Question

A particle is located at the vector position $\overrightarrow{r}=\left(4.00\hat{i}+2.00\hat{j}\right)\text{m}$, and a force exerted on it is given by $\overrightarrow{\text{F}}=\left(3.00\hat{i}+2.00\hat{j}\right)\text{N}$. (a) What is the torque acting on the particle about the origin? (b) Can there be another point about which the torque caused by this force on this particle will be in the opposite direction and half as large in magnitude? (c) Can there be more than one such point? (d) Can such a point lie on the y axis? (e) Can more than one such point lie on the y axis? (f) Determine the position vector of one such point.

Short answer

(a) $\overrightarrow{\tau }=\left(-10.0\mathrm{Nm}\right)\hat{\mathrm{k}}$ is the required torque. (b) Yes, Because of it the lever arm of the force along this required new axis will be in the amount half as large and the force which was generating clockwise torque will generate counter clockwise torque. (c) Yes there are possibility of infinite number of such points, along a line that is passing through the described point b and is acting parallel the line of acting force. (d) Yes and this point will lie at the intersection of the line that is described in the point b and c and the y axis. (e) No, because all the intersection of lines will produce only one point of intersection as described in d with the y axis. (f) The position of the new vector axis is $5.00\hat{\mathrm{j}}\mathrm{m}$ .

Step-by-step solution

Torque

Torque can be described as a rotational equivalent of any object’s linear force. It can be expressed as the product of the distance or the position of any particle and force acting on it.

$$\begin{gathered} \overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F} \\ \overrightarrow{\tau }=torque \\ \overrightarrow{r}=positionoftheparticle \\ \overrightarrow{F}=force \end{gathered}$$

Given

$$\begin{gathered} \overrightarrow{r}=\left(4.00\hat{i}+2.00\hat{j}\right)\mathrm{m} \\ \overrightarrow{\mathrm{F}}=\left(3.00\hat{\mathrm{i}}+2.00\hat{\mathrm{j}}\right)\mathrm{N} \end{gathered}$$

(a) The torque acting on the particle about the origin

The torque that is acting on any particle about the origin can be expressed as

$$\begin{gathered} \overline{\tau }=\overrightarrow{r}\times \overrightarrow{F}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& 6& 0\\ 3& 2& 0\end{array}\right| \\ \overline{\tau }=\hat{i}\left(0-0\right)-\hat{j}\left(0-0\right)+\hat{k}\left(8-18\right) \\ \overline{\tau }=\left(-10.0\mathrm{Nm}\right)\mathrm{k} \end{gathered}$$

$\overline{\mathrm{\tau }}=\left(-10.0\mathrm{Nm}\right)\mathrm{k}$is the required torque.

(b) The possibility of another point

The possibility of another point is present here as the given point or axis should be present on another side of the presented action of force line and at half distance as far from this present line along which the force is acting. Because of it the lever arm of the force along this required new axis will be in the amount half as large and the force which was generating clockwise torque will generate counter clockwise torque.

(c) possibility of more than one such point

Yes there are possibility of infinite number of such points, along a line that is passing through the described point b and is acting parallel the line of acting force.

(d) Can such a point lie on the y axis?

Yes and this point will lie at the intersection of the line that is described in the point b and c and the y axis.

(e) Can more than one such point lie on the y axis?

No, because all the intersection of lines will produce only one point of intersection as described in d with the y axis.

(f) Determine the position vector of one such point

We will consider any such point $\left(0,y\right)$representing the requires coordinate system of this given special axis of rotation that is situated at the y axis of Cartesian coordinates. Now the displacement of the particle from this point feeling the force is ${\overrightarrow{r}}_{new}=\left(4.00\hat{i}+\left(6.00-y\right)\hat{j}\right)\text{m and force}\overrightarrow{F}=\left(3.00\hat{i}+2.00\hat{j}\right)\text{N}$

The torque

$$\begin{gathered} \overrightarrow{{\tau }_{new}}=\overrightarrow{{\tau }_{new}}\times \overrightarrow{F}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& 6-y& 0\\ 3& 2& 0\end{array}\right| \\ \overrightarrow{\tau }=\hat{i}\left(0-0\right)-\hat{j}\left(0-0\right)+\hat{k}\left(8-18+3y\right)=\left(+5.0\text{N-m}\right)k \end{gathered}$$

Now,

$\begin{array}{rcl}\left(8-18+3y\right)& =& \left(+5.0\text{N-m}\right)\\ y& =& 5\end{array}$

So the position of the new vector axis is$5.00\hat{j}m$.