Arope of negligible mass passes over a uniform cylindrical pulley of 1.50 kg mass and 0.090 m radius. the bearings of the pulley have negligible friction, and the rope does not slip on the pulley. on one end of the rope hangs a 3.00 kg bunch of bananas, and on the other end hangs a 4.50 kg monkey. calculate the downward acceleration of the monkey and the tension in both ends of the rope.

1.8 m/s²

36 N

34.8 N

Explanation:

For the monkey :

m₁ = mass of monkey = 4.50 kg

T₁ = Tension force in the rope on monkey's side

a = acceleration

From the force diagram, force equation for the motion of monkey is given as

m₁ g - T₁ = m₁ a

(4.50 x 9.8) - T₁ = 4.5 a

T₁ = 44.1 - 4.5 a                                             eq-1

For the bunch of bananas :

m₂ = mass of bunch of bananas = 3 kg

T₂ = tension force in the rope on the side of banana

From the force diagram, force equation for the motion of bananas is given as

T₂ - m₂ g = m₂ a

T₂ - (3 x 9.8) = 3 a

T₂ = 29.4 + 3 a                                            eq-2

m = mass of the pulley = 1.50 kg

r = radius of the pulley = 0.090 m

α = angular acceleration of pulley = a/r

Torque equation for the pulley is given as

(T₁ - T₂ )r = I α

(T₁ - T₂ )r = I (a/r)

T₁ - T₂  = (0.5 m r²) (a/r²)

T₁ - T₂  = (0.5) ma

using eq-1 and eq-2

44.1 - 4.5 a  - (29.4 + 3 a) = (0.5) ma

44.1 - 4.5 a  - (29.4 + 3 a) = (0.5) (1.50) a

a = 1.8 m/s²

Using eq-1

T₁ = 44.1 - 4.5 a    

T₁ = 44.1 - 4.5 (1.8)    

T₁ = 36 N

using eq-2

T₂ = 29.4 + 3 a

T₂ = 29.4 + 3 (1.8)

T₂ = 34.8 N

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