€ = Problem2 Starting at rest, a 2500kg helicopter accelerates straight up at a constant 1.7m/s². The helicopter can be modeled as a 2.6m diameter sphere. [ C= 0.5, p = 1.2 kg/m³] a. Draw a free body diagram u= 0 29 is the helicopter's height at the moment its blades are providing an upward force of 29kN? 29000 a) v=183.2728 1-7 a = 1. v² = u²12aS (183.2720) 2. = 1-3m To A = R₁² = 4= 0 a=1-7 m/s² S = 9879 m b) Zry = Fg - D²³ Ex(3) 169 (66 70 29000 = mg - 1/2 CPAV² ✓29090 2 2 b==
€ = Problem2 Starting at rest, a 2500kg helicopter accelerates straight up at a constant 1.7m/s². The helicopter can be modeled as a 2.6m diameter sphere. [ C= 0.5, p = 1.2 kg/m³] a. Draw a free body diagram u= 0 29 is the helicopter's height at the moment its blades are providing an upward force of 29kN? 29000 a) v=183.2728 1-7 a = 1. v² = u²12aS (183.2720) 2. = 1-3m To A = R₁² = 4= 0 a=1-7 m/s² S = 9879 m b) Zry = Fg - D²³ Ex(3) 169 (66 70 29000 = mg - 1/2 CPAV² ✓29090 2 2 b==
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Is my working method correct? I am also confused ablut the free body diagrams. Do i have to show the net force for that?
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Problem2
Starting at rest, a 2500kg helicopter accelerates straight up at a constant 1.7m/s². The helicopter
can be modeled as a 2.6m diameter sphere. [C=0.5, p = 1.2 kg/m³ ]
a. Draw a free body diagram
b. What is the helicopter's height at the moment its blades are providing an upward force of
29000
a)
u= 0
v = 183.2728
a= 1-7
V *²=u²+2 as
1-3m
то
رادی
A=ær²
2
1
(66
6=0
= πC X (1-3) ²
(69
TC
a=1-7 m/s
b) zfy = Fg - D²³
Fy
- 29000 = mg - 1/2 CAV²
√29090 =
2
b = 12/²2 CPAV ²
24500-2 (05) (16²
(06
BE
-53500
:: 9880m 1/
сраи
- 29000=mg - = CPAV²
(1.2)
69
(!
- 29000 = 24500- 2 (0.5) (100 x) V²
169
+83500
2
V = 2
1/83528=2415427842
765
(10²3.27 202 = 0f 2C1-7)5 /53500 = /1.592781²
va (83.27 m/s
S = 9 879 m
2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F104d41eb-f02b-4e6d-903d-dd6e6944a268%2F5a6625e9-57cd-43d6-bccc-6a8d1b3e25ec%2Ftsgnc4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:S=
€=
Problem2
Starting at rest, a 2500kg helicopter accelerates straight up at a constant 1.7m/s². The helicopter
can be modeled as a 2.6m diameter sphere. [C=0.5, p = 1.2 kg/m³ ]
a. Draw a free body diagram
b. What is the helicopter's height at the moment its blades are providing an upward force of
29000
a)
u= 0
v = 183.2728
a= 1-7
V *²=u²+2 as
1-3m
то
رادی
A=ær²
2
1
(66
6=0
= πC X (1-3) ²
(69
TC
a=1-7 m/s
b) zfy = Fg - D²³
Fy
- 29000 = mg - 1/2 CAV²
√29090 =
2
b = 12/²2 CPAV ²
24500-2 (05) (16²
(06
BE
-53500
:: 9880m 1/
сраи
- 29000=mg - = CPAV²
(1.2)
69
(!
- 29000 = 24500- 2 (0.5) (100 x) V²
169
+83500
2
V = 2
1/83528=2415427842
765
(10²3.27 202 = 0f 2C1-7)5 /53500 = /1.592781²
va (83.27 m/s
S = 9 879 m
2
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