anonymous
  • anonymous
The equation below gives the maximum velocity in miles per hour that a vehicle can safely travel around a curve of radius r feet when friction is f. If the velocity is greater than Vmax, the tires will slip. Engineers find that under snowy conditions, Vmax = 15 miles per hour for a freeway off-ramp that has a radius of 50 feet. To the nearest tenth, what is the coefficient of friction for the off-ramp in these conditions?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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anonymous
  • anonymous
open up a new question
anonymous
  • anonymous
@mertsj can you do something about him?

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Mertsj
  • Mertsj
I can.
anonymous
  • anonymous
ok first step DRAW A FREE BODY DIAGRAM
anonymous
  • anonymous
always
anonymous
  • anonymous
4 years of physic classes taught me that :P
anonymous
  • anonymous
a free body diagram? Not familiar with that. This is from an Alg 2 class.
anonymous
  • anonymous
oh well, its a force equation Frictional force is defined as \[F_{friction}\le \mu N\] where mu is the coefficient of friction and N is the normal force normal force in this case would be the weight of the of the car
anonymous
  • anonymous
now, because the car is travelling around a curve with radius 50 ft it experiences what is known as centrifugal force and it can be written as \[F= m \frac{v^2}{r}\] where m is the mass of the object v is the velocity the object is travelling around the curve and r is the radius
anonymous
  • anonymous
now to draw the free body diagram this is a picture of the car from the top |dw:1370570858292:dw|
anonymous
  • anonymous
|dw:1370570909826:dw| this is just showing the motion of hte car and is not part of the free body diagram due to this motion, it causes the car to experience a force in this direction |dw:1370571028403:dw|
anonymous
  • anonymous
now, frictional force counteracts this centrifugal force the frictional force would be in this direction |dw:1370571093497:dw|
anonymous
  • anonymous
now, if the centrifugal force is greater than the force of friction, then the car will slip in order for the car to not slip, the force of friction must be equal to the centrifugal force
anonymous
  • anonymous
so basically, the sum of forces must be equal to zero in order for the car to not move/ slip giving us this equation \[F_{friction}=F_{centrifugal}\]
anonymous
  • anonymous
now, we know the equation for friction and for centrifugal force so we substitute that back into the equation \[\mu N = m \frac{v^2}{r}\]
anonymous
  • anonymous
now, before i said that the normal force is equal to the weight of the car the weight of the car can be defined as \[W= mg\] where m is mass and g is the gravitational constant - 32.2ft/s^2
anonymous
  • anonymous
and \[N= W\] this can also be determine using a FBD
anonymous
  • anonymous
there are alot of concepts you really need to look up so im just kinda quickly just going through the process of solving
anonymous
  • anonymous
subsituting weight into the equation we get \[\mu m g = m \frac{ v^2}{r}\]
anonymous
  • anonymous
Okay, but...I'm doing math...Algebra 2..I'm just trying to figure out the coefficient for friction, I'm sorry but all this physics stuff is kinda confusing me.
anonymous
  • anonymous
now then, looking at hte equation, we see that m can be divided out so we do that giving us \[\mu g = \frac{ v^2}{r}\]
anonymous
  • anonymous
just bear with me, we're almost at the end now since mu is coefficient of friction, you just need to solve for mu and this can be accomplished by dividing g to both sides to give us \[\mu= \frac{v^2}{g r}\]
anonymous
  • anonymous
now we have the equation to solve for the coefficent of friction in relation to speed of the car and radius of the curve
anonymous
  • anonymous
so from this point, you use the equation i derived to solve for mu and on a side note, watch your units, they should all cancel out
anonymous
  • anonymous
what i mean is, you need to convert mi/h to ft/s
anonymous
  • anonymous
How would I do that?
anonymous
  • anonymous
google?
anonymous
  • anonymous
or you could do unit conversions
anonymous
  • anonymous
3600s = 1 h 1m = 5280 ft
anonymous
  • anonymous
ahh okay
anonymous
  • anonymous
saving you the trouble 15mph =22 ft/s
anonymous
  • anonymous
g = 32.2 ft/s^2 gravitational constant, you could easily look this up
anonymous
  • anonymous
r = 50 ft
anonymous
  • anonymous
i hope i didnt make a mistake somewhere, i kind of speeded through all this
anonymous
  • anonymous
and its long, kinda the reason why i wanted you to open a new question oh well
anonymous
  • anonymous
So then my coefficient of friction is 32.2?
anonymous
  • anonymous
is it? i know g- gravitational constant is 32.2 ft/s^2 dont think mu would be 32.2 its usually less than 1
anonymous
  • anonymous
you need g to solve for mu- the little symbol thats in the equation that i was deriving
anonymous
  • anonymous
\[\mu \ne g\]
anonymous
  • anonymous
so 22/32.2? then to get my answer?
anonymous
  • anonymous
...
anonymous
  • anonymous
\[\mu = \frac{v^2}{g r}\] v= 22 ft/s g= 32.2 ft/s^2 r= 50 ft

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