Vector Method: T₁₂ 80° 60° Fg T₁ 40° <3 marks sin 80° sin 60° F g T₁ sin 60° T F sin 80° g -0.879-35 kg-9.8 m/s² -3.0 x 10² N 1 marks F mg (35 kg)(9.8 N/kg) - 343 N - 1 mark = sin 80° sin 40° F g Τ2 T₁-2.2 × 10² N 1 marks 001phk -6- February 24, 2000 3. A 35 kg traffic light is suspended from two cables as shown in the diagram. Cable 1 100° 140° 120° Cable 2 What is the tension in each of these cables? Component Method: 001phk T₁ T₂ 50° 30° - 2 marks cos 30° ΣΕ = 0 T₁ cos 50°-7, cos 30° T-T ΣΕ, = 0 cos 30° cos 50° T₁ sin 50° +T2 sin 30° = 35(9.8) cos 50° sin 50° +T₂ sin 30° = 343 343 T:- 1.03+0.5 T₂ = 224 N T₁ = 224 cos 30° cos 50° = 302 N (7 marks) -5- February 24, 2000
Vector Method: T₁₂ 80° 60° Fg T₁ 40° <3 marks sin 80° sin 60° F g T₁ sin 60° T F sin 80° g -0.879-35 kg-9.8 m/s² -3.0 x 10² N 1 marks F mg (35 kg)(9.8 N/kg) - 343 N - 1 mark = sin 80° sin 40° F g Τ2 T₁-2.2 × 10² N 1 marks 001phk -6- February 24, 2000 3. A 35 kg traffic light is suspended from two cables as shown in the diagram. Cable 1 100° 140° 120° Cable 2 What is the tension in each of these cables? Component Method: 001phk T₁ T₂ 50° 30° - 2 marks cos 30° ΣΕ = 0 T₁ cos 50°-7, cos 30° T-T ΣΕ, = 0 cos 30° cos 50° T₁ sin 50° +T2 sin 30° = 35(9.8) cos 50° sin 50° +T₂ sin 30° = 343 343 T:- 1.03+0.5 T₂ = 224 N T₁ = 224 cos 30° cos 50° = 302 N (7 marks) -5- February 24, 2000
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![Vector Method:
T₁₂
80°
60°
Fg
T₁
40°
<3 marks
sin 80°
sin 60°
F
g
T₁
sin 60°
T
F
sin 80°
g
-0.879-35 kg-9.8 m/s²
-3.0 x 10² N
1 marks
F mg (35 kg)(9.8 N/kg) - 343 N - 1 mark
=
sin 80°
sin 40°
F
g
Τ2
T₁-2.2 × 10² N
1 marks
001phk
-6-
February 24, 2000](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd08a6c19-dc79-4368-8e35-820b976eb472%2F6266dc25-0d55-4841-8434-8cecf9439f0b%2Fmppic1m_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Vector Method:
T₁₂
80°
60°
Fg
T₁
40°
<3 marks
sin 80°
sin 60°
F
g
T₁
sin 60°
T
F
sin 80°
g
-0.879-35 kg-9.8 m/s²
-3.0 x 10² N
1 marks
F mg (35 kg)(9.8 N/kg) - 343 N - 1 mark
=
sin 80°
sin 40°
F
g
Τ2
T₁-2.2 × 10² N
1 marks
001phk
-6-
February 24, 2000
![3. A 35 kg traffic light is suspended from two cables as shown in the diagram.
Cable 1
100°
140° 120°
Cable 2
What is the tension in each of these cables?
Component Method:
001phk
T₁
T₂
50°
30°
- 2 marks
cos 30°
ΣΕ = 0
T₁ cos 50°-7, cos 30°
T-T
ΣΕ, = 0
cos 30°
cos 50°
T₁ sin 50° +T2 sin 30° = 35(9.8)
cos 50° sin 50° +T₂ sin 30° = 343
343
T:- 1.03+0.5
T₂ = 224 N
T₁ = 224 cos 30°
cos 50°
= 302 N
(7 marks)
-5-
February 24, 2000](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd08a6c19-dc79-4368-8e35-820b976eb472%2F6266dc25-0d55-4841-8434-8cecf9439f0b%2Fquz9el_processed.jpeg&w=3840&q=75)
Transcribed Image Text:3. A 35 kg traffic light is suspended from two cables as shown in the diagram.
Cable 1
100°
140° 120°
Cable 2
What is the tension in each of these cables?
Component Method:
001phk
T₁
T₂
50°
30°
- 2 marks
cos 30°
ΣΕ = 0
T₁ cos 50°-7, cos 30°
T-T
ΣΕ, = 0
cos 30°
cos 50°
T₁ sin 50° +T2 sin 30° = 35(9.8)
cos 50° sin 50° +T₂ sin 30° = 343
343
T:- 1.03+0.5
T₂ = 224 N
T₁ = 224 cos 30°
cos 50°
= 302 N
(7 marks)
-5-
February 24, 2000
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