Stacked blocks & pulley system

[stunner5000pt]

Homework Statement

Consider the diagram given. All surfaces have friction (u(s) & u(k)). When the system is released from rest, the mass M is heavy enough to pull m1 and m2 up the slope. m2 remains at rest with respect to m1. Determine the acceleration of the system in terms of the given variables
M, m1, m2, u(k) and u(s)

Relevant Equations

Newton's 2nd law
Friction = coefficient of friction x normal force

I've posted my attempt at a solution but I haven't gone through the whole process of putting together equations 1 -4 yet as I wanted to clarify if I'm on the right path

My doubt lies in the formulation of equation 4 - the force equation for the stacked block. Since we don't know the acceleration of the masses and we don't know if mass M is heavy enough to cause m2 to slide, do we leave F_{12x} undetermined and not equate this to \mu_{s} F_{N} ?

Are all the equations considering all variables or did I miss anything in my formulation so far? I'm happy to retype all that is written on the solution attempt in tex if the writing is illegible - let me know!

Appreciate all your input in advance & thank you :)

 

[kuruman]

Since $m_2$ and $m_1$ accelerate as one, you don't need separate free body diagrams (FBDs) for the two masses. Solve the problem as if a single mass $(m_1+m_2)$ is placed on the block. The coefficient of static friction $\mu_s$ is irrelevant because we are not told that $m_2$ is on the verge of sliding on $m_1$.

You still have to draw the FBD for the hanging mass.

 

[vela]

Since we don't know the acceleration of the masses and we don't know if mass M is heavy enough to cause m2 to slide, do we leave F_{12x} undetermined and not equate this to \mu_{s} F_{N} ?

The problem statement explicitly says $M$ is heavy enough to cause the masses to slide up the incline.

 

[stunner5000pt]

Since $m_2$ and $m_1$ accelerate as one, you don't need separate free body diagrams (FBDs) for the two masses. Solve the problem as if a single mass $(m_1+m_2)$ is placed on the block. The coefficient of static friction $\mu_s$ is irrelevant because we are not told that $m_2$ is on the verge of sliding on $m_1$.

You still have to draw the FBD for the hanging mass.

This is great thank you for the advice here.

 

[stunner5000pt]

The problem statement explicitly says $M$ is heavy enough to cause the masses to slide up the incline.

my apologies i should have clarified. We don't know if the hanging mass M is the biggest mass possible to cause the boxes to slide
In fact, is this something that we'd have to solve for before we simplified & treated m1 and m2 are one unit?

 

[haruspex]

my apologies i should have clarified. We don't know if the hanging mass M is the biggest mass possible to cause the boxes to slide

Do you mean, you don’t know if it is the smallest mass that would cause them to slide up?
Why does that matter?

In fact, is this something that we'd have to solve for before we simplified & treated m1 and m2 are one unit?

What allows you to treat them as one unit is the information that "m2 remains at rest with respect to m1". That is, they behave exactly as if they were glued together.

 

[stunner5000pt]

Do you mean, you don’t know if it is the smallest mass that would cause them to slide up?
Why does that matter?

What allows you to treat them as one unit is the information that "m2 remains at rest with respect to m1". That is, they behave exactly as if they were glued together.

you're right it doesn't matter, I'm overthinking this (as usual)

Thank you again for your help

 

[Herman Trivilino]

my apologies i should have clarified. We don't know if the hanging mass M is the biggest mass possible to cause the boxes to slide

Please check the statement of the problem and make sure it matches what you posted.