Question

One force acting on a machine part is$\overrightarrow{\mathbf{F}}\mathbf{=}\left(-5.00 N\right)\mathbf{ }\hat{\mathbf{i}}\mathbf{+}\left(4.00 N\right)\hat{\mathbf{j}}$The vector from the origin to the point where the force is applied is $\overrightarrow{\mathbf{r}}\mathbf{=}\left(-0.450 m\right)\mathbf{ }\hat{\mathbf{i}}\mathbf{+}\left(0.150 m\right)\hat{\mathbf{j}}$.(a) In a sketchshow $\overrightarrow{\mathbf{r}}\mathbf{,}\overrightarrow{\mathbf{F}\mathbf{,}}$, and the origin. (b) Use right-hand rule to determine the direction of the torque.(c) Calculate the vector torque for an axis at the origin produced by this force. Verify that the direction of the torque is the same as you obtained in part (b).

Short answer

(a) The sketch of the position vector $\overrightarrow{r}$, the force $\overrightarrow{F}$and the origin is shown in the graph.

(b) The direction of the torque$\overrightarrow{\tau }$is into the page.

(c) The vector torque for an axis at the origin is, $\overrightarrow{\tau }=\left(-1.05\right) \hat{k} N·m$and

the torque vector is directed in the -z-direction.

Step-by-step solution

To mention the given data

We have the given data:

The force is given by,

$\overrightarrow{F}=\left(-5.00 N\right)\hat{i}+\left(4.00 N\right)\hat{j}$.

The position vector from the origin to the point is,

$\overrightarrow{r}=\left(-0.450 m\right)\hat{i}+\left(0.150 N\right)\hat{j}$.

Step 2:Concept

When a force acts on a body, the torque $\tau$of that force with respect to the point $O$is equal to the vector product of the position vector $\overrightarrow{r}$and the force

$\overrightarrow{F}$.

$\therefore \overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F} \cdots \cdots \left(1\right)$

The cross product of any two vectors $\overrightarrow{A}, \overrightarrow{B}$is given by,

localid="1667991221155" $$\begin{gathered} \overrightarrow{A}\times \overrightarrow{B}=\left|\hat{i} \hat{j} \hat{k} \\ {A}_x A_{y } A_z \\ {B}_x B_{y } B_z\right| \end{gathered}$$

$\overrightarrow{A}\times \overrightarrow{B}=\left(A_{y }{B}_z-A_{z }{B}_y\right)\hat{i}-\left(A_{x }{B}_z-A_{z }{B}_x\right)\hat{j}+\left(A_{x }{B}_y-A_{y }{B}_x\right)\hat{k} \cdots \cdots \left(2\right)$

(a)To show in a graph position vector, force and origin

The sketch of the position vector $\overrightarrow{r}$, the force $\overrightarrow{F}$and the origin is shown below:

(b)To determine the direction of torque

According to the right-hand rule, if you curl your fingers of the right hand from the direction of the position vector $\overrightarrow{r}$into the direction of force$\overrightarrow{F}$, then your right thumb point into the page.

That is nothing but the direction of the torque $\overrightarrow{\tau }$.

(c)To calculate the vector torque for an axis at the origin

The torque $\overrightarrow{\tau }$due to the force $\overrightarrow{F}$about the origin is given by from $\left(1\right)$,

$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$

Now, using the vector product of any two vectors from $\left(2\right)$and substituting the values, we get,

localid="1667991366200" $$\begin{gathered} \overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F} \\ =\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -0.450& 0.150& 0\\ -5.00& 4.00& 0\end{array}\right| \\ =\left(\left(0.150\right)·0-0·\left(4.00\right)\right)\hat{i}-\left(\left(-0.450\right)·0-0·\left(-5.00\right)\right)\hat{j} \\ +\left(\left(-0.450\right)·\left(4.00\right)-\left(0.150\right)·\left(-5.00\right)\right)\hat{k} \\ =\left(-1.05\right) \hat{k} N·m \\ \therefore \overrightarrow{\tau }=\left(-1.05\right) \hat{k} N·m \end{gathered}$$

Hence, the vector torque for an axis at the origin is, $\overrightarrow{\tau }=\left(-1.05\right) \hat{k} N·m$.

Since the value is negative, the torque vector is directed in the -z-direction or in the direction that we obtained in part (b).