How Does a Scale Determine the Mass of a Student on a Beam?

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SUMMARY

The discussion centers on calculating the mass of a student standing on a uniform 25 kg beam, with the scale reading 350 N. The mass of the student is conclusively determined to be 58 kg. Key concepts include Newton's 3rd Law, which states that the force of the beam on the scale is equal and opposite to the force of the scale on the beam. Additionally, understanding the balance of forces and torques is essential for accurate calculations in this scenario.

PREREQUISITES
  • Newton's 3rd Law of Motion
  • Basic principles of torque and equilibrium
  • Understanding of force measurement in Newtons
  • Concept of center of mass in uniform beams
NEXT STEPS
  • Study the principles of torque and how to calculate it in static systems
  • Learn about force distribution on beams and how it affects measurements
  • Explore advanced applications of Newton's Laws in real-world scenarios
  • Investigate the mechanics of scales and their calibration methods
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Physics students, educators, and anyone interested in mechanics and force analysis in static systems will benefit from this discussion.

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A student stands on a uniform 25 kg beam.
The scale on the right reads 350 N. What is the mass of the student?

Answer is 58Kg.

I don't understand what is the deal with the scale? I think I figured this out but I'm not sure.
The force of the scale is opposite the force of the beam, girl right? If so why?
Isn't the force on the centre of rotation equal to the force on the scale?
 
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brentwoodbc said:
4990811.gif


A student stands on a uniform 25 kg beam.
The scale on the right reads 350 N. What is the mass of the student?

Answer is 58Kg.

I don't understand what is the deal with the scale? I think I figured this out but I'm not sure.
The force of the scale is opposite the force of the beam, girl right? If so why?
Well, the force of the beam on the scale is opposite the force of the scale on the beam (Newton's 3rd Law), if that's what you are asking about.
Isn't the force on the centre of rotation equal to the force on the scale?
Not necessarily. That would be true only if the weight were distributed evenly between the centre of rotation and the scale.

A good approach here would be to identify all forces acting on the beam. Then:
  • Balance all the forces
  • Balance all the torques

The force due to the balance is simply the 350 N reading.
 

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