Question

A block of mass m1 = 2.00kg and a block of mass m2 = 6.00kg are connected by a mass less string over a pulley in a shape of a solid disk having radius radius R=0.250m and mass M = 10.0kg. The fixed wedgede shaped ramp makes a angle of theta = 30.0 degrees. The coefficient of kinetic friction is 0.360 for both blocks. (a) draw diagram(b) acceleration of the two blocks(c) tension of both sides.

Answer 1

We will (eventually) need the moment of inertia of the pulley. It is easy and straightforward to calculate so let us calculate it now. For a solid disk,

I = (1/2) M R²

I = (1/2) (10.0 kg) (0.25 m)² = 0.3125 kg m²

Apply Newton's Second Law to each of the three bodies:

First, for mass m₁:

F_{Net, 1} = T₁ - F_f1 = m₁ a

F_f1 = F_N1 = m₁ g

T₁ - m₁ g = m₁ a

Now look at the torques on the pulley. This is not the usual "lightweight pulley" we have encountered before. You might think of this as a flywheel. It has a moment of inertia that is large enough that it can not be neglected. This also means the tension in the rope on one side of it will be different than the tension on the other side. Hence, we have labeled the tensions at T₁ and T₂.

_net = R T₂ - R T₁ = I

T₂ - T₁ = ( I / R)

Now we are ready for the forces on m₂, the mass on the incline.

F_net = w₂ sin 30^o - F_f2 - T₂ = m₂ a

F_net = 0.866 w₂ - F_f2 - T₂ = m₂ a

F_f2 = F_N2 = w₂ cos 30^o = m₂ g cos 30^o

Recall that the linear acceleration of the blocks a is closely related to the angular acceleration of the pulley (or flywheel) ,

a = r

or

= a /r

Now we have three equations with three unknowns, T₁, T₂, and a.

T₁ - m₁ g = m₁ a

T₂ - T₁ = ( I / R) = ( I / R ) ( a / R ) = ( I / R² ) a

0.866 m₂ g - 0.50 m₂ g - T₂ = m₂ a

Now we can put in numbers and then solve for the unknowns.

T₁ - (0.36) (2.0 kg) (9.8 m/s²) = (2.0 kg) a

T₂ - T₁ = [ 0.3125 kg m² / (0.25 m)² ] a = 5 kg a

0.866 (6.0 kg) (9.8 m/s²) - 0.50 (0.36) (6.0 kg) (9.8 m/s²) - T₂ = (6.0 kg) a

50.92 N - 10.58 N - T ₂ = (6.0 kg) a

T₁ - 7.056 N = (2.0 kg) a

T₂ - T₁ = 5 kg a

40.34 N - T ₂ = (6.0 kg) a

T₁ = 7.056 N + (2.0 kg) a

T₂ - [7.056 N + (2.0 kg) a] = 5 kg a

T₂ = [7.056 N + (2.0 kg) a] + 5 kg a

T₂ = 7.056 N + (7.0 kg) a

40.34 N - T ₂ = (6.0 kg) a

40.34 N - [7.056 N + (7.0 kg) a] = (6.0 kg) a

40.34 N - 7.056 N = (7.0 kg) a + (6.0 kg) a

33.28 N = (13.0 kg) a

a = 33.28 N / 13.0 kg

a = 2.56 m / s²

T₂ = 7.056 N + (7.0 kg) a = 7.056 N + (7.0 kg) (2.56 m/s²) = 25 N

T₁ = 7.056 N + (2.0 kg) a = 7.056 N + (2.0 kg) (2.56 m/s²) = 12.2 N