Homework hellp. coefficent of friction.

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    Friction Homework
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Homework Help Overview

The discussion revolves around a physics problem involving a box on an incline, specifically focusing on the forces acting on the box, including friction. The original poster presents a scenario where an 18.0 kg box accelerates down a 37.0 degree incline, prompting questions about calculating the friction force and the coefficient of friction.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the scenario of the box accelerating down the incline, questioning how to approach the problem without friction initially. They discuss the principles of gravitational acceleration along the incline and attempt to relate the given acceleration to the forces acting on the box.

Discussion Status

The discussion has progressed with participants offering insights into the relationship between forces and acceleration, using Newton's laws. There is an ongoing exploration of how to calculate the frictional force and the coefficient of friction, with some participants providing calculations and others seeking clarification on the steps involved.

Contextual Notes

Participants express uncertainty about their understanding of the physics concepts involved, particularly in relation to the forces on an incline and the role of friction. The original poster indicates a lack of confidence in their physics skills, which may influence the discussion dynamics.

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Homework Statement


An 18.0 kg box is released on a 37.0 degree incline and accelerates down the incline at 0.270 m/s/s. Find the friction force impeding it's motion. How large is the coefficient of friction?


Homework Equations


Ff=uFn


The Attempt at a Solution


I have no idea how to do this.
 
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Can you do the problem if there were no friction, that is just asking for acceleration down an inclined plane?
 
ohh, maybe how can i do it without friction though?
 
Well you can't but the principles used apply:

In other words the acceleration from gravity along the surface of the ramp is sin(theta) * g. It is not 9.8. So if there were no friction a=9.8 sin(37). What is this come out to be?
 
5.9?
 
OK. Now we are told that the acceleration is not 5.9 but a measly 0.27.

So let's write an eqn using Newtons laws.

ma=sum of all forces ? is the frictional force

18.0*0.27=18*5.9-? Let's not worry about the components that make up ? yet. Can you compute ?
 
4.86=106.2-? so 101.34? or is there more than one other force?, Thank you for helping I'm sorry I'm really bad at physics.
 
Should be 18(5.9-0.27)

Ok now the tricky part. This frictional force is equal to the normal force * u where u is the coefficient of friction.

On level ground normal force is just opposite the weight. But on a plane it is
weight* cos(37)

So normal force is 18*9.8 cos (37). Let's call this N.

Then we have N*u=the same ? as you calculated before. can you get u?
 
176.4 cos(37)=140.9
N*u=101.34
140.9u=101.34
so u=.719
 
  • #10
By george I think you got it. It makes more sense with a free body diagram.
 
  • #11
Thank you so much!
 

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