Question

Find the magnitude of two forces such that, if they act at right angle, their resultant is √10N and when they act at an angle 60° their resultant is √13

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{two}\mathrm{forces}\mathrm{be}\mathrm{P}\mathrm{and}\mathrm{Q},\mathrm{when}\mathrm{they}\mathrm{act}\mathrm{at}\mathrm{right}\mathrm{angles}\mathrm{to}\mathrm{each}\mathrm{other},\mathrm{then} \\ \mathrm{R}=\sqrt{{\mathrm{P}}^2+{\mathrm{Q}}^2} \\ {\mathrm{R}}^2={\mathrm{P}}^2+{\mathrm{Q}}^2 \\ 10={\mathrm{P}}^2+{\mathrm{Q}}^2...............\left(1\right) \\ \mathrm{In}\mathrm{the}\mathrm{second}\mathrm{case},\mathrm{when}\mathrm{forces}\mathrm{act}\mathrm{at}\mathrm{an}\mathrm{angle}\mathrm{of}60°\mathrm{then}\mathrm{the}\mathrm{resultant}, \\ \mathrm{R}=\sqrt{{\mathrm{P}}^2+{\mathrm{Q}}^2+2\mathrm{PQ}\cos 60°} \\ {\mathrm{R}}^2={\mathrm{P}}^2+{\mathrm{Q}}^2+2\mathrm{PQ}\cos 60° \\ 13={\mathrm{P}}^2+{\mathrm{Q}}^2+2\mathrm{PQ}\times \frac{1}{2} \\ 13={\mathrm{P}}^2+{\mathrm{Q}}^2+\mathrm{PQ}.........\left(2\right) \\ \mathrm{from}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}, \\ 13=10+\mathrm{PQ} \\ \mathrm{PQ}=3.......\left(3\right) \\ \mathrm{Multiply}\mathrm{equation}\left(3\right)\mathrm{by}2\mathrm{and}\mathrm{add}\mathrm{to}\mathrm{equation}\left(1\right), \\ 10+6={\mathrm{P}}^2+{\mathrm{Q}}^2+2\mathrm{PQ} \\ 16={\mathrm{P}}^2+{\mathrm{Q}}^2+2\mathrm{PQ} \\ 16={\left(\mathrm{P}+\mathrm{Q}\right)}^2 \\ \mathrm{P}+\mathrm{Q}=\sqrt{16} \\ \mathrm{P}+\mathrm{Q}=4..........\left(4\right) \\ \mathrm{from}\mathrm{equation}\left(3\right)\mathrm{and}\left(4\right),\mathrm{we}\mathrm{get}, \\ \mathrm{P}\left(4-\mathrm{P}\right)=3 \\ {\mathrm{P}}^2-4\mathrm{P}+3=0 \\ {\mathrm{P}}^2-3\mathrm{P}-\mathrm{P}+3=0 \\ \mathrm{P}\left(\mathrm{P}-3\right)-1\left(\mathrm{P}-3\right)=0 \\ \left(\mathrm{P}-1\right)\left(\mathrm{P}-3\right)=0 \\ \mathrm{P}=1\mathrm{or}\mathrm{P}=3 \\ \mathrm{If}\mathrm{P}=1\mathrm{then}\mathrm{Q}=3 \\ \mathrm{If}\mathrm{P}=3\mathrm{then}\mathrm{Q}=1 \end{gathered}$$