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

A 1.4 kg block and a 2.7 kg block are attached to opposite ends of a light rope. The rope hangs over a solid, frictionless pulley that is 25 cm in diameter and has a mass of 0.72 kg. When the blocks are released, what is the acceleration of the lighter block?

I think I’ve figured out that the force equation for the 1.4 kg block is

T1-m1g=m1a,

and for the 2.7 kg block:

T2-m2g=m2a.

I believe the torque equation for the pulley is torque=rF, where F would be (T2+T1), I think. And that is set equal to I*alpha, or (mr^2)*(a/r).

But, I really have no idea what to do beyond that. I tried solving for a by rearranging the force equations to equal T1 and T2, and then plugged in those values into T1 and T2 in the torque equation, but my answer was wrong. Any help or clarification would be greatly appreciated, and will get the best answer.

• 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.