WEBVTT
00:00:00.000 --> 00:00:02.840
This is Giancoli Answers
with Mr. Dychko.
00:00:03.920 --> 00:00:08.020
Since this tree sapling is in equilibrium
it means all the forces balance out.
00:00:08.300 --> 00:00:10.420
So I have chosen a coordinate system
00:00:10.480 --> 00:00:13.320
so that one axis is along one of the forces
00:00:13.320 --> 00:00:16.480
we'll choose *x*-axis to be along force A
00:00:16.540 --> 00:00:20.740
and that will reduce our work a little bit
because we don't have to worry
00:00:20.740 --> 00:00:22.740
about any trigonometry when we are
dealing with force A—
00:00:22.740 --> 00:00:24.560
it's entirely in the *x*-direction.
00:00:25.140 --> 00:00:29.360
Now the force's balancing out means
they balance out separately
00:00:29.360 --> 00:00:32.360
in the *x*-axis and along the *y*-axis
00:00:32.440 --> 00:00:34.400
so we'll have two different equations
00:00:34.600 --> 00:00:39.020
and so we can consider this *x*-component
for force B up here
00:00:39.580 --> 00:00:43.740
pointing straight to the left
that's *F B x*
00:00:43.800 --> 00:00:50.040
and that added to the *x*-component
of force C, this is *F C x*
00:00:50.120 --> 00:00:57.300
those two have to equal the total *x*-forces
going to the right which is just *F A*.
00:00:57.840 --> 00:01:01.800
So that's where this equation comes from
and then we can solve for
00:01:01.800 --> 00:01:08.620
the *x*-component of force C by subtracting
*F B x* from both sides.
00:01:10.160 --> 00:01:15.460
And then we'll replace *F B x* with
*F BsinΘ B*
00:01:15.460 --> 00:01:18.680
where I have taken *Θ B* to be this
angle between
00:01:18.760 --> 00:01:21.520
the vertical *y*-axis and the force
00:01:21.740 --> 00:01:29.940
and so we are gonna calculate that as
being 105 minus 90, just 15 degrees.
00:01:32.560 --> 00:01:37.400
And *F B x* is the opposite leg of
this triangle so we take
00:01:37.400 --> 00:01:42.580
*sin* of *Θ B* to get the opposite leg
multiplied by the hypotenuse *F B*.
00:01:43.700 --> 00:01:44.720
Okay.
00:01:44.760 --> 00:01:47.160
And then the same idea for the *y*-direction
00:01:47.220 --> 00:01:49.080
and in the *y*-direction, there are only
00:01:49.080 --> 00:01:52.680
two forces to consider since force A is
entirely in the *x*-direction.
00:01:52.880 --> 00:02:00.040
So the component of force B that's vertical,
upwards, this is *F B y*
00:02:00.040 --> 00:02:05.040
has to equal the component of force C
which is downwards, that's *F C y*.
00:02:05.620 --> 00:02:11.020
So *F C y* is *F B* times *cos Θ B*—
00:02:11.240 --> 00:02:17.200
*cos* is what we use to find the adjacent
leg of the triangle—
00:02:17.380 --> 00:02:25.040
and here I mention again that *Θ B* is
105 degrees from force A
00:02:25.120 --> 00:02:28.080
minus the 90 degrees here
00:02:28.180 --> 00:02:32.480
and that gives us just a little sliver that's
in the corner that slice
00:02:32.480 --> 00:02:34.680
in this triangle here
00:02:36.960 --> 00:02:41.520
and now we can calculate each of these
components for force C.
00:02:41.600 --> 00:02:46.440
So force C in the *x*-direction is
force A, 385 newtons,
00:02:46.540 --> 00:02:49.320
minus 475 times *sin* 15
00:02:49.420 --> 00:02:53.660
so I just substituted into this formula
just now with numbers here
00:02:54.620 --> 00:02:57.700
and that works out to 262.06 newtons.
00:02:57.820 --> 00:03:00.040
And then force C in the *y*-direction
00:03:00.040 --> 00:03:05.780
is force B, 475 newtons, times *cos* 15
which is 458.81.
00:03:06.100 --> 00:03:08.700
So those are the two components for force C
00:03:08.780 --> 00:03:11.680
and then defined the hypotenuse...
00:03:11.920 --> 00:03:16.680
see what we figured out so far is,
we found out that force C
00:03:16.940 --> 00:03:22.040
goes downwards this much, 458.81 newtons,
00:03:22.200 --> 00:03:28.800
and it goes to the right this much,
*F C x*, 262.06 newtons
00:03:28.900 --> 00:03:33.700
that makes a right triangle with this
resultant being the hypotenuse
00:03:34.180 --> 00:03:36.600
and we use Pythagoras to calculate it.
00:03:36.600 --> 00:03:38.120
So force C is the square root of
00:03:38.120 --> 00:03:40.560
*F C x squared* plus *F C y squared*
00:03:40.940 --> 00:03:44.320
so that's the square root of
262.06 newton squared
00:03:44.320 --> 00:03:46.840
plus 458.81 newton squared
00:03:46.840 --> 00:03:49.760
that gives 528.38.
00:03:49.840 --> 00:03:55.000
And then to figure out the angle and
that is just the sliver in here, *Θ C*,
00:03:55.100 --> 00:04:00.380
that's the part between the vertical axis
and the force
00:04:01.700 --> 00:04:04.020
so we are gonna have to add
90 to our answer here
00:04:04.260 --> 00:04:09.540
is gonna be the inverse tangent
of the opposite which is *F C x*
00:04:09.640 --> 00:04:12.280
divided by the adjacent *F C y*.
00:04:13.080 --> 00:04:18.280
And that's inverse tangent of 262.06
divided by 458.81 newtons
00:04:18.280 --> 00:04:20.280
this gives 29.73.
00:04:20.540 --> 00:04:26.360
And we'll call that *Θ* or the answer
that the question is looking for
00:04:26.360 --> 00:04:31.580
is the angle all the way from force A
to force C so we'll just call that *Θ*
00:04:31.580 --> 00:04:35.120
and that's gonna be *Θ C* plus 90.
00:04:37.740 --> 00:04:43.860
So that's 90 plus 29.73, 129.73 should have
three significant figures
00:04:43.860 --> 00:04:47.360
because you know the forces
we were given are
00:04:47.360 --> 00:04:51.060
385 and 475 and they had
three significant figures so
00:04:51.160 --> 00:04:54.240
we'll have to round this to three significant
figures which makes it into 130.
00:04:54.440 --> 00:04:58.700
So our final answer is the force C is
528 newtons,
00:04:58.700 --> 00:05:01.060
130 degrees and then we have to
00:05:01.060 --> 00:05:04.320
say how this angle is measured because
00:05:04.320 --> 00:05:06.000
it's not in standard position.
00:05:06.000 --> 00:05:08.740
If you just said 130 degrees and
that's all you said
00:05:08.740 --> 00:05:11.760
then the assumption would have to be
that it's in standard position
00:05:11.760 --> 00:05:14.300
going counter-clockwise from
the positive *x*-axis
00:05:14.300 --> 00:05:15.660
which is not the case
00:05:15.700 --> 00:05:18.100
so we have to be specific and say
that our angle is
00:05:18.100 --> 00:05:21.680
measured clockwise from force A.