91Ó°ÊÓ

The rod AC is in equilibrium and is held by a rope and friction with the wall.

We use the concept of static equilibrium i.e., balanced torques and balanced forces to determine the magnitudes of the various torques acting on the rod.

Formulae:

The value of the net force at equilibrium, $F_{net}=0$ (i)

The value of the torque at equilibrium, ${\mathrm{Ï„}}_{net}=0$ (ii)

The torque acting on a point,$\mathrm{τ}=r×F$ (iii)

The rod AC is in equilibrium; that is, the condition of equations (i) and (ii) is satisfied, so the torques and the forces acting on it are balanced.

The forces acting on it are tension from the rope, friction from the wall, and the weight of the rod. All these forces are balanced.

If we consider the pivot point at C, it will eliminate the effect of frictional forces. So, the equation of torque will just be composed of component tension in the string that is perpendicular to the rod and weight. The parallel component of the tension will also apply zero torque since it is directed along the rod.

Hence, the position of the rotation axis is at point C.

We consider the counter-clockwise rotation by the torque as positive and the clockwise rotation as negative. With the rotation axis at point C, the weight will rotate the rod in a counter-clockwise direction considering the condition of equation (iii).

Hence the torque ${\mathrm{Ï„}}_w$will be taken as positive.

As we have seen in part (b), the rotation due to torque${\mathrm{Ï„}}_r$will be clockwise.

Hence it is taken as negative.

The torque equation, about point C, will be${\mathrm{Ï„}}_w-{\mathrm{Ï„}}_r=0$as the torque due to other forces is zero.

Hence the magnitude of ${\mathrm{Ï„}}_r$will be equal to the${\mathrm{Ï„}}_w$ .