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Dynamics- Friction

  • Thread starter hsphysics2
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  • #1
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Homework Statement



A box is halfway up a ramp. The ramp makes an angle, θ with the ground. What is the maximum value of θ before the mass will slip? μ[itex]_{s}[/itex]=0.25

Homework Equations



F[itex]_{x}[/itex]=ma[itex]_{x}[/itex]

The Attempt at a Solution


I drew a free body diagram to show the forces affecting the box

η-mgcos=0
η=mgcosθ (eq'n 1)


F[itex]_{x}[/itex]=ma[itex]_{x}[/itex]
μ[itex]_{s}[/itex]η-mgsinθ=ma[itex]_{x}[/itex] (sub eq'n 1 in)
μ[itex]_{s}[/itex](mgcosθ)-mgsinθ=ma[itex]_{x}[/itex]
mg(μ[itex]_{s}[/itex]cosθ-sinθ)=ma[itex]_{x}[/itex]


I'm not sure where to go from here, or even if this is the correct path for me to take
 

Answers and Replies

  • #2
PhanthomJay
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You have correctly identified the forces acting, but just as the box is on the verge of slipping, is it accelerating?
 
  • #3
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the box wouldn't be accelerating, so would it be

mg(μ[itex]_{s}[/itex]cosθ-sinθ)=0
mgμ[itex]_{s}[/itex]cosθ=mgsinθ
μ[itex]_{s}[/itex]cosθ=sinθ
μ[itex]_{s}[/itex]=tanθ
θ=14.036
 
  • #4
PhanthomJay
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Homework Helper
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the box wouldn't be accelerating, so would it be

mg(μ[itex]_{s}[/itex]cosθ-sinθ)=0
mgμ[itex]_{s}[/itex]cosθ=mgsinθ
μ[itex]_{s}[/itex]cosθ=sinθ
μ[itex]_{s}[/itex]=tanθ
θ=14.036
Yes, good, in degrees (don't forget units), but you should round your answer to 14 degrees ( 2 significant figures).
 

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