WEBVTT
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This is College Physics Answers
with Shaun Dychko.
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Welcome to Question 30!
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You should be congratulating yourself for
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clicking "Play" on this video
because it's a long one
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and it's very interesting especially
when we get to part (b)
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when we have friction so fasten
your seat belt, here we go!
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In part (a), we are told that
there's no friction
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and we are asked to find what
the ideal speed
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for this car to go around this curve—
it's a banked curve—
00:00:28.380 --> 00:00:32.080
and the ideal speed being
the speed at which
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there's no sliding up or down the ramp
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and there being no friction.
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So we have our coordinate system oriented
such that the y-axis is straight up
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and the x-axis is straight to the side...
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that's a little bit different than other
incline questions we have done
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where we have tilted the x and y axis
to be along the ramp
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and perpendicular to the ramp respectively
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but we are not doing that here;
00:00:55.160 --> 00:00:58.120
instead we have straight up and
straight to the side.
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So that means this normal force now has
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a y-component straight upwards and
a x-component straight to the side
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and there are no other forces
acting on this car.
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This car—I mean I drew it as a box but
you can imagine that it's a car—
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and it's going away from us in this picture
or towards us, hard to tell.
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Okay!
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So we know that the x-component of this
normal force is gonna be providing
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the centripetal force—
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it is the centripetal force, in other words—
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and the y-component of this normal force
has to exactly balance
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gravity which is straight down
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and so that's what we said here
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so the y-component of the normal force
equals gravity.
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So that is the normal force, F N,
multiplied by cosine of Θ
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because this is the adjacent leg of
this right triangle
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and we use cos Θ multiplied by
the hypotenuse
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to get this adjacent leg.
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And then gravity is mg
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and we can solve for the normal force
and say that it's mg over cos Θ
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and I did that just because it's probably
useful for the next part
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in the x-direction
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because it's gonna be another equation
involving the normal force
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and if we have two equation's and
two things that we don't know
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then we'll be able to combine
the two equation's
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and solve for our answer.
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Okay!
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So the x-component of the normal force is
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mass times acceleration—this is
Newton's second law.
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You could put a little subscript c here
to say it's centripetal acceleration
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but it is acceleration just the same
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and we have a formula for it though
in the centripetal case
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which is the speed squared divided by
the radius of curvature.
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So we also have
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a substitution for the x-component of
the normal force
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which is to say that it's F N
times sin Θ
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because it's the opposite leg of
this triangle.
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So we substitute both of these
parts in black
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and we do that in red in
this blue equation.
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So acceleration has been replaced with
v squared over r
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and the x-component of the normal force
has been replaced with F Nsin Θ.
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And then we solve for v squared
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by multiplying both sides by r over m
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and we get the v squared is
rF Nsin Θ divided by m.
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And so here's an equation with two unknowns:
we don't know v and we don't know F N
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and so we need to return to the work
we did here
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and make a substitution for F N
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and it's mg over cos Θ.
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So we have rmg over cos Θ times
sin Θ over m
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and these m's cancel
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and sin Θ divided by cos Θ is
the same as tangent Θ
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and so we have v squared is rgtan Θ
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and so we can take the square root
of both sides to solve for v.
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So v is square root rgtan Θ
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which is a square root of 100 meters times
9.80 meters per second squared times
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tan of 15 degrees which is
16.2 meters per second.
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Now in part (b), we say suppose
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a driver who is scared of the curve
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takes it at a speed less than
16.2 meters per second;
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they take it at 20 kilometers per hour
which after converting
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into meters per second by multiplying
by 1 hour for every 3600 seconds
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and by 1000 meters for every kilometer,
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it is 5.5555 meters per second which is
significantly less than this.
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And what that will mean is that, you know,
if there was no friction
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then this centripetal force,
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which you know is going to be the same
regardless of the speed of the car,
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it's gonna be enough to actually
move it too much
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towards the center of the circle and
it will accelerate it
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more than necessary to go in in its curve
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and so it will slide down the ramp
as a result.
