torque, pivot points, and friction
my bad, mooserman, you are correct in that the normal force will not always be m*g (for a horizontal plane). give me a bit and i'll try to think up a way (without reading those articles :) ) on how to get the normal force. i'm sure there is something in my engineering texts about this. :)
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no supercat I think you were right. If 1 corner in touching the table, the normal force on that corner is -mg. The net force however is not 0 since the other 5 corners still experience gravity. If the box is off-balance, it still fall.
michael, you're refering to the m of each vertex, not of the box. still, it is not mg. example: suppose the box is exactly balanced on a flat table, the center of mass is directly above the contact point. The object does not fall, the torque on it is zero. hence the net force on the box must be Mg, where M is the net mass of the box. By the way, how did you get 6 vertices?
supercat, let me know what you find!
supercat, let me know what you find!
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little sleepy, meant 8 lol. I think I was thinking faces, the way Im going to do this is with faces rather than with vertices. And in my last post I was talking about the m total of the entire box. the normal force on the contact vertex would be -mg. how the hell would u figure that out tho? the force of gravity would be mg/8 on each vertex...
okay, i had a really long post worked up in my head, but i need to clarrify what we are looking for here. are we after the force due to friction from an inclined slope?
if that is the case then it is just:
u * Fn
where Fn is the normal force on the point in contact with the ground. in this case the normal force won't be exactly m * g because the ground will be inclined and to find the normal force will be some basic trig. and u is the coefficeint of kinetic friction.
then we got onto the debate of what the normal force will be. well, i again think it will be m * g if on a flat plane and then for an inclined plane it is m * g times whatever trig is needed. this is because the box more than likely won't move through the plane, so all forces in that direction must be zero. that doesn't rule out rotation though (which is what my long post was about), and i'll look into and report on that tomorrow when i'm fully awake.
i haven't completely thrown out the idea that the normal force will be changing with the block's position, so i'll keep that in mind. i plan to speak to some of my professors tomorrow about this because it is quite an interesting problem. :)
if that is the case then it is just:
u * Fn
where Fn is the normal force on the point in contact with the ground. in this case the normal force won't be exactly m * g because the ground will be inclined and to find the normal force will be some basic trig. and u is the coefficeint of kinetic friction.
then we got onto the debate of what the normal force will be. well, i again think it will be m * g if on a flat plane and then for an inclined plane it is m * g times whatever trig is needed. this is because the box more than likely won't move through the plane, so all forces in that direction must be zero. that doesn't rule out rotation though (which is what my long post was about), and i'll look into and report on that tomorrow when i'm fully awake.
i haven't completely thrown out the idea that the normal force will be changing with the block's position, so i'll keep that in mind. i plan to speak to some of my professors tomorrow about this because it is quite an interesting problem. :)
Author
your wrong about the friction. That is simply the equation for the scalar friction. My question was about the direction of the friction but it was already answered. The main question is now how you would calculate the normal force on some number of vertices of a cube touching some plane. Im pretty sure your right about the single vertex, that it will always be (n/|n|) dot -mg. but what if 2 vertices are touching? would it be that/2?
Net force would always be the same, if that object were to sit on the table and not push away or into it. so saying that the force applied when one vertex is touching should be devided by two when 2 vertices are touching would be correct, if you're applying that force to each vertex.
Author
so to find the normal force I should just add up how many vertices are touching the table, divide mg by that number and then use trig to find the normal force due to the new mg/x for each vertex?
michael, the one thing i am absolutely sure of is that using mg/(number of contact vertices) will lead to incorrect results. the physics of that is simply wrong.
i had a little time to write a very simple program to see how the method i suggested works. i think the results are pretty good, even though the program itself is pretty crude. i can send it to you if you'd like. i'm not sure how i can post it here without destroying the formatting...
i had a little time to write a very simple program to see how the method i suggested works. i think the results are pretty good, even though the program itself is pretty crude. i can send it to you if you'd like. i'm not sure how i can post it here without destroying the formatting...
Author
try the [_code_][_/code_] command (take out the underscores). If not Ill pm u my email. Thanks for the help. Another thing is that I didnt mean just using mg/#. I mean plug in the force of gravity as mg/(#of contact vertices) for each contact vertex and then find what the normal force would be on each vertex (if the plane is flat then it would just be mg/# for each vertex, if the plane were vertical is would be 0 for every vertex). Are you sure that wouldnt work?
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