Question

Two forces act on a particle simultaneously as shown in the figure. Find net force in milli new on the particle. [Dyne is the CGS unit of force]

Answer 1

sqrt(3)

Answer 2

The two forces acting on the particle are $F_1 = 100 \text{ dyne}$ and $F_2 = 1.0 \text{ milli newton}$. The angle between the forces is $\theta = 60^\circ$.

First, we convert the force $F_1$ from dyne to milli newton. The relationship between dyne and newton is $1 \text{ newton} = 10^5 \text{ dyne}$. So, $1 \text{ dyne} = 10^{-5} \text{ newton}$. $F_1 = 100 \text{ dyne} = 100 \times 10^{-5} \text{ newton} = 10^{-3} \text{ newton}$. The unit milli newton is $1 \text{ milli newton} = 10^{-3} \text{ newton}$. Therefore, $F_1 = 10^{-3} \text{ newton} = 1 \text{ milli newton}$. The second force is given as $F_2 = 1.0 \text{ milli newton}$. So, we have two forces of equal magnitude, $F_1 = 1 \text{ mN}$ and $F_2 = 1 \text{ mN}$, acting at an angle of $\theta = 60^\circ$.

To find the net force (resultant force), we use the parallelogram law of vector addition. The magnitude of the resultant force $R$ of two forces $F_1$ and $F_2$ acting at an angle $\theta$ is given by: $R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos\theta}$ Substituting the values: $R = \sqrt{(1 \text{ mN})^2 + (1 \text{ mN})^2 + 2(1 \text{ mN})(1 \text{ mN})\cos(60^\circ)}$ $R = \sqrt{1^2 + 1^2 + 2(1)(1)\cos(60^\circ)} \text{ mN}$ We know that $\cos(60^\circ) = \frac{1}{2}$. $R = \sqrt{1 + 1 + 2(1)(1)\left(\frac{1}{2}\right)} \text{ mN}$ $R = \sqrt{1 + 1 + 1} \text{ mN}$ $R = \sqrt{3} \text{ milli newton}$.

The net force on the particle is $\sqrt{3}$ milli newton.