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So to compensate for that, we need
a friction force which
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is always parallel to the interface between
the two surfaces
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and so it's going to be up and
parallel to the ramp,
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or the banked curve I should say,
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and this friction force has a x-component
and a y-component.
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Now because it has a y-component,
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it's gonna be adding to the normal force
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and these two together are going to
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compensate for the gravity downwards.
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And this friction force in
the x-component...
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I haven't drawn things to scale here and
this x-component
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is significantly larger to the left
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than this one is to the right because
there is a net force
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horizontally here, there is acceleration—
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this car is accelerating towards the center
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but also moving along its curve at
the same time
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though there has to be a net
centripetal force
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in order for this car to go around
the curve.
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So aside from that the drawing's good but
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just keep in mind that this x-component
of normal force
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is meant to be longer than
the x-component of friction.
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Okay!
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So considering the y-direction, I mentioned
that the y-component of normal force
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plus y-component of friction upwards
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minus the gravity downwards,
they have to balance
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and so their total is going to be zero.
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You know this is Newton's second law
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and here's mass times acceleration but
there's no acceleration vertically
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and so we write zero here
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and then we can replace each of these terms
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by saying that the y-component of
the normal force is
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F N times cos Θ
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because it's the adjacent leg of
this right triangle
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and the friction force is
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the coefficient of static friction times
the normal force
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but we want the y-component of it
and so we multiply by sin Θ
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because this is the opposite leg
of this triangle
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and then minus mg downwards
all that equals 0.
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And that's about all we can do with this
consideration of the y-direction
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and so then we have to turn our attention
to the x-direction.
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Ultimately, we are solving for this
by the way, that's our goal
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and right now, we have two unknowns:
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one being the normal force and
the other being μ
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so that means we need a second equation.
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So we look at the x-direction and
we say that...
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I have x going positive to the left, it's
just the way I drew it and that's fine
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but just be aware that it's
not your convention
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of having x positive to the right.
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So we have this normal force x-component
positive to the left
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and then minus this x-component of
friction which is negative to the right
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and this being the adjacent leg,
you need to have cosine of Θ
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multiplied by the hypotenuse and
the hypotenuse is μ sF N.
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So we have μ sF Ncos Θ is
the x-component of friction
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and we have F N times sin Θ
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is the x-component of the normal force
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and all this equals mass times acceleration
00:08:00.720 --> 00:08:02.900
and there is acceleration in
the x-direction—
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it is centripetal acceleration towards
the center of the curve
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that the car is traveling in
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and that acceleration is
v squared over r—
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so we replace 'a' there as well.
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So now we are gonna solve this for μ
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and so we'll add μ sF Ncos Θ
to both sides
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and then we'll also subtract mv squared
over r from both sides
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and then switch the sides around and
we have this line here:
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μ sF Ncos Θ equals
F Nsin Θ
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minus mv squared over r.
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And then divide both sides by these factors
that are multiplying by the μ s
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and you get that the coefficient of
static friction then is
00:08:48.600 --> 00:08:50.100
the normal force times sin Θ
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minus mv squared over r
all divided by F Ncos Θ.
00:08:53.740 --> 00:08:57.060
Now this isn't the end of the story
00:08:57.060 --> 00:08:58.740
because we don't know what F N is
00:08:58.820 --> 00:09:01.160
so we are gonna make
a substitution for F N
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and in preparation for that, let's
clean things up a little bit—
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it's messy to have a fraction
within a fraction—
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so let's divide both terms by
this denominator
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and then that will make things
a little bit simpler.
00:09:15.740 --> 00:09:20.320
So F Nsin Θ divided by F Ncos Θ
is gonna be sin Θ over cos Θ
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because the F N's will cancel
00:09:21.840 --> 00:09:24.040
but then you have to divide this
second term by
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F Ncos Θ as well
00:09:25.620 --> 00:09:29.400
and we'll have mv squared over
F Nrcos Θ;
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nothing really changed except that the term
became separated from this term here.
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And then sin Θ over cos Θ can be written
more simply as tan Θ—
00:09:39.360 --> 00:09:42.780
that's a trigonometric identity that is
worth memorizing.
