Calculating Force Applied to Wall with 1st & 3rd Laws

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SUMMARY

The discussion focuses on calculating the force required to keep a particle with a mass of 3.0 kg from falling against a wall, considering static friction with a coefficient of friction (μ_s) of 0.7. The normal force (η) is determined to be equal to the gravitational force (mg), and the frictional force (F_f) is calculated as F_f = μ_s * η. According to Newton's third law, the push force (P) exerted on the wall is equal to the normal force, and for the particle to remain stationary, the gravitational force (F_g) must equal the frictional force (F_f), leading to the conclusion that P = η = mg.

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Alem2000
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I had a question about a particle with mass m=3.0kg if you take

this mass and push it up against a wall how much force would you have to

apply to keep it from falling if you had \mu_s = .7? where \eta<br /> <br /> is the normal force and P is the push force what I did was apply the first and 3rd laws my

equations came out to be

\eta=mg and f_k=P now after I calculate this I know I have to do something next b/c of the 3rd law. Do I draw in vector form the push exherted on the wall and the the push that the wall exherts back and add the (by taking the negative vector and reversing it to make it posative) I am a little confused?
 
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What keeps particle from falling is the friction force F_{f} which is equal to F_{f}=\mu_{s}\eta. On the other side, from 3rd law you know that P=\eta and that in order for particle not to fall must be F_{g}=F_{f}, where F_{f}=mg. When you push the body pushes the wall with force P, so vector P "attacks" wall. What Sir Isaac Newton says in his 3rd law is tha walls makes a "counter attack" on particle with force of the same intensity and direction as P, only in different way and this reactive force acts on a particle, and not on a wall. I hope this is what you wanted to know.
 

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