Tricky center of mass question

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SUMMARY

The discussion focuses on solving a physics problem involving the measurement of mass and the center of mass using two calibrated spring balances and a light, rigid plank. Participants emphasize the importance of understanding torque and equilibrium, stating that the sum of torques must equal zero for the system to be in equilibrium. The readings from the spring scales represent the normal forces acting on the plank, which can be used to calculate both the gravitational force (Fgrav) and the center of mass (Xcm) by applying the conditions for equilibrium.

PREREQUISITES
  • Understanding of torque and equilibrium principles in physics
  • Familiarity with spring scales and their function in measuring forces
  • Knowledge of gravitational force and its relation to mass
  • Ability to apply the equation Xcm=ƩMX/ƩM for center of mass calculations
NEXT STEPS
  • Study the principles of torque and how to calculate it using torque = distance × force
  • Learn about the conditions for equilibrium in static systems
  • Explore the relationship between normal force and gravitational force in physics
  • Practice solving problems involving center of mass calculations using multiple forces
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for effective methods to teach concepts of mass and center of mass measurement.

scoobysnacks
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Homework Statement


Explain how you could measure both your mass and the position of your body’s centre of mass if all you had as experimental equipment was two spring balances (calibrated, of course), and a light, rigid plank, set up as shown in the figure.

Figure: #1 http://s3.amazonaws.com/docuum/attachments/7166/2011 Final Exam.pdf?1355070977


Homework Equations



Xcm=ƩMX/ƩM
Torque=r*F
F=ma

The Attempt at a Solution



I approached it as a torque problem. Since the system is in equilibrium the sum of Torques is 0. The radius of one spring scale is Xcm while the other is L-Xcm. I'm just not sure what the reading of the spring scale represents. I know scales give the normal force, but is this equal to the torque or to Fgrav or what?

I really want to completely understand the problem so please point me in the right direction, thanks
 
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After reading the problem, I don't think the radius of the scale is part of the problem. I think the scales merely determine the weight at each point. Since you have a light plank, I would assume you would ignore the mass of the plank.
 
Oh sorry by radius I was referring to the distance from center of mass to each spring scale
 
scoobysnacks said:
I approached it as a torque problem. Since the system is in equilibrium the sum of Torques is 0.
Good.

The radius of one spring scale is Xcm while the other is L-Xcm.
I assume you mean that the center of mass is a distance Xcm from one spring and L-Xcm from the other. Good.

I'm just not sure what the reading of the spring scale represents. I know scales give the normal force, but is this equal to the torque or to Fgrav or what?
Those spring scale readings represent the normal forces acting upward on the plank. Identify all the forces acting on the plank (there are three). Then apply the conditions for equilibrium to solve for Xcm and Fgrav, given those normal forces from the springs.
 
Doc Al said:
Those spring scale readings represent the normal forces acting upward on the plank. Identify all the forces acting on the plank (there are three).

Are the three forces Fn from each scale and Fgrav from the person onto the plank?
 
scoobysnacks said:
Are the three forces Fn from each scale and Fgrav from the person onto the plank?
Yes!
 
Does the normal force of one spring scale equal Fgrav-normal force of the other spring?
 
scoobysnacks said:
Does the normal force of one spring scale equal Fgrav-normal force of the other spring?
Yes.
 
Ok so can we equate the torque on one scale as the product of its radius and normal force acting on it?
 
  • #10
scoobysnacks said:
Ok so can we equate the torque on one scale as the product of its radius and normal force acting on it?
Not sure what you mean. You can can compute the torque about some point (the ends of the plank are good points) by multiplying the force in question by its distance to that point.
 
  • #11
i'm still not sure how to approach this. Should I be using torque to solve for the person's mass?

Isnt the force in question gravity? so torque=d*Fgrav?
 
  • #12
scoobysnacks said:
i'm still not sure how to approach this. Should I be using torque to solve for the person's mass?
You'll use torque equations to solve for Fgrav and Xcm.

Isnt the force in question gravity? so torque=d*Fgrav?
There are three forces. Pick a point--say the left end of the plank. Figure out the torque about that point from all the forces. The clockwise torques must equal the counterclockwise torques in order for there to be equilibrium.
 

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