Question

$\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathit{k}\mathit{g}$ the object is subjected to three forces that give it an acceleration $\overrightarrow{\mathbf{a}}\mathbf{=}\mathbf{-}\left(8.00m/s^2\right)\hat{\mathbf{i}}\mathbf{+}\left(6.00m/s^2\right)\hat{\mathbf{j}}$ . If two of the three forces are role="math" localid="1656998692451" $\overrightarrow{{\mathbf{F}}_{\mathbf{1}}}\mathbf{=}\mathbf{(}\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{N}\mathbf{)}\hat{\mathit{i}}\mathbf{+}\mathbf{(}\mathbf{16}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{N}\mathbf{)}\mathbf{}\hat{\mathit{j}}$ and$\overrightarrow{{\mathbf{F}}_{\mathbf{2}}}\mathbf{=}\mathbf{-}\left(12.0N\right)\mathbf{}\hat{\mathbf{i}}\mathbf{+}\left(8.0N\right)\mathbf{}\hat{\mathbf{j}}$ find the third force.

Short answer

The third force is $\left(-36\mathrm{N}\right)\hat{i}+\left(-12\mathrm{N}\right)\hat{j}$

Step-by-step solution

The given data

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

Understanding the concept of Newton’s second law

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 ${\overrightarrow{\mathit{F}}}_{\mathit{N}\mathit{e}\mathit{t}}$ on a body with a mass m and the body’s acceleration as,

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

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

Calculation of third force

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

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

Substitute the values of mass, acceleration, and two forces to calculate the third force.

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

Hence, the third force is$\left(-34\mathrm{N}\right)\hat{i}+\left(-12\mathrm{N}\right)\hat{j}$ .