Question

Figure 11-24 shows two particles Aand Bat xyzcoordinates (1 m, 1 m, 0) and (1 m, 0, 1 m). Acting on each particle are three numbered forces, all of the same magnitude and each directed parallel to an axis. (a) Which of the forces produce a torque about the origin that is directed parallel to y? (b) Rank the forces according to the magnitudes of the torques they produce on the particles about the origin, greatest first.

Short answer

a) Forces 5 and 6 producea torque about the origin that is directed parallel to y.

b) Ranking of forces according to the magnitudes of the torques is ${\tau }_1={\tau }_4>{\tau }_2={\tau }_3={\tau }_5={\tau }_6$,

Step-by-step solution

Step 1: Given

The co-ordinate ofA is (1 m, 1 m, 0 m) and B is (1 m, 0 m, 1 m).

Determining the concept

Using the right-hand rule for torque find the forces which produce torque pallet to y. Also, rank the torques by calculating the moment arm between origin and these forces.

Fleming’s Left Hand Rule states that if we arrange our thumb, forefinger, and middle finger of the left-hand perpendicular to each other, then the thumb points towards the direction of the magnetic force, the forefinger points towards the direction of the magnetic field, and the middle finger points towards the direction of the current.

Formulae are as follows:

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

Where, r is radius ,F is force and$\tau$ is torque.

(a) Determining the forces which produce a torque about the origin that is directed parallel to y.

Write given points in vector form as

For A $\overrightarrow{r_1}=1i+1j+0k$and for B$\overrightarrow{r_2}=1i+0j+1k$

Now, calculate for torque produced along y-direction:

To find the direction of torquemoves the force vector at the origin without changing its direction.Theright-hand ruledetermines that torques due to forces 5 and 6 are directed along the y-axis.

(b) Determining the ranking of forces according to the magnitudes of the torques.

Now, calculatethe magnitude of torques:

$\left|\overrightarrow{{\tau }_1}\right|=|{\overrightarrow{r}}_1\times \overrightarrow{F}|$

Or

$\left|{\overrightarrow{\tau }}_2\right|=|\overrightarrow{r_2}\times \overrightarrow{F}|$

The angle between force 1 and position vectoris${\tau }_1$and the angle between force 4 and position vector$r_2$is 90⁰.

$\text{Therefore,themomentarmbetweenoriginandtheseforcesis\hspace{0.17em}}\sqrt{\text{2}}\text{\hspace{0.17em}m.}$

All forces are of the same magnitude.

From this it can be concluded that,

${\tau }_1={\tau }_4>{\tau }_2={\tau }_3={\tau }_5={\tau }_6$

Therefore, we can find the torques using the formula for torque.