Three forces act on a particle that moves with unchanging velocity $\overrightarrow{\mathbf{v}}\mathbf{=}\left(2m/s\right)\hat{\mathbf{i}}\mathbf{-}\left(7m/s\right)\hat{\mathbf{j}}$. Two of the forces arelocalid="1660906954720" ${\overrightarrow{\mathbf{F}}}_{\mathbf{1}}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{}\mathbf{N}\mathbf{)}\hat{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{3}\mathbf{}\mathbf{N}\mathbf{)}\hat{\mathbf{j}}\mathbf{+}\mathbf{(}\mathbf{‐}\mathbf{2}\mathbf{}\mathbf{N}\mathbf{)}\hat{\mathbf{k}}$and${\overrightarrow{\mathbf{F}}}_{\mathbf{2}}\mathbf{=}\mathbf{(}\mathbf{‐}\mathbf{5}\mathbf{}\mathbf{N}\mathbf{)}\hat{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{8}\mathbf{}\mathbf{N}\mathbf{)}\hat{\mathbf{j}}\mathbf{+}\mathbf{(}\mathbf{‐}\mathbf{2}\mathbf{}\mathbf{N}\mathbf{)}\hat{\mathbf{k}}$. What is the third force?

It is given that,

$$\begin{gathered} \overrightarrow{v}=2\hat{i}-7\hat{j} \\ {\overrightarrow{\mathrm{F}}}_1=2\hat{\mathrm{i}}+3\hat{\mathrm{j}}-2\hat{\mathrm{k}} \\ {\overrightarrow{\mathrm{F}}}_2=-5\hat{\mathrm{i}}+8\hat{\mathrm{j}}-2\hat{\mathrm{k}} \end{gathered}$$

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it.

Formula:

According to Newton’s second law,

$\sum _{}{F}_{net}=ma$

where,$F_{net}$ is the net force, mis mass and a is an acceleration.

If velocity is constant, acceleration would be zero.According to Newton’s second law, acceleration would be zero when net force is zero.

Thus,

$$\begin{gathered} \sum _{}{\overrightarrow{F}}_{net}=0 \\ {\overrightarrow{F}}_1+{\overrightarrow{F}}_2+{\overrightarrow{F}}_3=0 \end{gathered}$$

Consider,$F_3={\mathrm{F}}_{\mathrm{x}}\hat{\mathrm{i}}+{\mathrm{F}}_{\mathrm{y}}\hat{\mathrm{j}}+{\mathrm{F}}_2\hat{\mathrm{k}}$

Hence,

role="math" localid="1660907302155" $0=\left(2-5+F_x\right)\hat{i}+\left(3+8+F_z\right)\hat{k}$

Now, by comparing the coefficients of$\hat{\mathrm{i}},\hat{\mathrm{j}}\mathrm{and}\hat{\mathrm{k}}$on both sides,

$$\begin{gathered} F_x=3 \\ {F}_y=-11 \\ {F}_z=4 \end{gathered}$$

Hence, the unknown force $F_3$acting on the particle is$F_x=3\hat{\mathrm{i}}-11\hat{\mathrm{j}}+4\hat{\mathrm{k}}$