Static Equilibrium: Homework Questions

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The discussion focuses on homework questions regarding static equilibrium, specifically addressing the treatment of normal forces and torque calculations. Participants debate whether each hand should have a separate normal force or if it can be simplified to one. The confusion around torque calculations arises from misunderstanding the pivot point, which is clarified as being at the foot rather than the hands. The conversation also touches on the importance of understanding lever mechanics and static problems involving beams and supports. Overall, the thread emphasizes the need for a solid grasp of torque and equilibrium principles in physics.
Speedking96
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Homework Statement


I have a question about the following question:

upload_2014-12-15_20-2-23.png


Firstly, shouldn't there be a normal force for each hand (part a), but here they treat it as one normal force.

Also in part b, why isn't the torque for Fn1h = 0? Since that is the pivot point, isn't the moment arm zero?
 
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Firstly, shouldn't there be a normal force for each hand (part a), but here they treat it as one normal force.
How is the result different if you give each hand a separate normal force?

Also in part b, why isn't the torque for Fn1h = 0? Since that is the pivot point, isn't the moment arm zero?
##F_{N1} \neq 0## and ##h\neq 0## therefore ##F_{N1}h \neq 0## ... take a closer look: what are they using for the pivot point in that calculation?
 
Wow. The pivot point is the location of the foot and N1 is for the hands. I see. Thank you.
 
Well done. Remember all those horizontal mean static problems - beam between two supports?
Remember how you'd do a sum-of-torques for each support as the pivot?
 
I actually don't know what you're referring to as I am learning torque and rotational dynamics on my own (on my final exam tomorrow and my teacher rushed everything on rotation ) - I don't think I've come across a problem like that
 
Have a look at:
http://www.pstcc.edu/departments/natural_behavioral_sciences/Web%20Physics/Experiment%2002-1320.htm
... it counts as essential physics.

Have you done levers?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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