friction on a banked road

[xZero]

rofl all this banked curve questions got me thinking, say that an object is at rest on a banked curve, N=mgcos(theta) since its essentially a inclined plane. However once it starts moving, N=mg/cos(theta). This mean that the normal force actually increases if you starts running compared to at rest, but how does that work?

[jinny1]

where did you get the "middle" bit/term?
?

[costa94]

where did you get the "middle" bit/term?
?

that is the previous Fc

[Shark 774]

So, you would feel heavier when travelling greater than the design speed, and lighter when travelling less than the design speed right??

[golden]

Okay here's what I did. Point out any mistakes I may have made

How did you know to use cosine?

[davidle_10]

Okay here's what I did. Point out any mistakes I may have made

isn't it because you cancel the m

[Vincezor]

isn't it because you cancel the m

You are correct. Sorry, sometimes it hard to spot mistypes when you are typing in LaTeX

Hang on, I'll make an edit.

[Vincezor]

How did you know to use cosine?

I just followed the formula.



As the friction is perpendicular to the normal, both the friction and normal must be split into horizontal and vertical components. If you draw out a banked motion question and label the forces acting on it, you will notice that the friction is costheta instead of sinetheta. Hard to explain, looking at it visually is so much simpler.

[yawho]

[quote ]
a circular corner of radius 10 m is banked at 15 degree with the horizontal. find the speed that a car can make the turn without friction to assist the turn. If the 800-kg car travels at 5 m/s higher than that speed, what friction between the tyres and the road is required for the car to make the turn?
[/quote]
first answer 5.2 m/s
Second answer requires solving simultaneous equations:
vert comp, net force=Ncos-mg-fsin=0, Ncos15-8000-fsin15=0
hori comp, net force=Nsin+fcos=mv^2/r, Nsin15+fcos15=800x10.2^2/10
f=5930 N approx

[golden]

How did you know to use cosine?

I just followed the formula.



As the friction is perpendicular to the normal, both the friction and normal must be split into horizontal and vertical components. If you draw out a banked motion question and label the forces acting on it, you will notice that the friction is costheta instead of sinetheta. Hard to explain, looking at it visually is so much simpler.

Hmm I see, however when they ask you to actually label the forces acting on the person/car etc. you wouldn't label friction as being perpendicular but rather parallel to the slope?

[Vincezor]

How did you know to use cosine?

I just followed the formula.



As the friction is perpendicular to the normal, both the friction and normal must be split into horizontal and vertical components. If you draw out a banked motion question and label the forces acting on it, you will notice that the friction is costheta instead of sinetheta. Hard to explain, looking at it visually is so much simpler.

Hmm I see, however when they ask you to actually label the forces acting on the person/car etc. you wouldn't label friction as being perpendicular but rather parallel to the slope?

Yeah I guess usually you label the normal reaction force as well as the weight force - those are the 2 marks. I'm not sure if I've seen the friction being labelled.

[cranberry]

i don't get it... aren't those two forces in different directions? Fnet is horizontal, so is Frcos(theta), but Nsin(theta)/mgsin(theta) is down the slope.....
Are there any legit past exam questions that have asked this? need to look at one

[Vincezor]

i don't get it... aren't those two forces in different directions? Fnet is horizontal, so is Frcos(theta), but Nsin(theta)/mgsin(theta) is down the slope.....
Are there any legit past exam questions that have asked this? need to look at one

is actually the horizontal component of the normal force. I believe you are thinking of inclined planes. Trust me, I've had this heated discussion with my teacher which went on for over a double period! Try not to think about it too much, it just gets too confusing.

[golden]

i don't get it... aren't those two forces in different directions? Fnet is horizontal, so is Frcos(theta), but Nsin(theta)/mgsin(theta) is down the slope.....
Are there any legit past exam questions that have asked this? need to look at one

is actually the horizontal component of the normal force. I believe you are thinking of inclined planes. Trust me, I've had this heated discussion with my teacher which went on for over a double period! Try not to think about it too much, it just gets too confusing.

Could you please provide a diagram? You will get over 9000 respect from everyone for it!

[ttn]

I posted one in the other thread.

Here it is again.