Question

In Fig. 12-20, a stationary 5 kg rod ACis held against a wall by a rope and friction between rod and wall. The uniform rod is 1 m long, and angle

(a) If you are to find the magnitude of the force $\overrightarrow{\mathbf{T}}$
on the rod from the rope with a single equation, at what labeled point should a rotation axis be placed? With that choice of axis and counter-clockwise torques positive,

what is the sign of

(b) the torque${\mathit{\tau }}_{\mathbf{w}}$due to the rod’s weight and

(c) the torque${\mathit{\tau }}_{\mathbf{r}}$due to the pull on the rod the rope?

(d) Is the magnitude of ${\mathit{\tau }}_{\mathbf{r}}$greater than, less than, or equal to the magnitude of ${\mathit{\tau }}_{\mathbf{w}}$?

Short answer

a) The position of the rotation axis on the rod is point C

b) The sign of the torque${\tau }_w$ due to the rod’s weight is positive.

c) The sign of the torque${\tau }_r$ due to the pull on the rod by the rope is negative.

d) The magnitude of${\tau }_r$ is equal to the magnitude of.

Step-by-step solution

The given data

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

Understanding the concept of static equilibrium

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, ${\tau }_{net}=0$ (ii)

The torque acting on a point,$\tau =r\times F$ (iii)

a) Calculation of the position of the rotation axis

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.

b) Calculation of the sign of the torque

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 ${\tau }_w$will be taken as positive.

c) Calculation of the sign of the torque

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

Hence it is taken as negative.

d) Calculation of the magnitude of the torque

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

Hence the magnitude of ${\tau }_r$will be equal to the${\tau }_w$ .