00:09:44.040 --> 00:09:48.960
So now we have a formula for μ s here
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and F N appears only one place—
that's nice—
00:09:52.320 --> 00:09:53.860
because if we are gonna substitute for it,
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we would like to have it in only
00:09:55.860 --> 00:10:00.240
do that substitution in only one place and
we do know based on our work up here
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that the substitution's gonna be
a little bit big.
00:10:05.480 --> 00:10:09.980
So let's return our attention to
this y-direction
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and that is gonna reappear here
00:10:14.340 --> 00:10:16.140
where I did a little bit of work to it:
00:10:16.140 --> 00:10:21.100
I factored out the normal force and then
moved the mg to the right hand side.
00:10:21.520 --> 00:10:23.780
So mg's on the right hand side,
factored out the normal force
00:10:23.780 --> 00:10:27.320
and it's cos Θ plus μ ssin Θ
00:10:27.860 --> 00:10:29.920
and then divide both sides by this bracket
00:10:29.920 --> 00:10:34.040
and you get the normal force is mg
over cos Θ plus μ ssin Θ.
00:10:34.260 --> 00:10:37.160
So that is what we are gonna
substitute into F N
00:10:37.160 --> 00:10:40.700
in this formula for the coefficient
of static friction.
00:10:41.820 --> 00:10:46.460
Now instead of putting this fraction in
the denominator of this fraction—
00:10:46.460 --> 00:10:47.940
that's messy—
00:10:47.940 --> 00:10:53.280
we are gonna instead multiply this fraction
by the reciprocal of the denominator.
00:10:53.280 --> 00:10:56.300
So dividing by something is the same as
multiplying by its reciprocal
00:10:56.300 --> 00:10:58.480
so we can flip this fraction
00:10:58.480 --> 00:11:03.520
and multiply by it instead of
dividing by it.
00:11:03.880 --> 00:11:07.400
So we have μ s equals tan Θ
00:11:07.400 --> 00:11:09.580
minus mv squared over rcos Θ
00:11:09.580 --> 00:11:11.640
and now here we have the multiplying by
00:11:11.640 --> 00:11:14.300
the flipped version of the normal force
00:11:14.380 --> 00:11:17.000
so that's cos Θ plus μ ssin Θ
over mg
00:11:17.000 --> 00:11:21.220
and if you don't like this multiplying by
the reciprocal, you don't have to
00:11:21.220 --> 00:11:26.740
you could substitute in this fraction
00:11:26.740 --> 00:11:31.860
as is into there if you like but I just think
that's gonna be messy.
00:11:33.180 --> 00:11:34.360
Okay!
00:11:34.600 --> 00:11:41.840
So now our job is to collect the μ s
together on one side...
00:11:41.840 --> 00:11:46.160
you see that this substitution introduced
another instance of
00:11:46.160 --> 00:11:48.300
the variable we are trying to solve for
00:11:48.400 --> 00:11:50.500
so that means we have to do
a bunch of algebra work
00:11:50.580 --> 00:11:54.320
to again isolate this μ s alone
on one side.
00:11:54.960 --> 00:12:02.440
So let's divide the top's...
00:12:02.440 --> 00:12:04.000
there's two terms to consider here:
00:12:04.000 --> 00:12:05.600
there's mv squaredcos Θ
00:12:05.600 --> 00:12:08.120
and there's mv squaredμ ssin Θ.
00:12:08.120 --> 00:12:09.460
Those are two different terms
00:12:09.500 --> 00:12:13.480
you know, you get that after you distribute
this into the brackets here
00:12:13.840 --> 00:12:15.560
keeping in mind that there's a minus there
00:12:15.560 --> 00:12:20.740
so this makes minus mv squaredcos Θ
and minus mv squaredμ ssin Θ.
00:12:20.980 --> 00:12:23.960
Then you are dividing both those terms by
00:12:24.300 --> 00:12:28.340
mgrcos Θ but the m's cancel
by the way
00:12:28.340 --> 00:12:30.340
so you don't have to worry about
the m's actually.
