WEBVTT
00:00:00.300 --> 00:00:02.875
This is College Physics Answers
with Shaun Dychko.
00:00:03.750 --> 00:00:06.875
We’re going to calculate the tension
in each of these segments of rope,
00:00:07.450 --> 00:00:11.325
given that this woman is hanging
with a weight equal to her mass,
00:00:11.320 --> 00:00:12.900
times acceleration due to gravity.
00:00:13.650 --> 00:00:18.375
And this tension has to add up to
zero when combined with the weight.
00:00:18.370 --> 00:00:24.450
So the total force on this woman, because
she’s stationary, has to add up to zero.
00:00:25.150 --> 00:00:28.725
So we have this tension two pulling
in this direction along this rope.
00:00:29.450 --> 00:00:32.775
And we put the tail of tension one
on the head of tension two vector.
00:00:33.250 --> 00:00:35.000
And in that tension one is up like this
00:00:35.825 --> 00:00:39.725
with this angle *theta one*, 15
degrees with respect to the vertical.
00:00:41.550 --> 00:00:45.375
This here is 15 degrees as well, because
these are interior opposite angles
00:00:46.125 --> 00:00:47.825
between two parallel lines.
00:00:49.575 --> 00:00:53.025
So that’s 15 degrees here and
this one is 10 degrees.
00:00:54.175 --> 00:00:58.025
And we have then the tail of the
weight vector straight down,
00:00:58.800 --> 00:01:03.500
and ends up at the place where we
started. Because they add up to zero.
00:01:05.425 --> 00:01:07.625
So we’ll consider the y-direction
00:01:08.150 --> 00:01:11.250
and we’ll take the y-component
of the tension two force
00:01:11.500 --> 00:01:15.625
which is this opposite segment
here. This is *t two y*.
00:01:17.200 --> 00:01:20.875
And because it’s the opposite
segment, we will take sine of this
00:01:21.200 --> 00:01:23.400
angle and multiply it by the
hypotenuse *t two*.
00:01:24.900 --> 00:01:28.725
And then the y-component of
*t one* will be this leg here,
00:01:28.720 --> 00:01:32.400
which is adjacent to the angle
*theta one*.
00:01:34.475 --> 00:01:37.475
So, *t one y* gets multiplied
by cosine of *theta one*
00:01:37.800 --> 00:01:40.525
to get it’s y-component. Both of those
are positive because they’re upwards
00:01:41.300 --> 00:01:45.900
and then minus this weight which is entirely
in the y-direction downwards *m g*
00:01:46.325 --> 00:01:50.175
and all that equals zero. Because there’s
no acceleration, that equals *m a*,
00:01:50.170 --> 00:01:53.125
but I just substituted zero for
*a* to make this zero.
00:01:55.775 --> 00:01:59.125
Okay, and in the x-direction, we have
00:01:59.650 --> 00:02:02.100
the x-component of tension two
00:02:02.775 --> 00:02:04.875
which is the adjacent leg
of this right triangle.
00:02:05.875 --> 00:02:09.175
And, so we use cosine of *theta two*
times *t two* to find it.
00:02:10.150 --> 00:02:13.150
And then that's in the positive
direction. So that makes it
00:02:13.750 --> 00:02:20.400
a positive here and then tension one has
a x-component in the negative direction.
00:02:20.700 --> 00:02:23.675
So we put a minus *t one*
times sine *theta one*.
00:02:23.670 --> 00:02:27.350
Because this is the opposite
leg of this triangle.
00:02:30.150 --> 00:02:33.825
And all of that equals mass times
acceleration, but acceleration being zero
00:02:33.875 --> 00:02:34.750
and just put zero here.
00:02:36.100 --> 00:02:39.250
Now we have two equations and two
unknowns *t two* and *t one*.
00:02:39.900 --> 00:02:43.825
So well solve this x-direction
equation for *t two*,
00:02:44.425 --> 00:02:48.675
and we’ll add *t one* sine
*theta one* to both sides.
00:02:51.950 --> 00:02:54.850
And then divide both sides
by cosine *theta two*
00:02:55.300 --> 00:02:58.675
and we end-up with *t two* equals *t one*
sine *theta one* over cos *theta two*.
00:02:59.275 --> 00:03:01.775
And this is useful because now
we can substitute this
00:03:02.150 --> 00:03:05.675
into our y-direction equation and
replace *t two* with all of this.
00:03:06.325 --> 00:03:09.375
And now we have a single equation with
only one unknown, which is *t one*.
00:03:11.050 --> 00:03:15.625
So this is the y-direction equation
rewritten with *t two*
00:03:15.975 --> 00:03:18.675
replaced in red with this expression here.
00:03:20.300 --> 00:03:22.825
So we can factor out *t one*
from both of these two terms
00:03:23.575 --> 00:03:25.725
and we get *t one* times
bracket, sine *theta one*
00:03:25.800 --> 00:03:28.475
times sine *theta two*, over cos
*theta two* plus cos *theta one*.
00:03:29.075 --> 00:03:30.875
And then we add *m g* to both sides.
00:03:32.875 --> 00:03:34.575
And we get *m g* on
the right hand side here.
00:03:34.825 --> 00:03:37.525
And then we divide both sides by this
bracket to solve for *t one*.
00:03:38.550 --> 00:03:40.950
So, *t one* is *m g*
over all of the stuff;
00:03:41.750 --> 00:03:44.550
So that’s 76 kilograms times
9.8 newtons per kilogram
00:03:44.825 --> 00:03:48.025
divided by sine of 15 degrees.
Times sine of 10 degrees,
00:03:48.500 --> 00:03:51.950
divided by cosine of 10 degrees,
plus cosine of 15 degrees.
00:03:52.675 --> 00:03:57.150
So *theta one* is 15 and
*theta two* is 10.
00:03:58.325 --> 00:04:00.475
This works out to 736 newtons.
00:04:01.650 --> 00:04:05.475
Now tension two then we can
return to this expression here
00:04:05.475 --> 00:04:07.470
tension two is tension one
that we just found
00:04:07.750 --> 00:04:09.575
times sine *theta one*
over cos *theta two*.
00:04:10.100 --> 00:04:13.650
So we have this 736.287 newtons
00:04:13.650 --> 00:04:16.925
times sine 15 over cos 10, gives 194 newtons.