Finding the coefficient of Kinetic force with an angle

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Homework Help Overview

The problem involves determining the coefficient of kinetic friction for a crate being pushed across a level floor at a constant speed by a force applied at an angle. The crate's weight is given, and the applied force is specified, but the relationship between these forces and the normal force is under discussion.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of the normal force and its relationship to the weight of the crate. There are attempts to clarify the role of the normal force in the context of the problem, with some questioning the assumption that the normal force equals the weight of the crate.

Discussion Status

The discussion is ongoing, with participants exploring the implications of the normal force not necessarily being equal to the weight of the object. Some guidance has been offered regarding the use of free body diagrams to analyze the forces acting on the crate.

Contextual Notes

There is a noted lack of explicit information regarding the normal force in the problem statement, leading to confusion about its calculation. The original poster has referenced a solution from the textbook that differs from their own calculations, prompting further inquiry into the assumptions made.

KellyTarzuoty
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Homework Statement


A 1000N crate is being pushes across a level floor at a constant speed by a force (F) of 300N at an angle of 20 degrees below the horizontal, what is the coefficient of kinetic force?

Homework Equations


Fk= uk*N
uk= Fk/N[/B]
Fk=F2

The Attempt at a Solution


in the back of the textbook it comes up as 0.256, but when I try I do:
Fk = F2
F2 = 300cos20 = 281.9
uk= F2/n
uk= 281.9/1000
uk= 0.281

Can someone tell me where I've gone wrong? It seemed like it should be a simple question
 
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The normal force from the ground is not always the weight of the object.
The normal force is what ever it needs to be to keep the block from accelerating vertically.
 
Nathanael said:
The normal force from the ground is not always the weight of the object.
The normal force is what ever it needs to be to keep the block from accelerating vertically.
okay, so should I try and use that to try and find the weight? Because that was all the question gave me, it didn't mention any weights.
 
KellyTarzuoty said:
okay, so should I try and use that to try and find the weight? Because that was all the question gave me, it didn't mention any weights.
The weight of the crate is given, it is 1000N.
In your equations you seem to be saying that the normal force is equal to the weight of the crate (1000N).

If I push down on a piece of paper, the normal force on the paper (from the table or whatever) is going to be much more than the weight of the paper, right?
 
Nathanael said:
The normal force from the ground is not always the weight of the object.
The normal force is what ever it needs to be to keep the block from accelerating vertically.
To be precise and general, the normal force from a surface is the force of least magnitude required to prevent the object penetrating the surface. Necessarily, that will be at right angles to the surface.

@KellyTarzuoty, draw a free body diagram of the block. Three forces act on it. What are they? Use the usual statics equations to find the normal force.
 

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