How Does the Angle of Inclination Affect Motion in Rigid Body Dynamics?

[fedecolo]

Homework Statement

In the figure there are a block 1 of mass $M$, a pulley (disk) of mass $M$ and radius $R$ and a second block of mass $2M$ on an inclined plane with an angle $\alpha$ with respect to the horizontal. On this plane the static friction is $\mu$.
The wire does NOT slip on the pulley (so there is friction on the pulley too, but it says that the tension $T_1 \not= T_2$).

What is the condition that $\alpha$ must satisfy such that the block 1 moves downward?

Homework Equations

The Attempt at a Solution

I wrote the diagram of forces for all the elements (with the systems of reference in the figure) and the torques with respect to the center of mass of the pulley.

For the block 1: $T_1 - Mg= M(-a)$

For the pulley:
$$\begin{cases}
R_y -Mg -T_1 -T_2 \sin\alpha=0 \\ -R_x +T_2 \cos\alpha=0 \\
\end{cases}$$

$R$ is the vincolar reaction of the pulley support.

For the block 2 (the system of reference is different!)
$$\begin{cases}
T_2-f_s -2Mg\sin\alpha =2Ma \\ 2Mg\cos\alpha -N=0
\end{cases}$$

Now using the fact that $\sum \vec{M} = I_{CoM} \cdot \alpha_z$ ($\alpha_z=\frac{a}{R}$ is the angular acceleration of the pulley, sorry for the bad notation similar to the angle) I find the torques with respect to the center of mass of the pulley:
$$ \hat{u_z}: RT_1 -RT_2 \sin\alpha = 1/2 MR^2 \alpha_z$$

Now I can't solve this big system!
The condition is $T_2 -2Mg\sin\alpha +2Ma \geq f_{s,MAX} = 2Mg\cos\alpha \cdot \mu$.

It seems that I must use the $R$ to find $T_1$ or $T_2$ but I have two different $R_x$ and $R_y$...

Any help?

 

[haruspex]

What is the condition that α must satisfy such that the block 1 moves downward?

Are there more parts to the question? If it is only that then this is a statics problem, no? So at the critical value of α the tensions will be equal unless there is axial friction in the pulley, and we have no info on that.

 

[fedecolo]

Are there more parts to the question? If it is only that then this is a statics problem, no? So at the critical value of α the tensions will be equal unless there is axial friction in the pulley, and we have no info on that.

Yes, it's statics problem.
The other question asks what are the accelerations of the blocks if the condition is satisfied.

The pulley can rotate without friction around its axis, but there is friction between the wire and the pulley, so the 2 tensions are different.

 

[haruspex]

there is friction between the wire and the pulley, so the 2 tensions are different.

I have not seen a pulley problem yet that permits any slipping between rope and pulley. There is always friction there. Consider the torque balance about the pulley's axle. If the pulley is not accelerating, what does that tell you about the tensions?

 

[fedecolo]

I have not seen a pulley problem yet that permits any slipping between rope and pulley. There is always friction there. Consider the torque balance about the pulley's axle. If the pulley is not accelerating, what does that tell you about the tensions?

But the pulley is accelerating, because the block 1 must move downward!

 

[jbriggs444]

But the pulley is accelerating, because the block 1 must move downward!

If the pulley is accelerating, that tells you something about the difference in the tensions.

A useful starting approach would be to express the rotational acceleration of the pulley in terms of the linear acceleration of the block. The objective would be to then use this to find a formula for the tension difference in terms of the block acceleration.

It seems that you almost have that:

$$ \hat{u_z}: RT_1 -RT_2 \sin\alpha = 1/2 MR^2 \alpha_z$$

However, that does not look correct to me. What is the $\sin \alpha$ doing there?

 

[haruspex]

But the pulley is accelerating, because the block 1 must move downward!

It is asking for the boundary condition, i.e. the circumstance in which it barely moves downwards. That means its acceleration is arbitrarily small, so the difference in the tensions will be arbitrarily small.