Question

Two forces are acting at an angle 120°. Bigger force is of 8N. Resultant acts at right angles to the smaller force. Find the value of smaller force in newton.

Answer 1

4N

Answer 2

Let the smaller force be $F_1$ and the bigger force be $F_2$. Given:

1. Bigger force $F_2 = 8 \, \text{N}$.
2. The angle between the two forces is $\theta = 120^\circ$.
3. The resultant force $R$ acts at right angles to the smaller force $F_1$.

We can solve this problem using vector components or by constructing a vector triangle.

Method 1: Using Vector Components Let the smaller force $F_1$ lie along the positive x-axis. So, $\vec{F_1} = F_1 \hat{i}$.

The bigger force $F_2$ makes an angle of $120^\circ$ with $F_1$. The components of $\vec{F_2}$ are: $F_{2x} = F_2 \cos \theta = 8 \cos(120^\circ) = 8 \left(-\frac{1}{2}\right) = -4 \, \text{N}$ $F_{2y} = F_2 \sin \theta = 8 \sin(120^\circ) = 8 \left(\frac{\sqrt{3}}{2}\right) = 4\sqrt{3} \, \text{N}$ So, $\vec{F_2} = -4 \hat{i} + 4\sqrt{3} \hat{j}$.

The resultant force $\vec{R}$ is the vector sum of $\vec{F_1}$ and $\vec{F_2}$: $\vec{R} = \vec{F_1} + \vec{F_2} = (F_1 \hat{i}) + (-4 \hat{i} + 4\sqrt{3} \hat{j})$ $\vec{R} = (F_1 - 4) \hat{i} + 4\sqrt{3} \hat{j}$.

The problem states that the resultant force $\vec{R}$ acts at right angles to the smaller force $F_1$. Since $F_1$ is along the x-axis, $\vec{R}$ must be along the y-axis (i.e., its x-component must be zero). Therefore, the x-component of $\vec{R}$ must be zero: $F_1 - 4 = 0$ $F_1 = 4 \, \text{N}$.

Method 2: Using a Vector Triangle (Graphical Method)

1. Draw the smaller force $\vec{F_1}$ along the x-axis, starting from the origin O and ending at point A. So, $OA = F_1$.
2. The resultant force $\vec{R}$ is perpendicular to $\vec{F_1}$. So, draw $\vec{R}$ along the y-axis, starting from the origin O and ending at point C. So, $OC = R$.
3. According to the triangle law of vector addition, $\vec{F_1} + \vec{F_2} = \vec{R}$. This implies $\vec{F_2} = \vec{R} - \vec{F_1} = \vec{OC} - \vec{OA} = \vec{AC}$. So, the third side of the right-angled triangle OAC represents $\vec{F_2}$. Thus, $AC = F_2 = 8 \, \text{N}$.
4. The triangle OAC is a right-angled triangle with the right angle at O ($\angle AOC = 90^\circ$).
5. The angle between $\vec{F_1}$ and $\vec{F_2}$ (when placed tail-to-tail) is $120^\circ$. In our triangle, $\vec{F_1}$ is $\vec{OA}$ and $\vec{F_2}$ is $\vec{AC}$. The angle $\angle OAC$ is the angle between $\vec{AO}$ (which is in the opposite direction of $\vec{F_1}$) and $\vec{AC}$ ($\vec{F_2}$). Therefore, the angle between $\vec{F_1}$ and $\vec{F_2}$ is $180^\circ - \angle OAC$. Given this angle is $120^\circ$: $120^\circ = 180^\circ - \angle OAC$ $\angle OAC = 180^\circ - 120^\circ = 60^\circ$.
6. Now, in the right-angled triangle OAC: $OA = F_1$ (adjacent side to $\angle OAC$) $AC = F_2 = 8 \, \text{N}$ (hypotenuse) Using the cosine function: $\cos(\angle OAC) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{OA}{AC}$ $\cos(60^\circ) = \frac{F_1}{8}$ $\frac{1}{2} = \frac{F_1}{8}$ $F_1 = 8 \times \frac{1}{2}$ $F_1 = 4 \, \text{N}$.

Both methods yield the same result.

Let the smaller force be $F_1$ and the bigger force be $F_2 = 8 \, \text{N}$. The angle between them is $120^\circ$. The resultant force $R$ is perpendicular to $F_1$. Using the formula for the direction of the resultant, if the resultant $R$ makes an angle $\alpha$ with $F_1$: $\tan \alpha = \frac{F_2 \sin \theta}{F_1 + F_2 \cos \theta}$ Given $\alpha = 90^\circ$, $\tan 90^\circ$ is undefined, which means the denominator must be zero: $F_1 + F_2 \cos \theta = 0$ $F_1 = -F_2 \cos \theta$ Substitute $F_2 = 8 \, \text{N}$ and $\theta = 120^\circ$: $F_1 = -8 \cos(120^\circ)$ $F_1 = -8 \left(-\frac{1}{2}\right)$ $F_1 = 4 \, \text{N}$.