Two forces, while acting on a particle in opposite; directions, have the resultant of $10N$. If they act at right angles to each other, the resultant is found to be $50N$. Find the two forces.

Step 1: Given data

From the figure,

The value of resultant is $R$=$10N$ for $\theta =180°$.

The value of resultant is $R$=$50N$ for $\theta =90°$.

Step 2: Formula used

$\mathit{R}\mathbf{}\mathbf{=}\sqrt{{{\mathbf{F}}_{\mathbf{1}}}^{\mathbf{2}}\mathbf{+}{{\mathbf{F}}_{\mathbf{2}}}^{\mathbf{2}}\mathbf{+}\mathbf{2}{\mathbf{F}}_{\mathbf{1}}{\mathbf{F}}_{\mathbf{2}}\mathbf{C}\mathbf{o}\mathbf{s}\mathbf{\theta }}$

Where,

$R_1$= Resultant force.

$F_{1}and{F}_2$are the magnitudes of two forces.

Step 3: Resultant force when the forces are opposite.

Replace the values in the formula,

$$\begin{gathered} {\mathit{R}}_{\mathbf{1}}\mathbf{}\mathbf{=}\sqrt{{{\mathbf{F}}_{\mathbf{1}}}^{\mathbf{2}}\mathbf{+}{{\mathbf{F}}_{\mathbf{2}}}^{\mathbf{2}}\mathbf{+}\mathbf{2}{\mathbf{F}}_{\mathbf{1}}{\mathbf{F}}_{\mathbf{2}}\mathbf{C}\mathbf{o}\mathbf{s}\mathbf{\theta }} \\ \mathbf{}10=\sqrt{{{\mathrm{F}}_1}^2+{{\mathrm{F}}_2}^2+2{\mathrm{F}}_1{\mathrm{F}}_2\mathrm{Cos}{180}^{\circ }} \\ 10=\sqrt{{{\mathrm{F}}_1}^2+{{\mathrm{F}}_2}^2-2{\mathrm{F}}_1{\mathrm{F}}_2} \\ 10={\mathrm{F}}_1-{\mathrm{F}}_2 \\ {F}_{1=}10+F_2................\left(1\right) \end{gathered}$$

Step 4: Resultant force when the forces are at the right angle.

$R_2$=$50N$

$\theta =90°$

Replace the values in the formula:

$$\begin{gathered} {\mathit{R}}_{\mathbf{2}}\mathbf{}\mathbf{=}\sqrt{{{\mathbf{F}}_{\mathbf{1}}}^{\mathbf{2}}\mathbf{+}{{\mathbf{F}}_{\mathbf{2}}}^{\mathbf{2}}\mathbf{+}\mathbf{2}{\mathbf{F}}_{\mathbf{1}}{\mathbf{F}}_{\mathbf{2}}\mathbf{C}\mathbf{o}\mathbf{s}\mathbf{\theta }} \\ \mathbf{}\mathbf{5}0=\sqrt{{{\mathrm{F}}_1}^2+{{\mathrm{F}}_2}^2+2{\mathrm{F}}_1{\mathrm{F}}_2\mathrm{Cos}{90}^{\circ }} \\ 50=\sqrt{{{\mathrm{F}}_1}^2+{{\mathrm{F}}_2}^2}................\left(2\right) \end{gathered}$$

Step 5: Solve the two equation

From equation(1) we got,

$F_1=10+F_2$ substitute this value in the above equation (2).

$\begin{array}{rcl}{50}^2& =& {F_2}^2+{\left(10+F_2\right)}^2\\ & \Rightarrow & {F_2}^2+{F_2}^2+100+20{F_2}^{}=2500\\ & \Rightarrow & 2{F_2}^2+20{F_2}^{}-2400=0\\ & \Rightarrow & {F_2}^2+40F_2-30F_2-1200=0\\ & \Rightarrow & F_2\left(F_2+40\right)-30\left(F_2+40\right)=0\\ & \Rightarrow & \left(F_2+40\right)\left(F_2-30\right)=0\\ & \Rightarrow & F_2=30N\end{array}$

Force cannot be negative so $F_2\ne -40$

Step 6: Find $F_1$

Substitute $F_2$ value in equation(1).

$\begin{array}{rcl}{F}_1& =& 10+F_2\\ & & =10+30\\ & & =40N\end{array}$

Therefore the two forces are $\mathbf{40}\mathit{N}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathbf{30}\mathit{N}$.