Block A Reaches Equilibrium: 92N Force

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SUMMARY

The discussion centers on the equilibrium of forces acting on block A, which is determined to be 92N. The normal force (N) between block B and the slope is influenced by the weight of block A, calculated using the equation N = A cos(30°). The normal force between blocks A and B is directly dependent on the weight of block A, while the normal force between block B and the slope is derived from the combined effects of both blocks. The calculations confirm that N = 100 cos(30°) results in a normal force of 86.6N between block B and the slope.

PREREQUISITES
  • Understanding of equilibrium equations in physics
  • Knowledge of normal force calculations
  • Familiarity with trigonometric functions, specifically cosine
  • Basic principles of friction and its relationship with normal forces
NEXT STEPS
  • Study the derivation of equilibrium equations in static systems
  • Learn about the role of normal forces in friction calculations
  • Explore advanced applications of trigonometry in physics problems
  • Investigate the effects of multiple blocks on normal force interactions
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Physics students, engineers, and anyone involved in mechanics or statics who seeks to understand the interactions of forces in multi-block systems.

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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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