Chapter 8: Problem 91
A meterstick balances at its center. If an \(86-g\) necklace is suspended from one end of the meterstick, the balance point moves \(8.2 \mathrm{~cm}\) toward that end. (a) Is the mass of the meterstick greater than, less than, or equal to the mass of the necklace? Explain. (b) Find the mass of the meterstick.
Short Answer
Step by step solution
Understanding the Balance Point Shift
Relating Torque and Balance
Analyzing the Equal Torque Points
Calculating Mass of the Meterstick using Torque Equation
Concluding the Comparison of Masses
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Torque and Balance
Center of Mass
Torque Equation
\[ \text{Torque} = \text{Force} \times \text{Distance} \]
In our physics exercise, the torque from the meterstick must equal the torque from the necklace to maintain balance. This is expressed in the equation:
\[ m \times 8.2 = 86 \times 50 \]
Where \( m \) is the mass of the meterstick we are looking for. After solving this equation, we find that the mass of the meterstick is approximately 524.39 g. This demonstrates the power of the torque equation in identifying unknown forces or weights in balance problems.
Balancing Forces
The necklace exerts a downward force on one side, shifting the meterstick's balance point closer to it. To achieve a state of equilibrium, the torque generated by the meterstick's mass (as calculated by its weight and distance from the pivot) needs to counterbalance the torque exerted by the necklace.
Once balanced, we see that the position shift accurately reflects the differing masses and distances involved. Balancing forces ensures that objects remain stable and in equilibrium, providing a clear insight into how forces interact in various scenarios.