Find the coefficient of Friction with weight and force needed to move and object

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Homework Help Overview

The problem involves determining the coefficient of friction for a block weighing 9.3 N that requires a force of 3.7 N to move at constant velocity. The context is centered around frictional forces and their relationship to normal force.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the applied force, normal force, and the coefficient of friction, with one participant attempting to calculate the coefficient based on the given values. Questions arise regarding the interpretation of weight as a force and its implications for the normal force.

Discussion Status

The discussion includes differing interpretations of the normal force and its calculation. Some participants suggest reconsidering the definition of weight in this context, while others acknowledge the original poster's reasoning. There is no explicit consensus on the correct approach, but the dialogue is productive.

Contextual Notes

Participants are navigating the definitions of weight and normal force, with some confusion about whether the weight given is already in Newtons or if further calculations are necessary. This reflects the constraints of the problem as presented.

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Homework Statement


A block weighing 9.3 N requires a force of
3.7 N to push it along at constant velocity.
What is the coefficient of friction for the
surface


Homework Equations


Ff=μ*Fn


The Attempt at a Solution


So the 3.7 N to push is the Force of friction so 3.7=μ*9.3
μ=3.7/9.3=.37755

Is this correct
 
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well, i think you are close. The Normal Force is equal to the force going down, that is, mass times acceleration, because this is in one dimension. That means that you actually should do Normal Force = 9.3 * 9.8. Then you divide 3.7 by the answer, which should give you 0.406.
 
That is what i was thinking but since it said "weigh", i thought that that was already in Newtons... in other words, it was force and it was the value for normal force. Am i missing something??
 
oh, that is a good point. Sorry, I didn't notice that; in that case, you are correct. And my bad, had it been the force, the answer is actually 0.0406, not 0.406. But all is well that ends well!
 
thanks
 

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