt{3}N \end{gathered}$$

Thus the magnitude of the resultant force is $\sqrt{3}\mathrm{N}$

Hence Option C is the correct answer.