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?

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Asked on March 30, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-03-30T08:06:34.590Z

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 g (or 0.1 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 cm mark. The pivot point (balancing point) is at the 30 cm mark. The log is hung at the 5 cm mark. Assumed mass of the log, $M_l = 100

Accepted Answer · 2026-03-30T08:06:34.590Z

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 g (or 0.1 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 cm mark. The pivot point (balancing point) is at the 30 cm mark. The log is hung at the 5 cm mark. Assumed mass of the log, $M_l = 100

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?

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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?

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