Post by: naved_s9994 on June 06, 2009, 08:58:15 pm
Can anyone conclude all the types of questions ( style of questions), on Newtons seccond law EF=ma
Thanks VN,
naved_s9994 ;)
Post by: naved_s9994 on June 06, 2009, 10:12:58 pm
relative questions to it, or even banked curve?
Post by: kurrymuncher on June 07, 2009, 01:37:31 am
Post by: danieltennis on June 07, 2009, 01:38:03 am
What the fuck do you mean by EF=ma??
Net force?
Post by: dekoyl on June 07, 2009, 01:47:21 am
What the fuck do you mean by EF=ma??Lol.
Post by: kurrymuncher on June 07, 2009, 01:49:37 am
Post by: TrueTears on June 07, 2009, 01:49:51 am
Hey Im preety sure ( prediction) that in some way or form, theyre going to ask questions on it...as theres usually one EACH year.Your standard
Can anyone conclude all the types of questions ( style of questions), on Newtons seccond law EF=ma
Thanks VN,
naved_s9994 ;)
Or in circular motion,
Post by: Gloamglozer on June 07, 2009, 09:47:38 am
Or in circular motion,etc etc?
That's when you're only at the top of the loop, isn't it?
Post by: Flaming_Arrow on June 07, 2009, 10:24:17 am
Or in circular motion,
That's when you're only at the top of the loop, isn't it?
no thats the bottom, top is Fnet = N - mv^2/r
Post by: kurrymuncher on June 07, 2009, 11:51:47 am
Post by: Flaming_Arrow on June 07, 2009, 11:57:45 am
No, Gloamglozer is correct, that is the net force at the top.
er what? my teacher gave a handout that says Weight is the highest at the bottom.
Post by: Mao on June 07, 2009, 12:01:36 pm
Top: F net = N + mg (points towards the same direction)
Bottom: F net = N - mg (points upwards, N > mg)
Outside loop:
Top: F net = mg - N (points downwards)
Post by: kurrymuncher on June 07, 2009, 12:02:11 pm
Post by: Flaming_Arrow on June 07, 2009, 12:04:33 pm
Inside loop:
Top: F net = N + mg (points towards the same direction)
Bottom: F net = N - mg (points upwards, N > mg)
Outside loop:
Top: F net = mg - N (points downwards)
ye i meant when its outside the loop
Post by: Gloamglozer on June 07, 2009, 12:07:21 pm
Inside loop:
Top: F net = N + mg (points towards the same direction)
Bottom: F net = N - mg (points upwards, N > mg)
Outside loop:
Top: F net = mg - N (points downwards)
Thanks for the confirmation, Mao. :D
Post by: Mao on June 07, 2009, 12:11:50 pm
Inside loop:
Top: F net = N + mg (points towards the same direction)
Bottom: F net = N - mg (points upwards, N > mg)
Outside loop:
Top: F net = mg - N (points downwards)
ye i meant when its outside the loop
Interesting thing about circular motion in the vertical direction outside of the loop:
At top:
gravitational attraction downwards
normal reaction force by surface on object upwards
Net force = mg - N = Fc, points downwards towards center of circle
At bottom:
gravitational attraction downwards
normal reaction force by surface on object downwards
Net force = mg + N, points downwards AWAY from center of circle :. circular motion not possible.
Post by: dino on June 07, 2009, 01:34:16 pm
[img]http://img.skitch.com/20090607-cywdej8pyjf85yey5e3pre35uw.preview.jpg[/img]
is that wrong?
Post by: TrueTears on June 07, 2009, 01:35:40 pm
AWAY? I'm lost now. grrr. i need more vertical questions.That is correct.
[img]http://img.skitch.com/20090607-cywdej8pyjf85yey5e3pre35uw.preview.jpg[/img]
is that wrong?
Post by: dejan91 on June 07, 2009, 02:26:43 pm
So is:
Top: F net = N + mg, N subtract mg (because opposite direction).
F net = N - mg, N subtract the negative of mg. (i.e. adding mg to N)
Outside loop:
Top: F net = mg - N,..............(i don't even know what way to explain this lol)
I just tend to use intuition for these really......... but can anybody explain this?
Post by: TrueTears on June 07, 2009, 02:31:06 pm
Are these forces with or without a direction??When you are on the top:
So is:
Top: F net = N + mg, N subtract mg (because opposite direction).
F net = N - mg, N subtract the negative of mg. (i.e. adding mg to N)
Outside loop:
Top: F net = mg - N,..............(i don't even know what way to explain this lol)
I just tend to use intuition for these really......... but can anybody explain this?
Let up be negative and down be positive.
Since forces are vectors, they are all pointing down hence they are all positive: We have
When you are on the bottom.
Let up be positive and down be negative.
Normal force is acting up,
Post by: dejan91 on June 07, 2009, 02:34:10 pm
Post by: TrueTears on June 07, 2009, 02:36:24 pm
Ahhhh ok so you use those "formulas" with the absolute value of N and mg, correct?Yeap, then just put the +/- sign depending on the direction.
So if you are on the bottom:
Post by: dejan91 on June 07, 2009, 02:44:39 pm
But yes, I get it now.
Post by: naved_s9994 on June 07, 2009, 03:45:24 pm
umm, yea thanks everyone...CLEARED it all up
Post by: NE2000 on June 07, 2009, 04:03:37 pm
Inside loop:
Top: F net = N + mg (points towards the same direction)
Bottom: F net = N - mg (points upwards, N > mg)
Outside loop:
Top: F net = mg - N (points downwards)
ye i meant when its outside the loop
Interesting thing about circular motion in the vertical direction outside of the loop:
At top:
gravitational attraction downwards
normal reaction force by surface on object upwards
Net force = mg - N = Fc, points downwards towards center of circle
At bottom:
gravitational attraction downwards
normal reaction force by surface on object downwards
Net force = mg + N, points downwards AWAY from center of circle :. circular motion not possible.
So is that if you are outside of the circle when at the bottom, so you are upside down and there is no possibility of a centripetal force?
Post by: kamil9876 on June 07, 2009, 07:42:38 pm
Inside loop:
Top: F net = N + mg (points towards the same direction)
Bottom: F net = N - mg (points upwards, N > mg)
Outside loop:
Top: F net = mg - N (points downwards)
ye i meant when its outside the loop
Interesting thing about circular motion in the vertical direction outside of the loop:
At top:
gravitational attraction downwards
normal reaction force by surface on object upwards
Net force = mg - N = Fc, points downwards towards center of circle
At bottom:
gravitational attraction downwards
normal reaction force by surface on object downwards
Net force = mg + N, points downwards AWAY from center of circle :. circular motion not possible.
So is that if you are outside of the circle when at the bottom, so you are upside down and there is no possibility of a centripetal force?
Well, If the car was a N magnet and you placed some S magnet in the centre, or place some really large planet above the hoop, or maybe have a favourable ammount of upward wind on that day then maybe you could :P. But if the the only two forces are the Normal Reaction force of hoop and gravity of earth and car then no because they always act down in this situation and the centre is above you so obviously a centripetal force would have to be upwards.