Equilibrium & Weightlessness: Mechanics Questions

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A ball thrown in the air and momentarily at rest is not in equilibrium because it is still subject to the gravitational force acting downwards. Equilibrium requires that the sum of forces and torques equals zero, which is not the case here. Weightlessness occurs when an object is in free fall, where the only force acting is gravity, resulting in no net force felt by the object, such as an astronaut in orbit. This sensation of weightlessness is experienced when there is no normal force acting on the body, making the apparent weight zero. Thus, while the ball may appear stationary at its peak, it is not in equilibrium due to the ongoing gravitational pull.
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If a ball is thrown in the air and comes momentarily at rest is it at equilibrium?
I thought yes because it's stationary, but then again.. there is a resultant gravitational force.. ?


Also what is "weightlessness" exactly?
Is it when you're in freefall/only gravity acts? so that would be like an astronaut in a satellite orbiting the earth?
 
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briton said:
If a ball is thrown in the air and comes momentarily at rest is it at equilibrium?
I thought yes because it's stationary, but then again.. there is a resultant gravitational force.. ?

Equilibrium conditions are:

\sum F = 0, \sum \tau = 0 where \tau [/tex] is the torque. <br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Also what is &quot;weightlessness&quot; exactly? <br /> Is it when you&#039;re in freefall/only gravity acts? so that would be like an astronaut in a satellite orbiting the earth? </div> </div> </blockquote><br /> I want to say you feel weightless when your body feels no net force, but there is a net centripetal force during orbit.
 
When the ball is at rest at the top , it is being acted upon by the 'g' downwards , so it is not in equilibrium .

Weightlessness is 'free fall under gravity ' This is when your weight downwards (mg) is balanced by pseudo force uppwards when you are freely falling ...

Example: You are in a lift and the rope breaks and then you fall freely , you will lose contact with ground and you are under free fall.

For more:
http://science.howstuffworks.com/weightlessness.htm
 
briton said:
Also what is "weightlessness" exactly?
Is it when you're in freefall/only gravity acts? so that would be like an astronaut in a satellite orbiting the earth?
Yes. "Weightlessness" is when your apparent weight is zero. Your apparent weight, the "weight" that you feel, is due to the normal force of whatever surface is holding you up (or other force preventing you from falling). Take away the support, and you feel "weightless". (In physics, the term weight has a specific technical meaning: the force of gravity on an object. So, while you may experience the feeling of "weightlessness", your real weight is not zero!)
 
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Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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