A Car Is Traveling at 51.0 Mi/h on a Horizontal Highway.
Aid with Friction and Stopping Car
- Thread starter Husker70
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Homework Statement
A car is traveling at 50.0 mi/h on a horizontal highway. (a) If the coefficient of
static friction betwixt route and tires on a rainy mean solar day is .100, what is the minimum
distance in which the automobile will finish? (b) What is the stopping distance when
the surface is dry and coefficient of static friction is .600?
ii. Homework Equations [/]The Attempt at a Solution
I'm not sure how to get started as I don't know the normal force, and I'one thousand not
sure what to solve for. If someone can help get the ball rolling I should be able
to end. Any aid would be appreciated.
Kevin
Homework Statement
Homework Equations
The Attempt at a Solution
Answers and Replies
A car is traveling at l.0 mi/h on a horizontal highway. (a) If the coefficient of
static friction between road and tires on a rainy 24-hour interval is .100, what is the minimum
distance in which the motorcar will cease? (b) What is the stopping altitude when
the surface is dry and coefficient of static friction is .600?
…
I'm not sure how to get started as I don't know the normal force, and I'grand not
sure what to solve for.
Hi Husker70!
Call the mass of the car chiliad (it will abolish out afterward ).
Then the normal forcefulness (on a horizontal surface) is just mg.
Now use F = ma to find the acceleration … which will be constant, so you can so use the usual uniform accleration kinetic equations.
slove for the acceleration, I'm non sure how to find the force.
Cheers,
Kevin
… in order to
slove for the dispatch, I'grand not sure how to detect the force.
Hi Kevin!
Since information technology'due south a horizontal route, the only force slowing information technology down is the friction forcefulness …
then multiply the normal strength by µ.
50 mi/h = 80.4672 km/h = 22.352 thou/southward
a= .100n
ma = .100(k)
a = .98
solving for time
v=vo + at
22.352m/due south=0+.98(t)
t=22.81s
solving for distance
d=exercise+vot+at^2/2
d= 0+0+.98(22.815)^2/ii
d= 255m
Thanks for everyone'south help
Kevin
Right reply, long-winded method …
Does this seem right?
50 mi/h = 80.4672 km/h = 22.352 m/sa= .100n
ma = .100(grand)
a = .98
a = .98 is right, but the other lines should be:
ma= .100n
a = .100(g)
solving for time
five=vo + at
22.352m/s=0+.98(t)
t=22.81ssolving for altitude
d=do+vot+at^two/2
d= 0+0+.98(22.815)^two/2
d= 255m
hmm … you have fivef and vi and you need s but not t …
so it would have been a lot quicker just to use fivef 2 = vi 2 + 2as
I am working on the same trouble, but I practise not comprehend the work that was done to solve this trouble.
Problem statement and variables given:
Q: A car is traveling fifty.0 mi/h on a horizontal highway. (a.) If the coefficient of static friction betwixt road and tires on a rainy mean solar day is 0.100, what is the minimum altitude in which the auto will terminate? (b.) What is the stopping distance when the surface is dry and mu(southward)=0.600? (mu(due south) I think means static friction)
Attempt:
What I did was draw a gratis-trunk diagram where the weight force(earth is acting upon the motorcar) is pointing down, normal force(road interim upon the motorcar) is pointing up, and there is a friction force going to the left. I did catechumen 50.0mi/60 minutes to 22.35m/s, just later on that I am lost.
Cheers,
Joyce
Hello, Joyce! Welcome to PF!
(have a mu: µ )
What I did was depict a complimentary-torso diagram where the weight force(globe is acting upon the car) is pointing down, normal force(road interim upon the auto) is pointing upwardly, and there is a friction strength going to the left. I did convert 50.0mi/hr to 22.35m/s, but after that I am lost.
i] First, find the friction force.
two] So notice the dispatch.
Would the friction forcefulness be: F=-w+Due north (friction equals negative weight force plus normal strength)? I recollect I might be confused on how to obtain the friction force.
Thank you,
Joyce Kuang
Would the friction force exist: F=-w+N (friction equals negative weight force plus normal force)?
That is a valid equation, but information technology's the equation for the internet vertical strength …
in this instance the vertical acceleration is cypher,
so (from skillful ol' https://www.physicsforums.com/library.php?practice=view_item&itemid=26") mass times 0 = net force = -W + N.
so W = Due north …
in other words, when something stays on a flat surface (and so that its vertical dispatch is zero), so the normal force equals the weight.
(yous can get a similar equation for the normal force on a gradient, past doing 0 = net perpendicular force)
ie friction = µ times normal forcefulness
This works either for kinetic friction (µone thousand), or for maximum static friction (µsouthward) …
in this case, the static friction is maximum, since the question says so (because it asks for the minimum stopping altitude).
What I have so far is:
Summation of F=m(0)=net force=-w+N
In which you said due west=North
So you said to utilize F=[itex]\mu[/itex] N
I know that [itex]\mu[/itex] is 0.100
so, F=(.100)N
What is N? I know that it is equal to the weight force, but it does not give me a value for the weight or normal force. How tin can I obtain that value?
And subsequently finding the normal force, then I know the friction force, right?
After finding the friction strength why practise I have to find the acceleration? Is it considering nosotros demand acceleration to discover the minimum distance?
Thank you,
Joyce
The normal strength (N in newtons) is equal to the weight force.
Y'all said to use "grand" to correspond the mass, but what practise I do after that?
What do I do with my F=[itex]\mu[/itex]N equation? How does that relate to the mass?
Mayhap that is where my problem lies. I practise not meet a connexion between the mass and the equation, or I do non empathize.
Cheers,
Joyce
You know that F=[itex]\mu[/itex]N. You also know that Northward = W = mg.
Now, gravity is just an acceleration in the y-direction, correct?
And then, at present you know that N = ma[itex]_{y}[/itex]
Yous already knew that F=[itex]\mu[/itex]N. Now F, in this situation, is friction force. So, information technology is a strength in the x-direction. Then, given that, y'all could say that.....
ma[itex]_{ten}[/itex]=[itex]\mu[/itex]ma[itex]_{y}[/itex]
Since the mass is the same for both sides, the masses cancel out (since you would divide both sides by g). That leaves you with......
a[itex]_{x}[/itex]=[itex]\mu[/itex]a[itex]_{y}[/itex]
From that you can solve for a[itex]_{x}[/itex] and then apply kinematics to solve for [itex]\Delta[/itex]10.
Bask,
Alex
Cheers, I got it.
What I did was:
F=[itex]\mu[/itex]southNorthward
N=mg
N-may (g is ay)
max=[itex]\mu[/itex]s(may)
a=(.100)(-nine.8)= -0.98
Vi=22.35 chiliad/s
Vf=0 m/s
x=?
So accept the equation Vf 2=Vi two +2ax
Solved for ten, which gives you lot
(Vf 2-5i ii)/(2a)
ten=255m?
Cheers very much,
Joyce
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