Frictional force on a block held against a vertical wall

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Homework Help Overview

The problem involves a 5kg block held against a vertical wall by a horizontal force, focusing on determining the frictional force exerted by the wall and the minimum horizontal force needed to prevent the block from falling, given a static coefficient of friction.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the equilibrium of forces acting on the block and the role of the coefficient of friction in determining the frictional force. There is uncertainty about how to incorporate the friction coefficient into the calculations for the second part of the problem.

Discussion Status

Some participants have provided hints regarding the equilibrium condition and the relationship between the normal force and frictional force. There is ongoing exploration of how changes in the horizontal force affect the maximum available frictional force.

Contextual Notes

Participants are navigating the implications of static friction and the conditions under which the block remains at rest against the wall. There is a mention of a homemade equation for force balance, indicating a personalized approach to the problem.

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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