A uniform meter rule balances on a pivot at the 30 cm mark. A log of mass 100g is hung from the 5 cm mark. Where must a weight be placed to balance the rule?

The problem states "a meter of log" which is ambiguous regarding its mass. For this solution, we will assume that "a meter of log" refers to a standard mass of $100 \text{ g}$ (or $0.1 \text{ kg}$). If a different mass is intended, please substitute it into the formula.

Given:

• The meter rule is uniform, so its center of mass is at the $50 \text{ cm}$ mark.
• The pivot point (balancing point) is at the $30 \text{ cm}$ mark.
• The log is hung at the $5 \text{ cm}$ mark.
• Assumed mass of the log, $M_l = 100 \text