A guy-wire supports a pole that is 75 ft high. One end of the wire is attached to the top of the pole and the other end is anchored to the ground 50 ft from the base of the pole. Determine the horizontal and vertical components of the force of tension in the wire if its magnitude is 50 lb . (Round to the nearest integer.)
A guy-wire supports a pole that is 75 ft high. One end of the wire is attached to the top of the pole and the other end is anchored to the ground 50 ft from the base of the pole. Determine the horizontal and vertical components of the force of tension in the wire if its magnitude is 50 lb . (Round to the nearest integer.)
A guy-wire supports a pole that is
75
ft
high. One end of the wire is attached to the top of the pole and the other end is anchored to the ground
50
ft
from the base of the pole. Determine the horizontal and vertical components of the force of tension in the wire if its magnitude is
50
lb
.
A: Tan Latitude / Tan P
A = Tan 04° 30'/ Tan 77° 50.3'
A= 0.016960 803 S CA named opposite to latitude,
except when hour angle between 090° and 270°)
B: Tan Declination | Sin P
B Tan 052° 42.1'/ Sin 77° 50.3'
B = 1.34 2905601 SCB is alway named same as
declination)
C = A + B = 1.35 9866404 S CC correction, A+/- B:
if A and B have same name - add, If
different name- subtract)
=
Tan Azimuth 1/Ccx cos Latitude)
Tan Azimuth = 0.737640253
Azimuth
=
S 36.4° E CAzimuth takes combined
name of C correction and Hour Angle - If LHA
is between 0° and 180°, it is named "west", if
LHA is between 180° and 360° it is named "east"
True Azimuth= 143.6°
Compass Azimuth = 145.0°
Compass Error = 1.4° West
Variation 4.0 East
Deviation: 5.4 West
A: Tan Latitude / Tan P
A = Tan 04° 30'/ Tan 77° 50.3'
A= 0.016960 803 S CA named opposite to latitude,
except when hour angle between 090° and 270°)
B: Tan Declination | Sin P
B Tan 052° 42.1'/ Sin 77° 50.3'
B = 1.34 2905601 SCB is alway named same as
declination)
C = A + B = 1.35 9866404 S CC correction, A+/- B:
if A and B have same name - add, If
different name- subtract)
=
Tan Azimuth 1/Ccx cos Latitude)
Tan Azimuth = 0.737640253
Azimuth
=
S 36.4° E CAzimuth takes combined
name of C correction and Hour Angle - If LHA
is between 0° and 180°, it is named "west", if
LHA is between 180° and 360° it is named "east"
True Azimuth= 143.6°
Compass Azimuth = 145.0°
Compass Error = 1.4° West
Variation 4.0 East
Deviation: 5.4 West
Direction: Strictly write in 4 bond paper, because my activity
sheet is have 4 spaces. This is actually for maritime.
industry course, but I think geometry can do this.
use nautical almanac.
Sample Calculation (Amplitude- Sun):
On 07th May 2006 at Sunset, a vesel in position 10°00'N
0 10°00' W observed the sun bearing 288° by compass. Find
the
compass error.
LMT Sunset
07d
18h
13m
(+)00d
00h
40 м
LIT:
UTC Sunset:
07d
18h
53 m
added - since
longitude is
westerly
Declination Co7d 18h): N016° 55.5'
d(0.7):
(+)
00-6
N016 56.1'
Declination Sun:
Sin Amplitude Sin Declination (Los Latitude
- Sin 016° 56.1'/Cos 10°00'
= 0.295780189
Amplitude = WI. 2N (The prefix of amplitude is
named easterly if body is rising.
and westerly of body is setting.
The suffix is named came as
declination.)
True Bearing: 287.20
Compass Bearing
288.0°
Compass Error: 0.8' West