00:12:30.800 --> 00:12:34.840
So we expect to have an rgcos Θ
00:12:35.960 --> 00:12:37.840
below both of our terms though
00:12:37.840 --> 00:12:40.200
and here we have rgcos Θ and
we have rgcos Θ—
00:12:40.200 --> 00:12:41.980
so we are checking our work that's good—
00:12:41.980 --> 00:12:45.520
and then distribute the v squared
into both terms
00:12:45.520 --> 00:12:47.320
and we have v squaredcos Θ—that's good—
00:12:47.320 --> 00:12:50.780
and we have v squared
times μ ssin Θ
00:12:50.820 --> 00:12:52.800
and we have minuses in front of
both of them.
00:12:53.240 --> 00:12:57.360
Alright and then cancel this cos Θ there
00:12:57.360 --> 00:13:00.760
and turn the sin Θ over cos Θ
into a tan Θ
00:13:01.100 --> 00:13:07.200
and then move this term to the left side
by adding it to both sides
00:13:08.180 --> 00:13:12.980
and we have μ s plus μ sv squaredtan Θ
over rg equals
00:13:12.980 --> 00:13:15.880
tan Θ minus v squared over rg.
00:13:16.200 --> 00:13:21.640
And then I think another cleanup task
is to get rid of all the fractions
00:13:21.640 --> 00:13:24.460
because they are gonna make things
complicated looking
00:13:24.520 --> 00:13:30.160
so I'm going to multiply every single thing
or both sides, in other words, by rg
00:13:30.200 --> 00:13:32.920
and this is just an aesthetic change;
00:13:32.920 --> 00:13:37.740
it's a personal preference, strategy
to get rid of fractions
00:13:37.740 --> 00:13:39.740
because I don't wanna have
fractions within fractions.
00:13:40.140 --> 00:13:43.000
Because we are gonna factor out
the μ s here and then if we did it
00:13:43.000 --> 00:13:45.400
without doing the 'times rg',
we would have
00:13:45.460 --> 00:13:49.840
μ s times 1 plus v squaredtan Θ
over rg
00:13:49.840 --> 00:13:54.400
and then we would have to divide all this
by that strange fraction. Okay!
00:13:54.500 --> 00:13:58.160
Anyway so we multiply both sides by rg
that makes μ srg here,
00:13:58.160 --> 00:14:01.300
this term becomes μ sv squaredtan Θ
00:14:01.300 --> 00:14:03.100
because the rg's will cancel in that instance
00:14:03.100 --> 00:14:06.900
and here we have rgtan Θ minus
v squared.
00:14:07.340 --> 00:14:10.500
Now we factor out the μ s from
these two terms
00:14:10.580 --> 00:14:13.740
and so we have rg plus
v squaredtan Θ leftover
00:14:13.740 --> 00:14:16.060
and then the right hand side
is unchanged
00:14:16.060 --> 00:14:18.960
and then we divide both sides
by this bracket
00:14:19.320 --> 00:14:24.500
and lastly, we have a formula for
the coefficient of static friction:
00:14:24.640 --> 00:14:26.920
it's gonna be rgtan Θ minus v squared
00:14:26.920 --> 00:14:29.720
over rg plus v squaredtan Θ
00:14:29.880 --> 00:14:31.360
then we plug in numbers.
00:14:31.440 --> 00:14:34.020
So we have 100 meters times 9.80
meters per second squared
00:14:34.020 --> 00:14:36.180
times tan of 15.0 minus this
00:14:36.180 --> 00:14:39.860
slow speed of 5.5555 meters
per second squared
00:14:40.040 --> 00:14:42.600
divided by 100 meters times 9.80
meters per second squared
00:14:42.600 --> 00:14:46.000
plus that speed squared times
tan Θ, or tan 15.0,
00:14:46.080 --> 00:14:48.820
and we get 0.234
00:14:48.820 --> 00:14:51.340
is the coefficient of static friction needed
00:14:51.340 --> 00:14:52.980
to prevent the car from slipping
00:14:53.020 --> 00:14:57.560
when they go at this speed slower
than the ideal speed.