Block A Reaches Equilibrium: 92N Force

[goodOrBad]

Homework Statement

I have been given the weight of block B, an angle and friction coefficient

Relevant Equations

Fx=0

Fy=0

friction=μN

I tried to solve it via equilibrium equations and I got 92N for block A

 

[gneill]

What effect, if any, does the weight of $A$ have on the normal force between $B$ and the slope?

 

[goodOrBad]

Well A should be the same force as N if I account for the angle that is N=Acos30 for y direction, right?
And N influences both blocks.

 

[gneill]

Well A should be the same force as N if I account for the angle that is N=Acos30 for y direction, right?
And N influences both blocks.

Okay, so what does that imply for calculating the friction between $B$ and the slope? I don't see it taken into account here:

 

[goodOrBad]

So
N-Acos30=0
-N+Bcos30=0 -> N=100cos30=86.6N

A=100N ?

 

[gneill]

There are two normal forces that you need to evaluate. One is the force between blocks $A$ and $B$, and the other is between $B$ and the slope.

The normal force between $A$ and $B$ depends only on the weight of $A$ as you've shown, and you've already realized that block $A$ will contribute to the net normal force between $B$ and the slope. What expression can you write for this second normal force?