In the pulley system shown the movable pulleys A,B and C have mass m each, D and E are fixed pulleys. The strings are vertical, light and inextensible. Then Pulleys A and B have acceleration each in downward direction and pulley C have acceleration in upward direction Then the value of N is
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Answers
M = 5 Kg
m = 3 Kg
m' = 2 Kg
Mg - = 5a -----(1)
30 + = 3a -----(2)
- 20 = 2a -----(3)
= 50 - 5a
Putting this in equation (2)
30 + 50 - 5a - = 3a
80 - = 8a
= 80 - 8a
Putting this in equation (3)
80 - 8a - 20 = 2a
10a = 60
a = 6 ms-².
= 50 - 30
= 20 N
= 80 - 48
= 32 N
Answer:
The tension in the string is 6.67N.
Explanation:
Since all strings are connected to each other, it will have some tension at all points. Let it be T.
Considering the acceleration of the pulley A, B, and C is a₁, a₂, and a₃.
According to constrain equation,
ΣT.a = 0
-2Ta₃-Ta₂-Ta₁ = 0
2a₃ + a₂ + a₁ = 0
Applying Newton's law on each block, we get
10 - 2T = a₃ ...(i)
10 - T = a₂ ...(ii)
10 - T = a₁ ...(iii)
From equation (i) and (ii), we get
-T = a₃ - a₂
T = a₂ - a₃
Substituting in (ii)
10 - a₂ + a₃ = a₂
2a₂ = 10 + a₃
On solving, we get
a₁ = a₂ = 10/3, a₃ = -10/3
T = a₂ - a₃
T = 10/3 - (-10/3)
T = 10/3 + 10/3
T = 20/3
T = 6.67N
Therefore, the tension in the string is 6.67N.
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