A ruler is suspended on a pivot point and a mass is hanging on each side of the suspension point. The system is in equilibrium. How does the torque on the right side compare to the torque on the left side?
The correct answer and explanation is :
In an equilibrium condition, the sum of torques around the pivot point must be zero. This means that the torque exerted on the right side of the ruler must be equal in magnitude but opposite in direction to the torque exerted on the left side.
Mathematical Explanation
Torque ((\tau)) is given by the equation:
[
\tau = r \times F
]
where:
- ( r ) is the perpendicular distance from the pivot to the point where the force is applied,
- ( F ) is the force (which, in this case, is the weight of the hanging mass: ( F = mg ), where ( m ) is mass and ( g ) is gravitational acceleration).
For equilibrium:
[
\tau_{\text{left}} = \tau_{\text{right}}
]
[
r_{\text{left}} \cdot m_{\text{left}} \cdot g = r_{\text{right}} \cdot m_{\text{right}} \cdot g
]
Since gravitational acceleration ( g ) is constant and appears on both sides, it cancels out:
[
r_{\text{left}} \cdot m_{\text{left}} = r_{\text{right}} \cdot m_{\text{right}}
]
This equation confirms that the torques on both sides are equal in magnitude but act in opposite directions (clockwise vs. counterclockwise). This balance maintains equilibrium.
Conceptual Explanation
Imagine a seesaw in balance. If one child sits farther from the center than the other, they must be lighter to maintain balance, or the heavier child must sit closer to the pivot. The same principle applies to the suspended ruler with masses.
Thus, in an equilibrium state, the torque exerted on the left side is equal and opposite to the torque on the right side.
