Frictional force on a block held against a vertical wall

AI Thread Summary
A 5kg block is held against a vertical wall by a horizontal force of 100 Newtons, resulting in a frictional force of approximately 49.1 Newtons due to equilibrium. The static coefficient of friction between the wall and the block is μs = 0.4, which is crucial for determining the minimum horizontal force required to prevent the block from falling. The frictional force is calculated using the equation F_friction = μs * Normal force, where the normal force is the horizontal force applied. As the horizontal force decreases, the maximum available frictional force also decreases, leading to a critical point where the available friction equals the required friction to keep the block stationary. Understanding these relationships clarifies how to incorporate the coefficient of friction into the problem effectively.
freddyeddyjor
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1. A 5kg block is held at rest against a vertical wall by a horizontal force of 100 Newtons.
1-What is the frictional force exerted by the wall on the block
2-What is the minimum horizontal force needed to prevent the block from falling if the static coefficient of friction between the wall and the block is μs = .4

Homework Equations


MAx = Right forces- Left Forces ( This is a homemade equation, basically the forces pushing from the wall are subtracted from the forces pushing against the wall.

The Attempt at a Solution


I am unsure how to proceed with this problem. I know that for part 1, the frictional force should be about 49.1n, this comes from the equation shown above. If right forces- left forces= 0, then there is no movement, and therefore they must be equal. So 5kg*9.81m/s = 49.05N
The second part is what confuses me, I fail to see a way to incorporate the friction coefficient into the problem in a meaningful way.
 
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freddyeddyjor said:
The First part is where I am stuck. I am unsure how to put the coefficient of friction into the problem in a meaningful way.
You don't need the coefficient of friction for the first part. Hint: The block is in equilibrium.
 
freddyeddyjor said:
The second part is what confuses me, I fail to see a way to incorporate the friction coefficient into the problem in a meaningful way.
You know how large the frictional force needs to be from the first part. If the horizontal force is reduced, what does that do to the max frictional force available? At what point will it the available frictional force equal the required frictional force?
 
I think the part you're missing is that the frictional force is determined by:

Friction force = μs * Normal force

In this case you start out with the normal force pushing against the wall at 100N, but how low could it go before the block fell down?
 
So I should set μk*Fn= Frictional Force exerted by the wall.
.4*FN = 49.1
Fn = 122.8 (about)

That actually makes a lot of sense, thank you all.
 
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