Block A Reaches Equilibrium: 92N Force

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Homework Help Overview

The discussion revolves around a physics problem involving two blocks, A and B, in equilibrium on a slope. The focus is on understanding the forces acting on block A, particularly the normal force and its relationship to the weight of block A and the angle of the slope.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore equilibrium equations and the relationship between the normal force and the weight of block A. Questions are raised about the implications of these forces on friction and the calculations involved in determining the normal forces between the blocks and the slope.

Discussion Status

The discussion is active, with participants sharing their reasoning and questioning the assumptions regarding the forces involved. Some guidance has been offered regarding the evaluation of normal forces, but there is no explicit consensus on the calculations or interpretations yet.

Contextual Notes

Participants are considering the effects of angles and forces in a two-block system on a slope, with specific attention to how the weight of block A influences the normal forces. There is an acknowledgment of the need to evaluate multiple normal forces in the context of the problem.

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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
blocks.png
 
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What effect, if any, does the weight of ##A## have on the normal force between ##B## and the slope?
 
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.
 
goodOrBad said:
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:
1600354863655.png
 
So
N-Acos30=0
-N+Bcos30=0 -> N=100cos30=86.6N

A=100N ?
 
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?
 

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