Question

Question: A particle moves through an xyz coordinate system while a force acts on the particle. When the particle has the position vector $\overrightarrow{\mathbf{r}}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\text{m}}\mathbf{)}\widehat{\mathbf{\text{i}}}\mathbf{-}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{\text{m}}\mathbf{)}\widehat{\mathbf{\text{j}}}\mathbf{+}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\text{m}}\mathbf{)}\widehat{\mathbf{\text{k}}}$,the force is given by $\overrightarrow{\mathbf{F}}\mathbf{=}{\mathbf{F}}_{\mathbf{x}}\widehat{\mathbf{\text{i}}}\mathbf{+}\mathbf{(}\mathbf{7}\mathbf{.}\mathbf{00}\mathbf{\text{N}}\mathbf{)}\widehat{\mathbf{\text{j}}}\mathbf{-}\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{\text{N}}\mathbf{)}\widehat{\mathbf{\text{k}}}$and the corresponding torque about the origin is$\overrightarrow{\mathbf{\tau }}\mathbf{=}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{\text{Nm}}\mathbf{)}\widehat{\mathbf{\text{i}}}\mathbf{+}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\text{Nm}}\mathbf{)}\widehat{\mathbf{\text{j}}}\mathbf{-}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\text{Nm}}\mathbf{)}\widehat{\mathbf{\text{k}}}$. Determine.

Short answer

Answer

Force component along x-axis is- 5.00N

Step-by-step solution

Identification of given data 

$$\begin{gathered} \overrightarrow{r}=2\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}} \\ \overrightarrow{F}=F_x\hat{\text{i}}+7\hat{\text{j}}-6\hat{\text{k}} \\ \overrightarrow{\tau }=\left(4.0\text{Nm}\right)\hat{\text{i}}+\left(2.0\text{Nm}\right)\hat{\text{j}}-\left(1.00\text{Nm}\right)\hat{\text{k}} \end{gathered}$$

To understand the concept

With the help of vector cross product, find the Force component along X-axis.

Write the equations for the cross product of radius vector and force and equate that cross product with the given torque. Now find the value of unknown component of the force.

Formulae:

$$\begin{gathered} \overrightarrow{r}=x\text{i}+y\text{j}+z\text{k} \\ \overrightarrow{F}=F_x\text{i}+F_y\text{j}+F_z\text{k} \\ \overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}=\left|\begin{array}{ccc}i& j& k\\ x& y& z\\ F_x& F_y& F_z\end{array}\right| \\ =\left(yFz-z{F}_y\right)\hat{\text{i}}+\left(z{F}_x-x{F}_x\right)\hat{\text{j}}+\left(x{F}_y-y{F}_x\right)\hat{\text{k}} \end{gathered}$$

Determine Fx

$$\begin{gathered} \overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}=\left|\begin{array}{ccc}i& j& k\\ 2& -3& 2\\ F_x& 7& -6\end{array}\right|=\left(18-14\right)\hat{\text{i}}+\left(2F_x+12\right)\text{j}\hat{}+\left(14+3F_x\right)\hat{\text{k}} \\ \Rightarrow \overrightarrow{\tau }=\left(4\right)\hat{\text{i}}+\left(2F_x+12\right)\hat{\text{j}}+\left(14+3F_x\right)\hat{\text{k}} \end{gathered}$$

Now, this will compare with given torque vector.

$\Rightarrow \overrightarrow{\tau }=\left(4.0\text{Nm}\right)\text{i}+\left(2.0\text{Nm}\right)\text{j}-\left(1.00\text{Nm}\right)\text{k}$

Comparing,

$\left(2.0\text{Nm}\right)\text{j}=\left(2F_{\text{x}}+12\right)\text{j}$

And,

$-\left(1.00\text{Nm}\right)\hat{\text{k}}=\left(14+3F_x\right)\tilde{\text{k}}$

Solving both of these two equations,

$F_x=-5.00\text{N}$