What Is the Maximum Mass M3 Can Have Without the Block Sliding Off the Slab?

[Singdasorrow]

Homework Statement

A block of mass M1 = .430 kg is initially at rest on a slab of mass M2 = .845 kg, and the slab is initially at rest on a level table. A string of negligible mass is connected to the slab, runs over a frictionless pulley on the edge of the table, and is attached to a hanging mass M3. The block rests on the slab but is not tied to the string, so friction provides the only horizontal force on the block. The slab has a coefficient of kinetic friction .365 and a coefficient of static friction .566 with both the table and the block. When released, M3 pulls on the string and accelerates the slab, which accelerates the block. Find the max mass of M3 that allows the block to accelerate with the slab, w/o sliding on the top of the slab

Homework Equations

Ʃfx= T-f
Ʃfy = T - W
f=uN

The Attempt at a Solution

I treated the masses on the table as one single mass which was 1.275 kg. i multiplied that by gravity to find N and it was 12.495. I took the kinetic coefficient of friction and found friction to be -4.56

i plugged this into the Ʃfx= T-f equation with my other knowns to find T+4.56= 1.275a
Looking at the second equation i see that i have two variables for the weight will depend on the mass of the hanging mass. so Ʃfy = T - (mass3 x gravity)

i'm stuck at this point relating the two together. Am i correct in how i have things set up so far? or have i not solved for a variable that i should b able to?

 

[SammyS]

You need to treat the block separately from the slab, or you'll never find the max that M₃ can be without the block slipping on the slab.

Draw TWO Free Body Diagrams; one for the block, one for the slab.

 

[Singdasorrow]

Ok, so since I'm going to treat them individually. I have 3 force diagrams (1 for each object) and have the sum of the forces acting on each one.

I am tho confused how to relate them because i do know i can't exceed fs. so i need to find an acceleration that will allow me to do that, then a mass as a result. am i thinking of this right?

 

[SammyS]

Static friction: f_s ≤ μ₂ F_Normal

 

[Singdasorrow]

ok so since i know the coefficient of static friction and the normal force of the block. Fs ≤ .566 x .430(-9.8). Fs ≤ -2.385 N

 

[SammyS]

ok so since i know the coefficient of static friction and the normal force of the block. Fs ≤ .566 x .430(-9.8). Fs ≤ -2.385 N

So, that's the maximum force that the slab can exert on the block.

I think you may be able to finish the problem now.