A rope is stretched between 2 poles. A 50 N boy hangs on it. Find tension in 2 parts of rope .
Answers
Tension in T₁ and T₂ parts of ropes is accordingly 68.27 and 61.50.
Explanation:
=> According to the question,
Here, F = 50
T₁ sin 30⁰ acts vertically and T1 cos 30⁰ Horizontally upwards.
T₁ sin 15⁰ acts vertically and T1 cos 15⁰ Horizontally upwards.
=> ∴ T₁ cos 30⁰ = T₂ cos 15⁰ (Because the system is in equilibrium)
T₁ = cos 15⁰/cos 30⁰ T₂
= 0.9659/0.8660 T₂
∴ T₁ = 1.11 T₂
=> Mg = T₁ sin 30⁰ + T₂ sin 15⁰
(But, F = Mg = 50 N and T₁ = 1.11T₂)
∴ 50 = 1.11T₂Sin30⁰ + T₂ sin 15⁰
50 = T₂ (1.11*1/2 + 0.258)
50 = T₂ (1.11*0.5 + 0.258)
50 = T₂ (0.555+0.258)
50 = T₂*0.813
T₂ = 50/0.813
T₂ = 61.50
Now as, T₁ = 1.11 T₂
T₁ = 1.11 * 61.50
T₁ = 68.27
Thus, tension in two parts T₁ and T₂ is accordingly 68.27 and 61.50.
Learn more:
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Tension in T₁ and T₂ parts of ropes is accordingly 68.27 and 61.50.
Explanation:
=> According to the question,
Here, F = 50
T₁ sin 30⁰ acts vertically and T1 cos 30⁰ Horizontally upwards.
T₁ sin 15⁰ acts vertically and T1 cos 15⁰ Horizontally upwards.
=> ∴ T₁ cos 30⁰ = T₂ cos 15⁰ (Because the system is in equilibrium)
T₁ = cos 15⁰/cos 30⁰ T₂
= 0.9659/0.8660 T₂
∴ T₁ = 1.11 T₂
=> Mg = T₁ sin 30⁰ + T₂ sin 15⁰
(But, F = Mg = 50 N and T₁ = 1.11T₂)
∴ 50 = 1.11T₂Sin30⁰ + T₂ sin 15⁰
50 = T₂ (1.11*1/2 + 0.258)
50 = T₂ (1.11*0.5 + 0.258)
50 = T₂ (0.555+0.258)
50 = T₂*0.813
T₂ = 50/0.813
T₂ = 61.50
Now as, T₁ = 1.11 T₂
T₁ = 1.11 * 61.50
T₁ = 68.27
Thus, tension in two parts T₁ and T₂ is accordingly 68.27 and 61.50.