mechanic (1 Viewer)

facebooknerd123 · Jul 28, 2015

a car of mass 2 tonnes is rounding curve of radius 840m on level track at 90km/hr show that force of friction between wheels and ground is 1488N
isn't F = uN? u = friction coefficient.

Drongoski · Jul 28, 2015

A mechanic is the guy who fixes your car.

InteGrand · Jul 28, 2015

facebooknerd123 said:
a car of mass 2 tonnes is rounding curve of radius 840m on level track at 90km/hr show that force of friction between wheels and ground is 1488N
isn't F = uN? u = friction coefficient.

$The force of friction between the wheels and the ground provides the centripetal force, so is equal in magnitude to $\frac{mv^2}{r}$, where $m=2\text{ t}=2000\text{ kg}$, $v=90\text{ km/h}=90\times \frac{1000}{3600}\text{ m/s}=25\text{ m/s}$, and $r=840\text{ m}$. Substitution of these values shows that the force has magnitude $1488$ N (nearest integer).$

facebooknerd123 · Jul 28, 2015

InteGrand said:
$The force of friction between the wheels and the ground provides the centripetal force, so is equal in magnitude to $\frac{mv^2}{r}$, where $m=2\text{ t}=2000\text{ kg}$, $v=90\text{ km/h}=90\times \frac{1000}{3600}\text{ m/s}=25\text{ m/s}$, and $r=840\text{ m}$. Substitution of these values shows that the force has magnitude $1488$ N (nearest integer).$

why isnt it uN?

Drongoski said:
A mechanic is the guy who fixes your car.

btw that was hilarious

InteGrand · Jul 28, 2015

facebooknerd123 said:
why isnt it uN?

btw that was hilarious

$The maximum frictional force that can be provided \emph{is} $F_\text{max.}=\mu N$, but we don't need to know or worry about this to solve the problem as stated.$

facebooknerd123 · Jul 28, 2015

InteGrand said:
$The maximum frictional force that can be provided \emph{is} $F_\text{max.}=\mu N$, but we don't need to know or worry about this to solve the problem as stated.$

so F = uN is still correct? what do you mean by maximum friction force isn't mv^2 /r already the max? im confused

Librah · Jul 28, 2015

facebooknerd123 said:
so F = uN is still correct? what do you mean by maximum friction force isn't mv^2 /r already the max? im confused

Centripetal force is provided by the friction of the wheels.

InteGrand · Jul 28, 2015

facebooknerd123 said:
so F = uN is still correct? what do you mean by maximum friction force isn't mv^2 /r already the max? im confused

$The quantity $\mu N$ refers to the maximum frictional force that can be provided. So for example, if $N=20000\text{ N}$ (roughly the value for a 2-tonne car) and $\mu = 0.72$ (the value given for car tires on asphalt on \url{http://www.engineeringtoolbox.com/friction-coefficients-d\_778.html}), the maximum frictional force providable would be $F_\text{max.}=0.72\times 20000\text{ N} = 14400\text{ N}$ (anything lower can also be provided). So this means that if the car tries to take a turn that would require \emph{more} than this for the centripetal force (which would happen if the car is travelling too fast for the given radius of the turn), then the car would swerve out (thus increasing the radius of turn, making the required centripetal force lower because $F_c \propto \frac{1}{r}$), to a new radius $r_2$, such that the centripetal force required at this new radius is less than or equal to $F_\text{max.}$ (and hence can be provided by the friction between the tires and the road).$

