Question

a car is traveling at 50.0 mi/h on a horizontal highway. if the coefficient of static friction between road and tires on a rainy day is 0.100, what is the minimum dstance in which the car will stop?

how do i approach this problem?

Answer 1

Retarding force,
F=\(\mu\)N
or, ma = \(\mu\) mg
or, a=\(\mu\)g=0.100 \(\times\) 9.8=0.98 mi/h^2

So, the retarding acceleration is 0.98 mi/h^2

Now,
\(v^2=u^2-2as\) .....(-ve sign to indicate that the acceleration is retarding)
Final velocity v is zero, since the car finally comes to rest.
so,
0=15^2 -2*0.98* s
s=110.25 miles..

Hence, the car will stop in 110.25 miles

Answer 2

can you explain what you did @ujjwal ? i mean step-by step...it's pretty hard when you cram everything

Answer 3

first of all, friction force is being applied on the tires because of which the car is decelerating. then as explained by ujjwal, calculate the deceleration of the car. now substitute that value in v^2 = u^2 + 2as to find out s, which is what u need.

Answer 4

wait...we do FBD first right?

Answer 5

we don't need an FBD in everything...this sum was quite straightforward...and u don't need to resolve any forces. so u don't need to draw FBD.

Answer 6

F=μN

what does that mean?

Answer 7

that's the formula for frictional force. u = coefficient of friction and N = normal force.

Answer 8

remember one thing. we use FBD to figure out the net force acting on a body. its a vector diagram.

Answer 9

oh...is this the same as \[\large f_k = \mu_k F_N?\]

Answer 10

yup

Answer 11

so this is kinetic friction?

Answer 12

oh wait it's static...

Answer 13

that's right.

Answer 14

no...its kinetic.

Answer 15

but the formula for static friction is
\[\large f_s \le \mu_s F_N\]
how does he know it's eqaul and not less?

and the problem says static friction...

Answer 16

@ujjwal shouldn't it be kinetic? since the tires are in motion?

Answer 17

@ujjwal got it.
"An example of static friction is the force that prevents a car wheel from slipping as it rolls on the ground. Even though the wheel is in motion, the patch of the tire in contact with the ground is stationary relative to the ground, so it is static rather than kinetic friction." got it from WIKI.

Answer 18

Lost my internet for a while..

Static friction will work here since the wheels are rolling. They are not sliding!

Answer 19

what about @Igbasallote's question?

Answer 20

yeah...

Answer 21

i guess this answers it:
The maximum possible friction force between two surfaces before sliding begins is the product of the coefficient of static friction and the normal force: .

Answer 22

uhmm layman's terms?

Answer 23

and in this case, the tires are not sliding. hence static friction.

Answer 24

u know in about pure rolling?

Answer 25

The frictional force will be less than \(\mu_s\)N as long as the body is not in motion.

Answer 26

well...it is said in the problem that it is static hehe

so how do you know if you use
\[f_s = \mu_s F_N\]
or
\[f_s < \mu_s F_N\]

Answer 27

Once the body is in motion, F=\(\mu_s\)N

Answer 28

yea....that explains it.

Answer 29

that applies always when in motion?

Answer 30

yup

Answer 31

so what's the next step? it's determined that \[f_s = \mu_s F_N\]
now what?

Answer 32

but from Newton's 2nd law, F = ma.....sub this in place of frictional force. then proceed how ujjwal showed b4.

Answer 33

\[f_s = F = ma???\]

Answer 34

yes

Answer 35

@lgbasallote Answer to your above question:
No. That applies in case of motion where the body is rolling and you can use \(\mu_s\)N for frictional force. Else you have to use coefficient of kinetic friction.

Answer 36

but mass and acceleration are not given...how will that help?

Answer 37

you are supposed to find acceleration and mass cancels out as i have done it.

Answer 38

@Igbasallote: whenever the body is in motion, u use F = uN.....but w.r.t the question u need to understand which friction is in play - static or kinetic....as ujjwal pointed out.

Answer 39

wait..\[F_N = F_g??\]

Answer 40

nooo...Fn = normal force

Answer 41

In this case \(F_N=mg\)

Answer 42

@ujjwal quote "Once the body is in motion, F=μsN"
This is not correct.

Here, the tire is not slipping, so F<µsN but, as we want the most effective braking, we will brake so that F is maximum, so F will be nearly equal to µsN.

This is why the equations will be solved using µsN = µs mg for the retarding force.

Answer 43

@Vincent-Lyon.Fr Isn't rolling friction equal to static friction?

Answer 44

Main friction, when rolling is static.

"Rolling friction" is something ELSE that is always neglected in such problems.

Answer 45

I mean your remark is ok, but you should just avoid using the phrase "rolling friction" which is incorrect.

Answer 46

okay...i get the acceleration...explanation for the second equation?

Answer 47

@Vincent-Lyon.Fr When the body is rolling, the friction which comes into play is static friction but then you say F<\(\mu\)s N..
Shouldn't it be F=\(\mu\)s N

Answer 48

@Igbasallote then u just have to substitute value of a in v^2 = u^2 + 2as.....like how ujjwal showed above.

Answer 49

lol people need to stop tagging igbasallote..it's LGBASALLOTE

Answer 50

ohhh.. now wonder ur name was not showing up!

Answer 51

@ujjwal
it is F≤μs N in the general case, and F=μs N in this example, because we are looking for the shortest braking distance, which implies the maximum braking force.

Answer 52

I guess i need to revise my concepts on friction.. Thanks @Vincent-Lyon.Fr ..

Answer 53

that second equation is similar to \[V^2 = V_o^2 + 2a(x_2 - x_1)\]
right?

Answer 54

yes!

Answer 55

why 15^2 shouldnt it be 50^2?

Answer 56

right...that should be 50^2

Answer 57

yeah it should be 50^2.. sorry, my mistake.

Answer 58

okay i tink i got the process in this one..onto more practices

Answer 59

best of luck, buddy!

Answer 60

gr8....work hard!