Static and Kinetic Friction problem

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Homework Help Overview

The discussion revolves around a problem involving static and kinetic friction, specifically analyzing the forces acting on a box weighing 100N on a horizontal surface with given coefficients of friction. Participants explore the implications of applying a 20N force to the box and its relationship with the static friction force.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of static friction and question the conditions under which the box would remain stationary or begin to move. There are inquiries about the nature of frictional forces and the interpretation of applied forces versus frictional resistance.

Discussion Status

The discussion is ongoing with various interpretations being explored. Some participants offer insights into the relationship between applied forces and friction, while others express uncertainty about the correct answer and the reasoning behind their calculations. There is no explicit consensus, but guidance has been provided regarding the nature of static friction.

Contextual Notes

Participants mention potential discrepancies in the problem statement or answer options, indicating that different versions of the question may exist. There is also a focus on understanding the maximum static friction force and its implications for the box's motion.

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1. Homework Statement [/b]
A box weighing 100N is at rest on a horizontal surface. The coefficient of static friction between the box and the surface is 0.5, and the coefficient of kinetic friction is 0.25. If the crate is pushed with a force of 20N parallel to the floor, which of the following is true?

2. Homework Equations [/b]
Force of Friction=μ(normal force)

3. The Attempt at a Solution [/b]
The static friction force is μ(normal force) and normal F=100N. Static Ff=0.5(100N)=50N. And the box is pushed with less force than the max static Ff so "the crate will remain stationary, due to the static Ff of 50N which opposes the intended motion"... but I got it wrong. I don't know why. Help?
 
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What were the other answer options?
 
The right answer choice was "the crate will accelerate across the floor at 3m/s^2." But at could be wrong because that was my friend's answer, and we might have different versions.
 
What would Newton say about a 50N force in one direction and a 20N force in the other?
 
Ummm... the box would move to the direction of 50N force?
 
So the box wouldn't move because 50N static Force of friction is greater than the 20N pushing force? That's what I put and I got the wrong answer. I mean, the static friction isn't causing the box to move in the other direction..
 
Ibix said:
What would Newton say about a 50N force in one direction and a 20N force in the other?

Think about this suggestion, and draw a force diagram if it helps. What is the static friction force when the block is at rest?
This is a little like if I told you that a force of 3000N is required to break down my front door. If it exerted 3000N whenever someone touched it, they would go flying backwards. This obviously isn't the case, and simply from F=ma we know that if we have a = 0, then F = Fn.
 
So the static Ff is 50N at rest? Then what's the maximum static Ff? And even if the box moved, the kinetic friction force would be 25N which is still less than 20N so it can't accelerate (I think). Could somebody at least tell me if the "right answer" is correct? If it is, I still have to show work so I'm going to ask my teacher anyways, and if it' not, well then, I have to move on cause I have a lot to do.
 
No, my point was that there is no "friction force" when we're not pushing on the box. There is always a static coefficient, but I was trying to get you to think of it like a wall that requires a certain force to break through. In my example, if I push on the door with 400 N, it pushes back with a normal force of 400 N, all the way up to 3000 N, at which point the net force becomes unbalanced and the door accelerates as it breaks.

I believe your answer was correct, other than stating that Ff = 50N. Ff is in fact just equal and opposite to your applied force, up to 50 N, at which point it "breaks" and then kinetic friction takes over. So the answer should be something like, ""the crate will remain stationary, due to the static Ff of 20N which opposes the intended motion"

Edit: Incase it wasn't clear, your friend must have a different version of the question.
 
  • #10
Yes, I got it now, that was exactly the answer. Thank you all, especially bossman27!
 

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