WEBVTT

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This is Giancoli Answers
with Mr. Dychko.

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The free body diagram of the block A

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has this tension force A directed
up the ramp,

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gravity straight down and the normal force
going perpendicular to the ramp surface.

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Block B has a normal force this way

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gravity straight down and tension force
along the ramp going up into the left

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and then this pulley—I just wrote the forces

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that are involved in our calculation's here,
I didn't bother writing...

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well, I guess I suppose I could,
I mean why not...

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we have gravity straight down the pulley

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and we have—you might call it—
a normal force upwards

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but it's this force applied by this rod that's
attached to the axle on the pulley

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the force of the rod is going upwards

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force of gravity on the pulley downwards
but those forces are

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don't appear in any of our workings
so they are not that important

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so let's just consider these
two tension forces

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applied at some points here,
tangent to the pulley

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and the <i>F</i> tension force B
going down to the right

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and tension force A going down
into the left.

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So to calculate the tension forces,
let's consider mass A first

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then the sum of the forces on mass A

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in the x-direction where we
are gonna consider our

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axis to be tilted like this where
x is positive up into the right

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and y is positive perpendicular
to the surface;

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we are just gonna consider
the x direction because

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that's where the acceleration is happening.

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So we have force tension A
up into the right

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and then we have the component of gravity

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on A in the x direction which is to the left.

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So if we resolve this gravity force
into components we have:

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<i>F g A</i> in the y direction downwards there

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and then we have <i>F g A</i> in the x direction
going to the left—

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that's why we have a minus sign there
because it's to the left—

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and these terms are just magnitudes.

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And this <i>F g A x</i> is gonna be
the force of gravity

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<i>m A</i> times <i>g</i>

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multiplied by <i>sin</i> of <i>Θ A</i> so that's
32 degrees here—

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that's <i>Θ A</i> up in this corner here—

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and <i>F g A x</i> is the opposite leg

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of this triangle here

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so that's why we use <i>sin Θ</i>
to calculate it.

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And this total is gonna be mass of A
times its acceleration

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and we are told the acceleration so
the only thing we don't know

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is the tension force.

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So we'll add this term to both sides
or move it to the right hand side

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however you like to say it

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and it becomes plus.

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So we have force tension A is <i>m A</i>
times acceleration plus

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<i>m A</i> times <i>gsin Θ A</i>

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and then we have <i>m A</i> factored out
and that's gonna be

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8 kilograms times 1 meter per second
squared acceleration plus

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9.8 meters per second squared
<i>g</i> times <i>sin</i> 32

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and that will be about 15 newtons for
the tension force on block A.

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And in the other segment of the line,
we'll have a different tension force B.

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And in chapter 4, we typically had
the same tension force along

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all segments of a rope

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regardless of whether there
was a pulley or not

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but now in this chapter 8 that's a new thing
where we have different tension forces

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on the different sides of the pulley

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because now we consider the pulley
to actually have some mass

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and so it has a moment of inertia and so on
so makes things a little bit different.

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So for this case we have

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we are gonna take the coordinate
system to be like this now

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where x is down into the right

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and y direction is like that

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and so the component of gravity in
the x direction will be positive

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so this part here that's <i>F g B x</i>

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and this is <i>F g B y</i> but we don't care
about that part.

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So that component of gravity along the ramp
<i>m B gsin Θ</i> is positive

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and then minus the tension force B
which is going

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in the negative x direction up the ramp

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and that total is gonna equal mass of B
times its acceleration

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and we can solve this for <i>F T B</i> by moving
it to the right hand side,

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moving this to the left hand side and
then switch the sides around

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and force tension B is <i>m Bg sin Θ</i>
minus <i>m Ba</i> and then

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factor out the mass of B

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<i>gsin Θ</i> minus <i>a</i>

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and then we have 10 kilograms times

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9.8 newtons per kilogram times <i>sin</i> 61
minus 1 meter per second squared

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which gives about 76 newtons will be
the tension force on block B.

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And then for the total torque—

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the total torque will be the counter
clockwise torque minus

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the clockwise torque—

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counter clockwise torque is
a result of <i>F T A</i>

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and so it's <i>F T A</i> times <i>r</i> to get the
total counter clockwise torque

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and then clockwise direction is <i>F T B</i>
times the same radius <i>r</i>.

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So factor out the <i>r</i> and we have
0.15 meters times

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49.5457 newtons minus
75.7127 newtons

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which gives negative 3.9 newton meters;

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the negative sign means that the torque
is clockwise which we expect

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because this whole system is accelerating
this direction we are told

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and so we expect the pulley to be rotating
this way which is clockwise

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so we expected a negative
answer for our torque.

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Now to find the moment of inertia
of the pulley,

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we know that the total torque equals
the moment of inertia

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times the angular acceleration

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and we have this formula here

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relating angular acceleration to
the tangental acceleration;

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we know what the tangental acceleration is

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for a point on the rim of the pulley,

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it's gonna be the same as the acceleration
of the blocks and of the rope

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because the rope is not slipping
on the pulley

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so everything's moving together

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with the same acceleration to the right
including the rim of the pulley.

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So we'll solve for <i>α</i> here by
dividing both sides by <i>r</i>

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and so now we have <i>α</i> in terms of
the tangental acceleration.

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So angular acceleration is tangental
acceleration divided by <i>r</i>

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and so we'll substitute for that here

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and this total torque also is gonna be
the sum of the two torques

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one due to each of these forces.

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Now I have switched the order around here
because we know that it's gonna be

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positive going this direction

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and that's the direction of our acceleration

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and so I have <i>F T B</i> being positive and
<i>F T A</i> being minus

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because it's in the negative x direction.

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The reverse of what I did up here kind of
following two different conventions:

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up here following the usual
convention of having

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counter clock negative and
clockwise positive

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and then down here, I'm thinking more about
this specific situation where we have

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acceleration positive to the right

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and so I put <i>F T B</i> positive and
<i>F T A</i> negative.

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So we can solve this for <i>I</i> by multiplying

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both sides by <i>r</i> over <i>a tan</i>

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and that makes <i>r squared</i> times

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the difference in forces divided by
tangental acceleration.

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So we have 75.713 newtons minus
49.546 newtons

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times 0.15 meter—radius of the pulley—
squared

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divided by 1 meter per second squared

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which gives 0.59 kilogram meter squared
for the moment of inertia of the pulley.