$You may have experienced this yourself or understand it intuitively if turning in a vehicle too fast on a turn. But all this knowledge about friction etc. falls more in the realms of Physics and is not really required for HSC maths exams these days (but I recall it being quite useful for a banked turn HSC Q from the pre-2000's or 2000, so it's better if you do know it, though the exams are easier these days).$

facebooknerd123 · Jul 28, 2015

InteGrand said:
$The quantity $\mu N$ refers to the maximum frictional force that can be provided. So for example, if $N=20000\text{ N}$ (roughly the value for a 2-tonne car) and $\mu = 0.72$ (the value given for car tires on asphalt on \url{http://www.engineeringtoolbox.com/friction-coefficients-d\_778.html}), the maximum frictional force providable would be $F_\text{max.}=0.72\times 20000\text{ N} = 14400\text{ N}$ (anything lower can also be provided). So this means that if the car tries to take a turn that would require \emph{more} than this for the centripetal force (which would happen if the car is travelling too fast for the given radius of the turn), then the car would swerve out (thus increasing the radius of turn, making the required centripetal force lower because $F_c \propto \frac{1}{r}$), to a new radius $r_2$, such that the centripetal force required at this new radius is less than or equal to $F_\text{max.}$ (and hence can be provided by the friction between the tires and the road).$

$You may have experienced this yourself or understand it intuitively if turning in a vehicle too fast on a turn. But all this knowledge about friction etc. falls more in the realms of Physics and is not really required for HSC maths exams these days (but I recall it being quite useful for a banked turn HSC Q from the pre-2000's or 2000, so it's better if you do know it, though the exams are easier these days).$

so mv^2 /r is still the max? or is just the required? does this show that even if you travel around a corner with like 1000N, increasing your speed by like 0.001 m/s wouldn't cause u to swerve away?

InteGrand · Jul 28, 2015

facebooknerd123 said:
so mv^2 /r is still the max? or is just the required? does this show that even if you travel around a corner with like 1000N, increasing your speed by like 0.001 m/s wouldn't cause u to swerve away?

$The quantity $\frac{mv^2}{r}$ is equal to the magnitude of the \emph{required} centripetal force for a turn of radius $r$ by an object of mass $m$ travelling at speed $v$. Since we are given that this turn is occurring, we know that the centripetal force is given by $\frac{mv^2}{r}$, using $m$, $v$ and $r$ values pertaining to the question, and that the frictional force between the road and tyres is able to provide this (if it weren't able to, then the turn wouldn't be being successfully completed, contradicting the fact that it \emph{is} being successfully completed). It is irrelevant to the question at hand to consider whether or not this is near the ``breaking point'' (i.e. max. providable force), nor do we have enough information to consider this (since we aren't given $\mu$ and have no way to calculate it here).$

facebooknerd123 · Jul 28, 2015

InteGrand said:
$The quantity $\frac{mv^2}{r}$ is equal to the magnitude of the \emph{required} centripetal force for a turn of radius $r$ by an object of mass $m$ travelling at speed $v$. Since we are given that this turn is occurring, we know that the centripetal force is given by $\frac{mv^2}{r}$, using $m$, $v$ and $r$ values pertaining to the question, and that the frictional force between the road and tyres is able to provide this (if it weren't able to, then the turn wouldn't be being successfully completed, contradicting the fact that it \emph{is} being successfully completed). It is irrelevant to the question at hand to consider whether or not this is near the ``breaking point'' (i.e. max. providable force), nor do we have enough information to consider this (since we aren't given $\mu$ and have no way to calculate it here).$

could we find a possible value for 'u' though? like
F max = umg
umg >= mv^2 /r
u >= v^2 / gr

InteGrand · Jul 28, 2015

facebooknerd123 said:
could we find a possible value for 'u' though? like
F max = umg
umg >= mv^2 /r
u >= v^2 / gr

$What you've done there is show what $\mu$ must be greater than or equal to. We can't actually find what precise value $\mu$ is, but we can find a lower bound, which is what you've done.$

