Block and disk on double inclined plane

[naianator]

Homework Statement

https://courses.edx.org/asset-v1:MITx+8.MechCx_2+2T2015+type@asset+block/rolling_quiz_3.svg

A block of mass mb and a disk of mass md and radius r are placed on a symmetric triangular slope connected with a massless string over a massless pulley as shown above. The string is connected to a center axle of the disk so that the disk is free to rotate. The moment of inertia of the disk about its axle is I=1/2*m_d*r^2. The coefficient of static friction between the slope and the block/disk is 0.05 and the coefficient of kinetic friction between the slope and the block/disk is 0.15. The angle θ is 30∘.

Reminder: sin(30∘)=cos(60∘)=12 and cos(30∘)=sin(60∘)=3√2

Find the maximum ratio m_b/m_d such that the disk still rolls without slipping up the hill.

Homework Equations

Just to be clear I'm using:
t = torque and
T = tension

t_net = I*alpha
F_net = m*a

The Attempt at a Solution

Since its the maximum ratio m_b/m_d, the static friction must be at its max: mu_s*m*g*cos(theta) so

t_net = r*mu_s*m_d*g*cos(theta) = I*alpha = 1/2*m_d*r^2*a/r = 1/2*m_d*r*a

This simplifies to:

2*mu_s*g*cos(theta) = a

Then for the disk the addition of N+F_g = m_d*g*sin(theta) and:

F_net = T - m_d*g*cos(theta)*mu_s - m_d*g*sin(theta) = m_d*a
T = m_d*g*cos(theta)*mu_s + m_d*g*sin(theta) + m_d*a

and for the block (I'm not sure if I've messed up the signs here, I guessed that the positive axis should point towards the acceleration):

F_net = m_b*g*sin(theta) - T - m_b*g*cos(theta)*mu_k = m_b*a
T = m_b*g*sin(theta) - m_b*g*cos(theta)*mu_k - m_b*a

m_d*g*cos(theta)*mu_s + m_d*g*sin(theta) + m_d*a = m_b*g*sin(theta) - m_b*g*cos(theta)*mu_k - m_b*a

then I plugged in 2*mu_s*g*cos(theta) = a and simplified:

m_d(mu_s + tan(theta) + 2*mu_s) = m_b(tan(theta) - mu_k - 2*mu_s)

and finally:

m_b/m_d = (mu_s + tan(theta) + 2*mu_s)/(tan(theta) - mu_k - 2*mu_s) = 2.22193

But I just realized I forgot a mu_s in my original solution so is this correct?

 

[TSny]

Overall, your work looks correct and I think your final answer is correct. I noticed a couple of trivial typos:

Reminder: sin(30∘)=cos(60∘)=12 and cos(30∘)=sin(60∘)=3√2

Also, I don't follow your statement:

Then for the disk the addition of N+F_g = m_d*g*sin(theta)

But I don't think you used this.

 

[naianator]

Overall, your work looks correct and I think your final answer is correct. I noticed a couple of trivial typos:

Also, I don't follow your statement: But I don't think you used this.

Yes it was correct, thank you.

 

[haruspex]

It is physically impossible for the kinetic friction coefficient to exceed the static friction coefficient. Did you quote them correctly?

 

[naianator]

It is physically impossible for the kinetic friction coefficient to exceed the static friction coefficient. Did you quote them correctly?

Yes, I copied and pasted it... that's strange