Static equilibrium and torque calculation

  • Thread starter Rasine
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  • #1
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Misako is about to do a push-up. Her center of gravity lies directly above a point on the floor which is d1 = 0.96 m from her feet and d2 = 0.71 m from her hands. If her mass is 50.0 kg, what is the force exerted by the floor on her hands?

F(feet)+F(head)=F(center of gravity0) becuase she is not moving

so what i was trying to do is to calculate the torque about the center of mass

so if she is not moving then T(feet)=T(head) and that would be
Ff(.96)=Fh(.71)

so where do i go from here ?
 

Answers and Replies

  • #2
radou
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If she is not moving, then the sum of all torques with respect to any point must vanish. Which would be a wise choice for such a point?
 
  • #3
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well if i am calculating the force at her head i should choose the point of rotation at her head so i only have to take in account her feet right?
 
  • #4
radou
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well if i am calculating the force at her head i should choose the point of rotation at her head so i only have to take in account her feet right?
Why not set the sum of all torques with respect to her feet to equal zero? In that case the only unknown is the force exerted by the floor on her hands.
 
  • #5
Doc Al
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well if i am calculating the force at her head i should choose the point of rotation at her head so i only have to take in account her feet right?
Just the opposite. If you choose her hands as the pivot, what torque will the force on her hands exert? And what about the force on her feet? (Ideally, you only want one unknown--the force you are trying to find.)

But there's nothing wrong with your original approach:

so what i was trying to do is to calculate the torque about the center of mass

so if she is not moving then T(feet)=T(head) and that would be
Ff(.96)=Fh(.71)
Hint: In addition to torques being zero about any point, what else must be true for her to be in equilibrium? (What is the sum of Ff and Fh?)
 
  • #6
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so i would have tf=0=tcm+th which is 0=Mg(d1)+f(d1+d2)
 
  • #7
radou
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Looks right.
 

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