Block on an Incline with Varying Elevation

  • Thread starter Becca93
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  • #1
84
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Homework Statement
Part one: A block is at rest on an inclined plane whose elevation can be varied. The coefficient of static friction is μs= 0.390, and the coefficient of kinetic friction is μk= 0.190. The angle of elevation θ is increased slowly from the horizontal. At what value of θ does the block begin to slide (in degrees)?

(Hint after getting it incorrect: mass*acceleration = sum of net forces in one direction. Use mu_s to calculate the frictional force for the moment when it starts to slide)

Part two: Evaluate the acceleration of the block.



The attempt at a solution

So,

Ff = μFn
Fn= mgcosθ
Fparallel = mgsinθ
Ff = μ(mgcosθ)

When the block begins to slide, the force of friction has to be equal to the force parallel, right? You have to overcome the static μ.

So,
Fparallel = Ff
mgsinθ = μmgcosθ

mg's would cancel, so

sinθ = μcosθ
sinθ/cosθ = μ
tanθ = μ

θ= tan^-1(0.390)

However, the answer I get is 0.372°, and that isn't correct.

For the second half,
ƩFnet = ma
Fparallel - Ff = ma
mgsinθ - mgcosθ = ma
gsinθ - gcosθ = a


There's a flaw in my method, I just don't know what it is. Anyone know how to help?
 
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Answers and Replies

  • #2
798
1
Your answer was right. It's just that you are in radians, rather than degrees.
 
  • #3
84
1
Thank you. I feel like a bit of an idiot, but I guess that's what I get for working with a new calculator.

Thank you!
 

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