Drongoski · Jul 29, 2015

InteGrand said:
$The quantity $\mu N$ refers to the maximum frictional force that can be provided. So for example, if $N=20000\text{ N}$ (roughly the value for a 2-tonne car) and $\mu = 0.72$ (the value given for car tires on asphalt on \url{http://www.engineeringtoolbox.com/friction-coefficients-d\_778.html}), the maximum frictional force providable would be $F_\text{max.}=0.72\times 20000\text{ N} = 14400\text{ N}$ (anything lower can also be provided). So this means that if the car tries to take a turn that would require \emph{more} than this for the centripetal force (which would happen if the car is travelling too fast for the given radius of the turn), then the car would swerve out (thus increasing the radius of turn, making the required centripetal force lower because $F_c \propto \frac{1}{r}$), to a new radius $r_2$, such that the centripetal force required at this new radius is less than or equal to $F_\text{max.}$ (and hence can be provided by the friction between the tires and the road).$

$You may have experienced this yourself or understand it intuitively if turning in a vehicle too fast on a turn. But all this knowledge about friction etc. falls more in the realms of Physics and is not really required for HSC maths exams these days (but I recall it being quite useful for a banked turn HSC Q from the pre-2000's or 2000, so it's better if you do know it, though the exams are easier these days).$

Another masterly piece of unstinting help from InteGrand!

This is a simple but subtle idea that most people without a little background in physics/mechanics find confusing. The 'uN' as Grand has explained is the maximum frictional force that can be generated. So slippage occurs when the frictional force, depended upon to provide some or all of the required centripetal force, is insufficient (i.e. exceeds the 'uN'). The other simple idea causing confusion is: the frictional force seems to be able to act in 2 directions, say up a slope and down the slope. In fact friction has no fixed direction; it acts in the direction opposing motion; i.e. it acts in the direction opposing (opposite to) the direction of resultant (nett) velocity.

Drongoski · Jul 31, 2015

InteGrand said:
$What you've done there is show what $\mu$ must be greater than or equal to. We can't actually find what precise value $\mu$ is, but we can find a lower bound, which is what you've done.$

But $\mu$ is a characteristic of the surface; a rough surface will have a higher value than a smoother one. It's value is found empirically, I'd imagine.

There are also the coeff. of static friction and the coeff. of kinetic friction, if this rusty brain of mine remembers correctly. They are supposed to differ.

Edit

On reflection, the friction between 2 surfaces ought to be a function of the coeffs. of both surfaces rather than that of the $\mu$ of one surface alone. Wonder what the actual facts are. Must be well-established by now.

InteGrand · Jul 31, 2015

I don't mean it's impossible to determine a value for μ, I mean that we can't find its value using the data in the question.

braintic · Jul 31, 2015

Drongoski said:
But $\mu$ is a characteristic of the surface; a rough surface will have a higher value than a smoother one. It's value is found empirically, I'd imagine.

There are also the coeff. of static friction and the coeff. of kinetic friction, if this rusty brain of mine remembers correctly. They are supposed to differ.

Edit

On reflection, the friction between 2 surfaces ought to be a function of the coeffs. of both surfaces rather than that of the $\mu$ of one surface alone. Wonder what the actual facts are. Must be well-established by now.

Just to add to that, as long as the car is travelling in a circle, it is the coefficient of static friction that is operating.

But the moment that the car begins to slide, the coefficient of kinetic friction takes over.

But the coefficient of kinetic friction is lower than the coefficient of static friction.

This means that what Integrand said before is not quite right. The car would slide out to a radius at which the coefficient of kinetic friction is just enough to supply the centripetal force. This radius is greater than that suggested by the coefficient of static friction. So when the car stops sliding and static friction again takes over, the greatest possible value of static friction is not being supplied. Which means that there is some leeway for the car to speed up without sliding.

Also, I'm not sure if "the friction between 2 surfaces ought to be a function of the coeffs. of both surfaces" is correct. The coefficient of friction is a value that already depends on the properties of both surfaces in contact. I don't think it is meaningful to talk of the coefficient of friction of say bitumen. It is like one hand clapping. But there would be a coefficient of friction for bitumen in contact with rubber, and another one for when this surface is lubricated with water. I doubt there is some generic function you could apply to the coefficients of each surface (if they existed) to get the combined coefficient. But I haven't looked it up, so I am prepared to be corrected.

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