The horizontal uniform rod shown below has length 0.60 m and mass 2.0 kg . The left end of the rod is attached to a vertical support by a frictionless hinge that allows the rod to swing up or down. The right end of the rod is supported by a metal wire that makes an angle of $$ 30^{\circ} $$ with the rod. A 0.50 kg block is also attached to the right end of the rod as shown below. Calculate the tension in the metal wire.

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Let’s analyze the equilibrium of the rod about the hinge. Since the rod is in static equilibrium, the net torque about any point is zero. Choosing the hinge as the pivot eliminates the reaction forces at that point. The forces causing torque about the hinge are:

1. The weight of the rod.
2. The weight of the block.
3. The tension $$ T $$ in the metal wire.

The rod is horizontal, and the wire makes an angle of $$30^\circ$$ with the rod. Therefore, at the point where the wire is attached (the right end of the rod), the tension $$ T $$ acts at a $$30^\circ$$ angle above the horizontal.

### Step 1. Compute the torques due to the weights

- **Weight of the rod:**  
  Mass of rod $$ m_r = 2.0~\mathrm{kg} $$  
  Weight of rod $$ W_r = m_r g = 2.0 \times 9.8 = 19.6~\mathrm{N} $$  
  Since the rod is uniform, its center of mass is at its midpoint, $$d_{r} = 0.60/2 = 0.30~\mathrm{m}$$ from the hinge.  
  The torque due to the rod’s weight is  
  $$
  \tau_r = 19.6~\mathrm{N} \times 0.30~\mathrm{m} = 5.88~\mathrm{Nm}
  $$
  (acting clockwise).

- **Weight of the block:**  
  Mass of block $$ m_b = 0.50~\mathrm{kg} $$  
  Weight of block $$ W_b = m_b g = 0.50 \times 9.8 = 4.9~\mathrm{N} $$  
  The block is attached at the right end of the rod, $$d_b = 0.60~\mathrm{m}$$ from the hinge.  
  The torque due to the block’s weight is  
  $$
  \tau_b = 4.9~\mathrm{N} \times 0.60~\mathrm{m} = 2.94~\mathrm{Nm}
  $$
  (also acting clockwise).

The total clockwise torque is:
$$
\tau_{\text{clockwise}} = \tau_r + \tau_b = 5.88 + 2.94 = 8.82~\mathrm{Nm}
$$

### Step 2. Express the torque due to the tension

The tension $$ T $$ acts at the right end of the rod, at a distance of $$ 0.60~\mathrm{m} $$ from the hinge. Only the component of $$ T $$ perpendicular to the rod produces a torque. Since the wire makes an angle of $$30^\circ$$ with the horizontal rod, the perpendicular (vertical) component of $$ T $$ is
$$
T_\perp = T \sin 30^\circ
$$
Thus, the torque due to the tension is:
$$
\tau_T = T_\perp \times 0.60 = T \sin 30^\circ \times 0.60
$$
Since $$\sin 30^\circ = 0.5$$, this simplifies to:
$$
\tau_T = T \times 0.5 \times 0.60 = 0.30\,T
$$
This torque acts counterclockwise.

### Step 3. Set up the equilibrium condition for torques

For equilibrium, the counterclockwise torques must balance the clockwise torques:
$$
\tau_T = \tau_{\text{clockwise}}
$$
Substituting the expressions we have:
$$
0.30\,T = 8.82~\mathrm{Nm}
$$

### Step 4. Solve for $$ T $$

$$
T = \frac{8.82~\mathrm{Nm}}{0.30} = 29.4~\mathrm{N}
$$

Thus, the tension in the metal wire is:
$$
T = 29.4~\mathrm{N}
$$
The tension in the metal wire is 29.4 N.

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