Question

A uniform rod OA of mass 10 kg is pivoted at O on a vertical wall and kept horizontal with the help of a light cable AB. Find the tension in the cable and reaction force applied by the pivot. Linear dimension of pivot can be neglected in comparison with the length of the rod.

Answer 1

Tension in the cable = 100 N Reaction force applied by the pivot = 100 N

Answer 2

To solve this problem, we need to apply the conditions for static equilibrium of a rigid body:

1. The net force in any direction must be zero ($\Sigma F_x = 0$, $\Sigma F_y = 0$).
2. The net torque about any point must be zero ($\Sigma \tau = 0$).

1. Free Body Diagram (FBD):

Let the length of the uniform rod OA be $L$.

• Weight of the rod (W): Since the rod is uniform, its weight acts at its center of mass, which is at the midpoint of OA (distance $L/2$ from O).
  $W = Mg = 10 \, \text{kg} \times 10 \, \text{m/s}^2 = 100 \, \text{N}$ (assuming $g=10 \, \text{m/s}^2$). This force acts vertically downwards.
• Tension in the cable (T): The cable AB pulls the rod at point A. The cable makes an angle of 30° with the horizontal rod.
  • Horizontal component: $T_x = T \cos 30^\circ$ (acting to the left).
  • Vertical component: $T_y = T \sin 30^\circ$ (acting upwards).
• Reaction force at the pivot (R): The pivot at O can exert both horizontal and vertical reaction forces. Let these be $R_x$ (horizontal) and $R_y$ (vertical). We assume $R_x$ acts to the right and $R_y$ acts upwards.

2. Rotational Equilibrium (Torque about O):

To find the tension T, it's convenient to take torques about the pivot point O. This eliminates the unknown reaction forces $R_x$ and $R_y$ from the torque equation, as they pass through O and thus produce no torque about O.

• Torque due to weight (W): This force acts downwards at $L/2$ from O, creating a clockwise torque.
  $\tau_W = W \times (L/2)$
• Torque due to tension (T): The vertical component of tension, $T \sin 30^\circ$, acts upwards at point A (distance $L$ from O), creating a counter-clockwise torque. The horizontal component $T \cos 30^\circ$ passes through O if extended vertically, so it does not produce torque about O.
  $\tau_T = (T \sin 30^\circ) \times L$

For equilibrium, the net torque is zero:

$\Sigma \tau_O = 0$
$\tau_T - \tau_W = 0$
$(T \sin 30^\circ) \times L = W \times (L/2)$
$T \times (1/2) \times L = W \times (L/2)$
$T \times (1/2) = W \times (1/2)$
$T = W$

Substitute the value of W:

$T = 100 \, \text{N}$

3. Translational Equilibrium:

Now, we apply the force equilibrium conditions to find the reaction forces at the pivot.

• Horizontal forces ($\Sigma F_x = 0$):
  $R_x - T \cos 30^\circ = 0$
  $R_x = T \cos 30^\circ$
  $R_x = 100 \, \text{N} \times (\sqrt{3}/2)$
  $R_x = 50\sqrt{3} \, \text{N}$

• Vertical forces ($\Sigma F_y = 0$):
  $R_y + T \sin 30^\circ - W = 0$
  $R_y = W - T \sin 30^\circ$
  $R_y = 100 \, \text{N} - 100 \, \text{N} \times (1/2)$
  $R_y = 100 \, \text{N} - 50 \, \text{N}$
  $R_y = 50 \, \text{N}$

4. Total Reaction Force at Pivot:

The total reaction force $R$ is the vector sum of its horizontal and vertical components:

$R = \sqrt{R_x^2 + R_y^2}$
$R = \sqrt{(50\sqrt{3})^2 + (50)^2}$
$R = \sqrt{(2500 \times 3) + 2500}$
$R = \sqrt{7500 + 2500}$
$R = \sqrt{10000}$
$R = 100 \, \text{N}$

Conclusion:

The tension in the cable is 100 N.
The reaction force applied by the pivot is 100 N.

Explanation of the solution:

1. Identify forces: Weight (W) acts at the center of the rod. Tension (T) acts at the end A, resolved into horizontal ($T \cos 30^\circ$) and vertical ($T \sin 30^\circ$) components. Pivot reaction (R) at O has horizontal ($R_x$) and vertical ($R_y$) components.
2. Apply torque equilibrium: Take torque about pivot O. Weight causes clockwise torque ($W \times L/2$). Vertical component of tension causes counter-clockwise torque ($T \sin 30^\circ \times L$). Equating these gives $T = W$.
3. Calculate Tension: $W = Mg = 10 \times 10 = 100 \, \text{N}$. So, $T = 100 \, \text{N}$.
4. Apply force equilibrium:
   • Horizontal: $R_x = T \cos 30^\circ = 100 \times \sqrt{3}/2 = 50\sqrt{3} \, \text{N}$.
   • Vertical: $R_y + T \sin 30^\circ = W \implies R_y = W - T \sin 30^\circ = 100 - 100 \times 1/2 = 50 \, \text{N}$.
5. Calculate total reaction force: $R = \sqrt{R_x^2 + R_y^2} = \sqrt{(50\sqrt{3})^2 + (50)^2} = \sqrt{7500 + 2500} = \sqrt{10000} = 100 \, \text{N}$.