anonymous
  • anonymous
you are falling off the edge.what should you do to avoid falling..?
Physics
chestercat
  • chestercat
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shamim
  • shamim
Need equal nd oposit force of my weight
anonymous
  • anonymous
Depends how you're falling. If you're tipping over the edge, you'd need a moment to counteract the tipping force. If you're free falling, you'd need something to oppose your weight like the drag force from a parachute, thrust from a jet pack, tension from a rope... etc etc
anonymous
  • anonymous
Okay got it.. thanks..

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shamim
  • shamim
U r most welcome !!!!
anonymous
  • anonymous
Hey can you guys explain it in detail?? i am not getting the answer.. :(
anonymous
  • anonymous
I can try and sketch it out if it helps.. |dw:1432962941227:dw|
anonymous
  • anonymous
Okay sure.. i am waiting..
anonymous
  • anonymous
so this beautiful drawing is you falling off "the edge" Do you want the "tipping off the edge" solution, or the "free falling over the edge" solution?
anonymous
  • anonymous
Solution fot the Free falling over the edge..
anonymous
  • anonymous
Sooo lets do a free body diagram for free falling|dw:1432963803490:dw|
anonymous
  • anonymous
so when you're free falling, the only force acting on you is gravity (air resistance too... but lets not complicate things..) the free body diagram (only looking at the person) looks something like this: |dw:1432963903189:dw|
anonymous
  • anonymous
now normally, if you were standing on the ground, there would be a normal force acting opposite to your weight. that's the ground pushing back on you to keep you from sinking through it. In free fall, you aren't touching the ground, so there's only a downward force due to gravity. Newton's laws of motion state that unbalanced forces will result in accelerations. So at the moment, our falling friend is going to accelerate towards the ground. Which is bad for him.
anonymous
  • anonymous
To keep him from falling, we'd need a force going upwards to counter act the force of gravity. This force could present itself in many different ways, but essentially, you just need an arrow pointing up to balance the one pointing down: |dw:1432964135012:dw| If the arrows are the same size (i.e. the forces have the same magnitude) then the forces balance and the person won't fall towards the ground!
anonymous
  • anonymous
you could have a rope to pull on (the rope would provide a Tension force upwards) |dw:1432964231726:dw| you could have a jet pack creating upwards thrust: |dw:1432964282830:dw| the possibilities are endless. you just need an upwards force
anonymous
  • anonymous
Thanks aloot.. it really helped me alot.. God bless you!
anonymous
  • anonymous
Thanks aloot.. it really helped me alot.. God bless you!
anonymous
  • anonymous
@LifeEngineer.. i am feeling embarrassed on saying that i also want a solution for tipping one.. :( i have a test after sometime.. and my syllabus included the second method also.. will you explain it with sketches??
anonymous
  • anonymous
i'll make it easier on myself this time. The "person" is just going to be represented by a block: |dw:1432966055551:dw|
anonymous
  • anonymous
now lets draw a proper free body diagram and identify the forces acting on the person: |dw:1432966157241:dw| As you can see, there's the weight as always, but now since the person is still touching the ground, there is a normal force. Tipping occurs when there is an imbalance of moments. A moment is just a force times a distance from a rotation point. Think of levers, the longer the lever arm, the easier it is to turn something, because a longer distance gives you more moment. Another example could be a door. If you've ever tried to open a door by pushing it really close to the hinged side, it feels a lot harder to push. Thats because you have less torque/moment Anyways, this poor fellow is tipping because the moments are imbalanced. Considering the back side that is still on the cliff: |dw:1432966406105:dw|
anonymous
  • anonymous
the circle represents the pivot point of the body, and the x is the distance to the center of gravity. Because the normal force is acting *ON* the pivot point, its moment arm is 0. M_N=F*0 = 0. There is no moment generated by the normal force, but since the center of gravity is over the edge (hence the tipping), there is a clockwise moment generated by the weight of M_w=W*x You can sum moments just like you sum forces. In this case, your net moment is W*x which is not zero. Newtons laws still apply for moments. Since there is an unbalanced moment, there will be an acceleration (moments are paired with angular accelerations, forces are paired with linear accelerations) that causes the person to rotate off the edge: |dw:1432966709558:dw|
anonymous
  • anonymous
U r done with the figures??
anonymous
  • anonymous
just like before, to keep the person from falling, you need to balance the unbalanced moment. In this case it would be as easy as putting the center of gravity over ground instead of hanging over the ledge (i.e. stepping backwards), but you could also apply what is called a couple moment to the system. A couple moment is essentially just a moment without any associated forces. (think of two perfectly balancing forces that are separated by a distance e.g. two hands pulling in opposite directions on a steering wheel. The wheel doesnt move, it just turns. Thats because the forces are balanced but the moments are not). Anyways, stick a couple moment, that opposes the moment cause by gravity, on the guy and he wont fall.
anonymous
  • anonymous
Okay thanks aloot.. God bless you forever.. now i am done With it.. you make me study easier..

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