Block sliding down plane

  • Thread starter reaperkid
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  • #1
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Homework Statement



A block of mass 19.0 kg is sliding down an inclined plane. Due to friction between the block and the plane the block is moving at constant speed. If the coefficient of kinetic friction is 0.24), what is the angle of the plane with respect to the horizontal ?

m = 19kg
mu = .24
theta = ?

Homework Equations



Fk = mu*Fn
Fn = g sin theta

The Attempt at a Solution



I'm confused about the hint of the block moving with a constant speed due to friction.. Because that means that the net force is 0, so how can I calculate this if one of the forces is equal to 0?? I can't get the normal force without the angle and I can't get the friction force without the normal force.. Can anybody help me out with what that clue implies?

Thanks!
 

Answers and Replies

  • #2
tiny-tim
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I'm confused about the hint of the block moving with a constant speed due to friction.. Because that means that the net force is 0, so how can I calculate this if one of the forces is equal to 0?? I can't get the normal force without the angle and I can't get the friction force without the normal force.. Can anybody help me out with what that clue implies?

Hi reaperkid! :smile:

(have a mu: µ and a theta: θ :wink:)

There are three forces … normal, friction, and gravitational …

so just balance them to add to zero

(either by using components, or a vector triangle ) :smile:
 
  • #3
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Hi reaperkid! :smile:

(have a mu: µ and a theta: θ :wink:)

There are three forces … normal, friction, and gravitational …

so just balance them to add to zero

(either by using components, or a vector triangle ) :smile:

Soo does that mean the normal force is equal to the gravitational force?
 
  • #4
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no, draw a FBD
 
  • #5
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no, draw a FBD

Ughhh, I don't know how to get the normal force or friction force. If I could get either of those then it would be easy to balance them. And isn't friction the only negative force here?

Nevermind I got it, thanks!
 
Last edited:
  • #6
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Ughhh, I don't know how to get the normal force or friction force. If I could get either of those then it would be easy to balance them. And isn't friction the only negative force here?

Nevermind I got it, thanks!



I think you should sketched the vector forces. I will give you a hint. There will be two components that make up the forces: The x-component and the y-component. Identify the forces in the x-directions and identify the forces in the y direction. What do you think the value of the acceleration in the y-direction will be?
 
  • #7
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Assume that angle with horizontal is theta.
mu = 0.24
You have written Fn = g sin theta
I think you missed m. Moreover, it should be cos, not sin.
Draw the diagram. You will find that correct is
Fn = mg cos(theta)--------------------(1)

You have written Fk = mu * Fn
i.e. Fk = 0.24 Fn-----------------------(2)

You got two equations (1) and (2). By resolving forces parallel to the incline and making net force in this direction as zero, you will get one more equation:-
Fk = mg sin(theta)-----------------(3)

You should be able to find theta from these three equations.
 

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