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Answer:
Explanation:
mass of pulley, m3 = 1.5 kg
Radius of pulley, R = 0.09 m
mass of monkey, m2 = 4.5 kg
mass of banana bunch, m1 = 3 kg
Let a is teh acceleration ans T1 and T2 be the tension in the rope.
The moment of inertia of the pulley
I = 0.5 x m3 x R² = 0.5 x 1.5 x 0.09 x 0.09 = 0.006075 kgm²
According to Newton's second law
T1 - m1 g = m1 x a .... (1)
m2 g - T2 = m2 x a ..... (2)
(T2 - T1 ) x R = I x α
where, α is the angular acceleration
α = a / R
(T2 - T1)R = 0.5 x m3 x R² x a / R
T2 - T1 = 0.5 x m3 x a ..... (3)
from (1), (2) and (3)
[tex]a = \frac{m_{2}-m_{1}}{m_{1}+m_{2}+\frac{m_{3}}{2}}\times g[/tex]
[tex]a = \frac{4.5-3}{3+4.5+0.75}\times 9.8[/tex]
a = 1.78 m/s²
from equation (1)
T1 = m1 ( g + a) = 3 ( 9.8 + 1.78) = 34.77 N
from equation (2)
T2 = m2 (g - a) = 4.5 (9.8 - 1.78) = 36.13 N
Downward Acceleration of The monkey is = 36.13 N
The tension in both ends of the rope is = 34.77 N
How to Calculation Acceleration?
The mass of pulley, m3 is = 1.5 kg
Then the Radius of the pulley, R is = 0.09 m
After that mass of monkey, m2 is = 4.5 kg
Then the mass of banana bunch, m1 = 3 kg
Now, Let be the acceleration T1 and also T2 be the tension in the rope.
The moment of inertia of the pulley
Then = [tex]0.5 x m3 x R² = 0.5 x 1.5 x 0.09 x 0.09 = 0.006075 kgm²[/tex]
According to Newton's second law
[tex]T1 - m1 g = m1 x a .... (1)[/tex]
[tex]m2 g - T2 = m2 x a ..... (2)[/tex]
(T2 - T1 ) x R = I x α
where, α is the angular acceleration
α = a / R
[tex](T2 - T1)R = 0.5 x m3 x R² x a / R[/tex]
T2 - T1 = 0.5 x m3 x a ..... (3)
from (1), (2) and (3)
a = m₂ - m₁ / m₁+m₂ + m₃/₂ ₓ g
a = 4.5 - 3/ 3+4.5+0.75 х 9.8
a = 1.78 m/s²
from equation (1)
[tex]T1 = m1 ( g + a) = 3 ( 9.8 + 1.78) = 34.77 N[/tex]
from equation (2)
[tex]T2 = m2 (g - a) = 4.5 (9.8 - 1.78) = 36.13 N[/tex]
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