Need help to solve pulley problem

Question

A massive, frictionless pulley consists of two 310 g solid disks with 3.8 cm radii connected by a lightweight axel with a 2 cm radius. A string is wound around the axel of the pulley and attached to a 130 g block which slides down an incline shown in the figure with θ = 31∘. The coefficient of kinetic friction between the block and the incline is 0.27.

Calculate the magnitude of the acceleration of the block, in m/s2.

Answer 1

Assume the +x direction is down the ramp (rotated coordinate system) and the -x direction is up the ramp.

Assume the +y direction is upward perpendicular to the ramp (up at a diagonal) and the -y direction is downward perpendicular to the ramp (into the ramp at a diagonal).

BLOCK:

Forces/components pulling down the ramp:

x Component of gravity = mgsinθ

Forces/components pulling up the ramp:

Tension = T

Kinetic friction = Fk = μ* Fn

Forces pointing in the +y direction:

Normal force = Fn

Forces point in the -y direction:

y Component of gravity = mgcosθ

Fnetx: mgsinθ - T - Fk = ma

(0.130)(9.8)sin(31) - T - 0.27Fn = 0.130a

0.65616 - T - 0.27Fn = 0.130a

Fnety: Fn - mgcosθ = 0

Fn - 1.09203 = 0

Fn = 1.09203 N

Put Fn into first equation:

0.65616 - T - 0.27(1.09203) = 0.130a

Combined Forces Equation:

0.36131 - T = 0.130a

PULLEY

Torque caused by Tension

τ = T* r

τ = 0.02T

//NOTE: r is the distance from the location of the force to the pivot point

τnet = Iα

Net torque requires an expression for moment of inertia (I) of pulley disks

I solid disk = 1/2MR^2 --> 2 solid disks will add their moments of inertia together

I total = 1/2MR^2 + 1/2MR^2 = MR^2

I total = (0.310)(0.038)^2 = 0.0004476 kgm^2

//NOTE: R is the radius of the disks, not the radius of the axel

τnet = 0.0004476α

Make an expression for alpha (α) in terms of acceleration (a)

a = α*r --> α = a/r --> α = a/0.02

Put a/0.02 in for alpha in the torque equation:

τnet = 0.0004476(a/0.02)

τnet = 0.02238a

Combined Torque equation:

0.02T = 0.02238a

Use the combined torque equation and the combined forces equation together!

0.02T = 0.02238a --> T = 1.119a substitute into other EQ and solve for a:

0.36131 - T = 0.130a

0.36131 - 1.119a = 0.130a

0.36131 = 1.249a

ANSWER: a = 0.28928 m/s^2