Find distance from COM using torque

  • #1

Homework Statement


Word for word, from the problem:
"A person’s center of mass is easily found by having the person lie on a reaction board. A horizontal 2.5-m-long, 6.1 kg reaction board is supported only at the ends, with one end resting on a scale and the other on a pivot. A 60 kg woman lies on the reaction board with her feet over the pivot. The scale reads 25 kg. What is the distance from the woman’s feet to her center of mass?"

Homework Equations


T=Iα (torque = moment of inertia times angular acceleration)
T=rFsin∅ (torque = radius times the force times sine of the angle between the radius and the force)
T=rFt (torque = radius times component of force perpendicular to the moment arm)
l=rmvsin∅ (angular momentum = radius times mass times velocity times the sine of the angle between the radius and the force)
?????

The Attempt at a Solution


So the scale reads 25kg, and I thought maybe that force could be found using F = ma. So I did that and got a force of 245N. I figure that force is perpendicular to the moment arm, so I multiplied it by 2.5m and got a torque of 612.5N⋅m. From there I'm kind of stuck though. I think it maybe also has something to do with angular momentum?
 

Answers and Replies

  • #2
SteamKing
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Homework Statement


Word for word, from the problem:
"A person’s center of mass is easily found by having the person lie on a reaction board. A horizontal 2.5-m-long, 6.1 kg reaction board is supported only at the ends, with one end resting on a scale and the other on a pivot. A 60 kg woman lies on the reaction board with her feet over the pivot. The scale reads 25 kg. What is the distance from the woman’s feet to her center of mass?"

Homework Equations


T=Iα (torque = moment of inertia times angular acceleration)
T=rFsin∅ (torque = radius times the force times sine of the angle between the radius and the force)
T=rFt (torque = radius times component of force perpendicular to the moment arm)
l=rmvsin∅ (angular momentum = radius times mass times velocity times the sine of the angle between the radius and the force)
?????

The Attempt at a Solution


So the scale reads 25kg, and I thought maybe that force could be found using F = ma. So I did that and got a force of 245N. I figure that force is perpendicular to the moment arm, so I multiplied it by 2.5m and got a torque of 612.5N⋅m. From there I'm kind of stuck though. I think it maybe also has something to do with angular momentum?
You're flailing at this. The woman lies on a board which is supported at two ends. A scale is fitted at one end to record the support reaction there.

Knowing the mass of the board and its length, you know where the center of mass of the board is located from one end. You know the mass of the woman, but you don't know where her center of mass is located.

This is a simple problem in static equilibrium. You don't need angular momentum or the special theory of relativity to find the answer.

Draw a free body diagram of the reaction board first to help with your analysis.


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[\CENTER]​
 
  • #3
Okay, so I tried the free-body diagram. It's on a fixed pivot point, and supported on both ends, so the acceleration would be 0? I just thought it would be torque because the problem was posed in the notes for the chapter about torque.

Also, my professor emailed back...the answer is .91m. Which is great, but I would really like to understand how to do the problem.
 
  • #4
SteamKing
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Okay, so I tried the free-body diagram. It's on a fixed pivot point, and supported on both ends, so the acceleration would be 0? I just thought it would be torque because the problem was posed in the notes for the chapter about torque.

Also, my professor emailed back...the answer is .91m. Which is great, but I would really like to understand how to do the problem.
Do you know what the equations of static equilibrium are?
 
  • #5
54
10
You just need to balance the torque and you'll be good, SteamKing gives a good visualisation of how a reaction board works. Drawing the FBD youll only find one unknown force and balancing torque about that point will eliminate the need of that force. You can concentrate the mass of the woman at her COM and let it be at a distance x from that point. Solving the equation youll get x
 
Last edited:
  • #6
I looked up static equilibrium in my textbook. We skipped that chapter. I'm actually really, really stupid when it comes to physics. Biology, chemistry, any other science, I'm good. Even calculus, I'm okay. Physics, not so much.
 
  • #7
SteamKing
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I looked up static equilibrium in my textbook. We skipped that chapter. I'm actually really, really stupid when it comes to physics. Biology, chemistry, any other science, I'm good. Even calculus, I'm okay. Physics, not so much.
It's hardly a physics course if you skipped static equilibrium.
 
  • #8
Regardless of whether or not it's a physics course, it has professors and a test, the nearest of which is in twelve hours.
 
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  • #9
SteamKing
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Regardless of whether or not it's a physics course, it has professors and a test, the nearest of which is in twelve hours.
Yes, but there are good professors and bad professors, and it appears that the one you drew is toward the bad end of the spectrum.

BTW, make sure that "spectrum" is one of the topics your physics course covers when it comes to studying light.
 

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