91Ó°ÊÓ

• Object of mass,$m=2.00kg$.
• Acceleration of object, $\stackrel{→}{a}=\left(-8.00\mathrm{m}/{\mathrm{s}}^2\right)\hat{i}+\left(6.00\mathrm{m}/{\mathrm{s}}^2\right)\hat{j}$.
• Two forces are given,localid="1657001038844" $\stackrel{→}{F_1}=\left(30\mathrm{N}\right)\hat{i}+\left(16\mathrm{N}\right)\hat{j}\&\stackrel{→}{F_2}=\left(-12.0\mathrm{N}\right)\hat{i}+\left(8.00\mathrm{N}\right)\hat{j}$.

If there is more than one force acting on the body, then the net force is found by vector addition of the object. Newton’s law gives the relation between the net force ${\stackrel{\mathbf{→}}{\mathit{F}}}_{\mathit{N}\mathit{e}\mathit{t}}$ on a body with a mass m and the body’s acceleration as,

${\stackrel{\mathbf{→}}{\mathit{F}}}_{\mathit{N}\mathit{e}\mathit{t}}\mathbf{=}\mathit{m}\stackrel{\mathbf{→}}{\mathit{a}}$ (i)

Using the formula for Newton’s second law, we can find the third force.

From equation (i), we have the net force as follows:

$$\begin{gathered} {\stackrel{→}{F}}_{Net}=\stackrel{→}{F_1}+\stackrel{→}{F_2}+\stackrel{→}{F_3} \\ m\stackrel{→}{a}=\stackrel{→}{F_1}+\stackrel{→}{F_2}+\stackrel{→}{F_3} \\ \stackrel{→}{F_3}=m\stackrel{→}{a}-\stackrel{→}{F_1}-\stackrel{→}{F_2} \end{gathered}$$

$$\begin{gathered} \stackrel{→}{F_3}=2.00kg\left[\left(-8.00\mathrm{m}/{\mathrm{s}}^2\right)\hat{i}+\left(6.00\mathrm{m}/\mathrm{s}\right)\hat{j}\right]-\left[\left(30\mathrm{N}\right)\hat{i}+\left(16\mathrm{N}\right)\hat{j}\right]-\left[\left(-12.0\mathrm{N}\right)\hat{i}+\left(8.00\mathrm{N}\right)\hat{j}\right] \\ =\left(-34\mathrm{N}\right)\hat{i}+\left(-12\mathrm{N}\right)\hat{j} \end{gathered}$$