When the blocks are released, what is the acceleration of the lighter block?

m₁ = mass 1 = 1.4 kg

m₂ = mass 2 = 2.7 kg

m₃ = mass of the pulley = 0.72 kg

R = radius of the pulley = 0.25 m

g = accelerayion by gravity = 9.8 m/s²

First, you need to ‘choose’ a direction in which the system will accelerate. As m₂ is bigger than m₁, it seems logic to choose m₂ moving down, and m₁ moving up.

(Note: If you choose wrong, that’s not a problem. You will end up with a negative value for acceleration so you will know that the system moves in the other direction than the one you chose)

The forces acting on m₁ are:

* tensile force in the rope, pulling up

* force of gravity, pulling down

As m₁ is moving up, you know that the tensile force is bigger than the gravitational force. The net force on m₁ will be:

Fnet₁ = T₁ – m₁×g

By Newton’s second law, you can say:

T₁ – m₁×g = m₁×a

T₁ = m₁×g + m₁×a . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . equation 1

The forces acting on m₂ are:

* tensile force in the rope, pulling up

* force of gravity, pulling down

As m₂ is moving down, you know that the gravitational force is bigger than the tensile force. The net force on m₂ will be:

Fnet₂ = m₂×g – T₂

By Newton’s second law, you can say:

m₂×g – T₂ = m₂×a

T₂ = m₂×g – m₂×a . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . equation 2

The moment of inertia of the pulley (assuming it is a solid disk) is:

I = m₃×R²/2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . equation 3

The equation for torque on the pulley is:

T₂×R – T₁×R = I×a/R

Rearranging this a bit, you get:

(T₂ – T₁)×R = I×a/R

T₂ – T₁ = I×a/R² . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . equation 4

Substituting equation 3 into equation 4, you get:

T₂ – T₁ = (m₃×R²/2)×a/R²

T₂ – T₁ = m₃×a/2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . equation 5

(note how the radius R cancels out)

Now substituting equations 1 and 2 into equation 5, you get:

(m₂×g – m₂×a) – (m₁×g + m₁×a) = m₃×a/2

m₂×g – m₂×a – m₁×g – m₁×a = m₃×a/2

m₂×g – m₁×g = m₃×a/2 + m₂×a + m₁×a

(m₂ – m₁)×g = (m₃/2 + m₂ + m₁)×a

a = g × (m₂ – m₁) / (m₃/2 + m₂ + m₁)

And finally, entering the numbers, you get:

a = (9.8 m/s²) × [(2.7 kg) – (1.4 kg)] / [(0.72 kg)/2 + (2.7 kg) + (1.4 kg)]

a = 2.8565… m/s²

a = 2.9 m/s² (to 2 sf)

(T2+T1)… This is causing the issue.

for 1.4 Kg block I have F2=m2g so we have T2 – m2g = ma

for 2.7 Kg block I have F1=m1g and m1g – T1 = ma

The torque is caused F1=m1g because its has more mass which caused the other block to accelerate upward.