Chapter 14: Problem 5
(Optional) A uniform meterstick is in static rotational equilibrium when a mass of \(220 \mathrm{~g}\) is suspended from the \(5.0-\mathrm{cm}\) mark, a mass of \(120 \mathrm{~g}\) is suspended from the \(90-\mathrm{cm}\) mark, and the support stand is placed at the \(40-\mathrm{cm}\) mark. What is the mass of the meterstick?
Short Answer
Step by step solution
Understanding the Problem
Define Torque
Determine Lever Arms
Write Torque Equation
Calculate Forces for Torque
Set Up Torque Balance Equation
Solve for the Mass of the Meterstick
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Torque Calculation
Lever Arms in Physics
- For the 220 g mass, the lever arm is 35 cm (calculated as 40 cm - 5 cm).
- For the 120 g mass, the lever arm is 50 cm (calculated as 90 cm - 40 cm).
- For the meterstick itself, with its center of mass located at the 50 cm mark, the lever arm is 10 cm (50 cm - 40 cm).
Rotational Equilibrium Concepts
- The sum of torques from the 220 g mass and the meterstick's mass must exactly balance the torque from the 120 g mass.
- This balance is expressed in the equation \[ (220 \cdot 9.8 \cdot 35) + (mass_{stick} \cdot 9.8 \cdot 10) = (120 \cdot 9.8 \cdot 50) \]