Answer:
0
Explanation:
m = Mass of person
g = Acceleration due to gravity = 9.81 m/s²
d = Vertical height from the ground
F = Force = Weight = mg
Net work done would be
Hence, the work done on the person by the gravitational force is 0
The valve in the exit pipe is closed . The density of water is 1000kg / m and the gravitational force on free fall of water is 1
2 answers:
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Refer to the diagram shown below.
The hoist is in static equilibrium supported by tensions in the two ropes.
For horizontal force balance, obtain
T₃ cos 50 = T₂ cos 38
0.6428T₃ = 0.788T₂
T₃ = 1.2259T₂ (1)
For vertical force balance, obtain
T₂ sin 38 + T₃ sin 50 = 350
0.6157T₂ + 0.766T₃ = 350 (2)
Substitute (1) into (2).
0.6157T₂ + 0.766(1.2259T₂) = 350
1.5547T₂ = 350
T₂ = 225.124 N
T₃ = 1.2259(225.124) = 275.979
Answer:
T₂ = 225.12 N
T₃ = 275.98 N
The hoist is in static equilibrium supported by tensions in the two ropes.
For horizontal force balance, obtain
T₃ cos 50 = T₂ cos 38
0.6428T₃ = 0.788T₂
T₃ = 1.2259T₂ (1)
For vertical force balance, obtain
T₂ sin 38 + T₃ sin 50 = 350
0.6157T₂ + 0.766T₃ = 350 (2)
Substitute (1) into (2).
0.6157T₂ + 0.766(1.2259T₂) = 350
1.5547T₂ = 350
T₂ = 225.124 N
T₃ = 1.2259(225.124) = 275.979
Answer:
T₂ = 225.12 N
T₃ = 275.98 N