Physics - Vectors Question with Solution | TestHub

Let us apply the parallelogram law of vectors to find the resultant of the forces. ${\mathrm{F}}_{\text{net }}=\sqrt{{\mathrm{F}}_1^2+{\mathrm{F}}_2^2+2\times {\mathrm{F}}_1{\mathrm{F}}_2\cos \theta }$ Given: ${\mathrm{F}}_1=1,{\mathrm{F}}_2=1$ and $\theta =60^{\circ }$ So, ${\mathrm{F}}_{\text{net }}=\sqrt{(1{)}^2+(1{)}^2+2\times 1\times 1\times \cos 60^0}=\sqrt{1+1+2\times 1/2}=\sqrt{3}\mathrm{N}$ Now $\mathrm{F}=\mathrm{ma}$ or, $a=\frac{F}{m}=\frac{\sqrt{3}}{2}m/s^2$ Thus, the net acceleration of the body is $\frac{\sqrt{3}}{2}\mathrm{m}/{\mathrm{s}}^2$