Maximum Force on Inclined Plane to keep object stationary

[wainker]

Homework Statement

A block of mass m1 is placed on a wedge of mass m2. If the wedge is held at rest, it is observed that the block will automatically slide down the incline from rest. The coefficients of kinetic and static frictions between the block and the wedge are the same as those between the wedge and the table. They are given by uk(mu k) and us(mu s), respectively. The angle of inclination of the incline of the wedge is B(Beta).
If we push the wedge along a horizontal surface with a horizontal force F, such that the block would not slide up or down the incline, what is the maximum possible value of Fmax?

Homework Equations

For system of Wedge and block: Net Force=Fmax-Fk
F=ma

The Attempt at a Solution

First off, my xy plane is x is parallel to the table and y is perpendicular to the table.

The four forces acting on the system are 1-Fn up, 2-Fg down, 3-Fmax to the right, 4-Fk to the left. Fn and Fg cancel each other out, so
Net Force=Fmax-Fk
Fmax=Net Force+Fk=(m1+m2)a-uk(m1+m2)g=(m1+m2)(a-ukg)

On the block, there are three forces acting 1-Fn at an angle perpendicular to the inclined surface, 2-Fg straight down and 3-Fs pointing down the inclined surface (parallel). Now it gets sticky for me.
Sum Forces in x direction:
Fx=FnsinB+FscosB
ma=FnsinB+usFncosB
To find Fn, I solved for sum of forces in y direction:
Fy=0(no acceleration)=FncosB-mg-FssinB=FncosB-mg-usFnsinB
FncosB-usFnsinB=mg
Fn=(mg)/(cosB-ussinB)

So ma=(mgsinB+usmgcosB)/(cosB-ussinB)
a=(gsinB+usgcosB)/(cosB-ussinB)

I plugged this "a" back into the Fmax equation's "a" and I get some crazy long answer...which I was told is wrong, but it is:
Fmax=g(m1+m2)((sinB+uscosB)/(cosB-ussinB)-uk)

I tried another way but I don't know if it is right or not:

Same Fmax equation as above except:
Fmax=Net Force+Fk=(m1+m2)a2-uk(m1+m2)g=(m1+m2)(a2-ukg) <-----Equation 1

For block, to keep the block from moving, there is a force that is traveling up the plane, but I don't know how to justify it, that counters the downward movement of the block, so:
ma1=mgsinB+usmgcosB
a1=gsinB+usgcosB

To connect the two:
a1=a2cosB
a2=a1/cosB=gtanB+usg

Plug a2 into Equation 1 and get:
Fmax=g(m1+m2)(tanB+us-uk)

But I think this is also wrong because I can't justify the force moving up the block parallel to the incline.

 

[PhanthomJay]

I don't know, I'm a numbers sort of guy, so when I see problems with variables given as letters instead of numbers, I generally am turned off, since I lose a 'feel' for the problem and get confused with the algebraic manipulations.

At a glance, your original method looks good and well thought out, so I think your error may be in your first equation,

Fmax=Net Force+Fk=(m1+m2)a - uk(m1+m2)g=(m1+m2)(a-ukg)

Looks like you made a silly error, the '+ Fk" is correct, but then you changed it to a minus sign as I have highlighted in red.

:grumpy:

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