# Homework Help: Pulley, Ramp with Moment of Inertia

1. Apr 2, 2012

### kc0ldeah

1. The problem statement, all variables and given/known data

http://imgur.com/ZkjXv

The given answers are (b) .309 m/s^2 (c) T1= 7.67N, T2=9.22N

2. Relevant equations

I_disc = .5(mass_disc)r^2
Sum of Torque = I α
a = r α

3. The attempt at a solution

We will define the system moving right and down as positive to get a positive acceleration.

The tension between mass 1 (2kg) and the pulley is T_1
The tension between mass 2 (6kg) and the pulley is T_2

Summation of Torque = I α = rT_2 - rT_1

plug in: α = a/r and I_disc = .5(mass_disc)r^2

into

I α = rT_2 - rT_1

to get

a(.5)(mass_disc) = T_2 - T_1

The summation of the forces in the x-direction for m1 is: F_x = m1(a) = T_1 - μ(m1)g

T_1 = (m1)a + μmg

How do you define the summation of the forces for m2 and get an expression for T_2?

My first attempt:

Sum of Forces for M2 in x direction:

F_x = (m2)a = T_2 - μ(m2)gcos30

T_2 = (m2)a + μ(m2)gcos30

is incorrect. This was done through rotating the system 30 degrees but the summation for mass 1 was not rotated therefore the equations are not consistent and apparently, wrong. What is the best way to do this?

Last edited: Apr 2, 2012
2. Apr 2, 2012

### azizlwl

$T_1-μm_1g=m_1a .......1$

$m_2Sinθ-μm_2Cosθ-T_2=m_2a .......2$

$r(T_2-T_1)=Iα........3$

3 equations with 3 unknowns- a,T1 and T2
Rotating the system make it more complicated.
You should be consistent with direction of acceleration.

Last edited: Apr 2, 2012
3. Apr 2, 2012

### kc0ldeah

The equation you gave for T_2 seems to be incorrect. Plugging in .309 for a yields a T_2 of -.725 which is incorrect

4. Apr 2, 2012

### kc0ldeah

Bump? Can anyone give some insight?

5. Apr 2, 2012

### tiny-tim

hi kc0ldeah!

(try using the X2 button just above the Reply box )
that's only tension and friction …

$m_2gSinθ-μm_2gCosθ-T_2=m_2a .......2$