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posted by  sean5 on 11/11/2009 11:20:18 PM  |  status: Live  |  Earned Karma: 60

physics

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a block is at rest on an incline of 37 degrees. the coefficients are μs=.81 and μk=.69, resepectively. the acceleration of gravity is 9.8 meters per second squared. what is the largest angle which the incline can have so that the mass does not slide down the incline? answer in units of degrees.
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posted by Altair on 11/11/2009 11:24:48 PM  |  status: Live
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Response Details:

let the angle be θ, we have:

mgsinθ - mgcosθμs = 0
=>tanθ = μs
θ = tan-1s) = tan-1(0.81) = 390
the greatest angle is 390 for the block to not slide

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posted by Ramned on 11/11/2009 11:25:41 PM  |  status: Live
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Response Details:
Static Friction is defined as


F(static) = F(normal) * μs

=mgcosθ * μs.
---

Must be strong enough to cancel out the gravitational force, which is

mgsinθ.  Therefore set them equal.

mgsinθ = mgcosθμ

sinθ / cosθ = μ

tanθ = μ

θ = tan^-1 (μ) = 39 degrees.


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posted by jerry2009 on 11/11/2009 11:37:00 PM  |  status: Live
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Response Details:
Let θ = Maximum angle of inclined plane for which the block does not slide
Coefficient of static friction, μs = 0.81
Coefficient of kinetic friction, μk = 0.69
Angle of inclined plane = 37 degrees
Static friction = fs
Component of the weight along the plane = M g sin θ
Component of the weight perpendicular to the plane = M g Cosθ
Condition for the book not to slide,
   Static friction = M g Sin θ
   fs = M g sin θ ...............(1)
   N = M g Cos θ  .............(2)   ( where N = Normal force )
By dividing ( 1) with (2)
   fs / N = tan θ
   But, fs / N = μs
   Hence, tan θ = μs
   Maximum angle of inclined plane for which the block does not slide, θ = arc tan ( μs )
                                                                                                             = arc tan ( 0.81 )
                                                                                                             = 39o
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