Finding the Center of Mass of a Person me Out

AI Thread Summary
To find the center of mass of a person using the given forces and distances, the net torque must be zero. The problem involves two scales reading forces of 380N and 320N, separated by 2.00m. A diagram should be drawn with the pivot at the center of mass, allowing for the calculation of torque from each scale. The relationship between weight (N) and mass (kg) is crucial, as the forces acting on the plank must be analyzed to determine the correct distance from the woman's feet to her center of mass. Systematic problem-solving and understanding of torque principles are essential for arriving at the solution.
NudeLove66
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Finding the Center of Mass of a Person...Please Help me Out!

Homework Statement



In excersise physiology studies, it is sometimes important to determine the location of a persons mcenter of mass. ( It shows a picture of a girl laying on a plank with 2 scales under it)
A light plank rest on two scales, which read F_g1_ = 380N and F_g2_ =320N. A distance of 2.00m separates the scales. How far from the womans feet is her center of mass?

F_g1_ = 380N
F_g2_ =320N
Person is 2.00m

Homework Equations



Center of Mass: Sum of (m x x)/M
Sorry I don't know how to type these symbols):blushing:

The Attempt at a Solution



I thought this would be a straightfoward problem but I cannot figure it out. I'm not even sure which equation of Center of Mass to use. Also your dealing with N as opposed to kilagrams, are you suppose to convert N to kg, is that even possible?

It's driving me crazy, If anyone could help Id be in great debt!
 
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You are going to have to use the concept of torque. From what you offered, I can't tell if you are familiar with that. The net torque must be zero, this will allow you to solve for the location of the CM. Mass (kg) and weight (N) are related, if you are trying to solve this type of problem, this should be a familar concept.
 
That makes sense since this section takes place right after the section on Torque (10.6 The Rigid Object In Equilibrium from Principals of Physics 4th Edition by Serway & Jewett)

But what equation would I plug this into. Torque is shown in units of Newton x meter so is the answer as easy as just:

(380N + 320N) x 2.00m
 
NudeLove66 said:
That makes sense since this section takes place right after the section on Torque (10.6 The Rigid Object In Equilibrium from Principals of Physics 4th Edition by Serway & Jewett)

But what equation would I plug this into. Torque is shown in units of Newton x meter so is the answer as easy as just:

(380N + 320N) x 2.00m

No. What you have written says that both forces are acting on the same end of the plank, 2 meters from the pivot point. That is not what is going on.
Draw a diagram, with the pivot placed at the centre of mass and a scale at each end. Since you know the length of the plank, you can relate the distance from each scale to the pivot point in terms of one variable. So now try to write out the torque due to each end of the plank. The net torque about the pivot must be zero. Note that torque = force*distance.
 
i was looking at this problem as well, so after you calculated the both the torques, since the net torque is suppose to be zero, i just substracted them then set the number i got equal to force times distance. THen divided by f to get the distance but i ran into a problem, i couldn't figure out what force value to use... is this the correct?
 
the correct way to go at the problem?
 
Solve it systematically. What forces act on the girl? About what point will you calculate the net torque?
 
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