Acceleration of masses connected with string, non massless pulley

[AHinkle]

Homework Statement

Homework Equations

\SigmaF = ma
\Sigma\tau=I\alpha
?

The Attempt at a Solution

m₁ = 1 Kg
m₂ = 8 Kg
\theta = 33 degrees
m_pulley = 7 Kg
r_pulley = .11 m
\mu = 0.27
g = 9.8 m/s²

My attempt, I'm falling behind in class due to an illness and I can figure this out as long as the pulley is massless and frictionless but I'm not sure what do with the pulley. I missed that day. If I can get an idea as to what do with it the rest of the problems should make sense, thanks for the help.

for m₁
\SigmaF_x= T - f₁ = m₁a
\SigmaF_y= N - m₁g = 0
N = mg

f₁ = \muN = (0.27)(1.0Kg)(9.8 m/s²) = 2.646 N

for m₂
I assume that because they are connected by an string a₁ = a₂?
is my assumption still correct with a non-massless pulley? also for the block on the incline
I put the x-axis along the incline

\SigmaF_x = m₂gsin\theta- T - f₂
Also is my assumption correct that the tension in the string (T) should still be the same on both sides of the pulley?
\SigmaF_y = N - mgcos\theta
N = mgcos\theta

f₂ = (0.27)(8.0Kg)(9.8m/s²) = 21.168 N

I added the equations together (T's cancel)

m₂gsin\theta - T + T - f₂ - f₁ = (m₁ + m₂)a

(8.0)(9.8)sin(33) - (21.168) - (2.646) = 9.0a

a = 6.0643 m/s²
This is obviously not the correct answer, plus I didnt use the pulley information at all..

for then I thought.. maybe i'll find the torque on the pulley
\Sigma\tau=I\alpha

I \approx m_pulleyr²
I \approx (7.0Kg)(0.11m)² = .0847 Kg*m²

\Sigma\tau = Tr_pulley
T - f₁ = m₁a

I don't know a or I don't trust the value I have up there are least.
but a = r\alpha
so maybe
T = m₁a - f₁
T = m₁(r\alpha)-2.646 N

I could just go around in circles forever..
Can someone help me figure out.
1) How does the pulley factor into this?
2) Is the tension in the string still the same on both sides of the pulley?
3) Is the acceleration of both blocks still the same because they're connected by the string?
4) am I even going in the right direction with how to solve this?

 

[p21bass]

Ignore this, as I missed the information for the pulley.

Duh, me...

 

[ehild]

I could just go around in circles forever..
Can someone help me figure out.
1) How does the pulley factor into this?
2) Is the tension in the string still the same on both sides of the pulley?
3) Is the acceleration of both blocks still the same because they're connected by the string?
4) am I even going in the right direction with how to solve this?

1) The pulley needs some torque to accelerate its rotation.
2) No, the tension is different on the both sides of the pulley and the difference of the torques will accelerate the rotation of the pulley
3) Yes.

As there is no mention about the friction between the string and pulley, it can be assumed big enough so as the string does not slip. If so, the linear speed of the rim of the pulley is the same as the speed of the string, so v=ωR.

4) For m₂, the component of gravity along the slope (and string) is m₂gsinθ, and the normal component is N= m₂g cosθ.

You have two write equations also for the pulley.

ehild

 

[AHinkle]

how (in what form) would i write equations for the pulleys..

would it be F = ma
or would it be in the form \tau=I\alpha?
i've got to leave for work but
would they look like

\tau = T₁ - (something) = I\alpha

i mean would they be analagous to F = ma equation but with diff. vars? with rotational instead of translational variables?

 

[ehild]

Both tensions produce a torque, they are of opposite sign and the net torque is equal to Iα.

ehild