A body of mass $2 \mathrm{~kg}$ is acted upon by two forces each of magnitude $1 \mathrm{~N}$ and inclined…
A body of mass $2 \mathrm{~kg}$ is acted upon by two forces each of magnitude $1 \mathrm{~N}$ and inclined at $60^{\circ}$ with each other. The acceleration of the body in $\frac{\mathrm{m}}{\mathrm{s}}$ is $\left[\cos 60^{\circ}=0.5\right]$
$\sqrt{0.35}$
$\sqrt{0.65}$
$\sqrt{0.75}$
$\sqrt{0.20}$
Solution
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 $\frac{\sqrt{3}}{2} \mathrm{~m} / \mathrm{s}^{2}$
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