Concept:Using rotational equilibrium to balance torque about a pivot.
Formula:$$\Sigma \tau = 0 \implies \tau_{\text{clockwise}} = \tau_{\text{anticlockwise}}$$
Solution:The pivot is at the 50 cm mark. Let's calculate the moment arms:
- The 5 N force is at the 0 cm mark. Its moment arm is:
$$d_1 = 50\text{ cm} - 0\text{ cm} = 50\text{ cm}$$ - This force produces an anticlockwise torque:
$$\tau_{\text{anticlockwise}} = 5\text{ N} \times 50\text{ cm} = 250\text{ N}\cdot\text{cm}$$ - The 10 N force must produce an equal clockwise torque of \(250\text{ N}\cdot\text{cm}\) on the opposite side of the pivot:
$$\tau_{\text{clockwise}} = 10\text{ N} \times d_2 = 250\text{ N}\cdot\text{cm}$$
$$d_2 = 25\text{ cm}$$
Since the clockwise torque must be on the right side of the pivot, the position of this force is:
$$\text{Position} = 50\text{ cm} + 25\text{ cm} = 75\text{ cm mark}$$
Why other options are incorrect:- A, C, and D do not satisfy the condition of rotational equilibrium. If placed at these marks, the clockwise and anticlockwise torques would be unbalanced, causing the rod to tilt.
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