Question

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?

Short answer

The unknown force ${\mathrm{F}}_3$acting on the particle is${\overrightarrow{\mathrm{F}}}_3=3\hat{\mathrm{i}}-11\hat{\mathrm{j}}+4\hat{\mathrm{k}}$

Step-by-step solution

Given information

It is given that,

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

Determining the concept

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.

Determining the unknown force F3 acting on the particle

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$F_x=3\hat{\mathrm{i}}-11\hat{\mathrm{j}}+4\hat{\mathrm{k}}$