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Determining maximum coefficients of friction

  • Thread starter sleepcity
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1. Homework Statement
A block of weight 20 N (m = 2 kg) sits on an plane inclined at 37°. g = 10 m/s2 (for simplicity).

a) Calculate the value of the weight components.
b) Calculate the acceleration assuming no friction.
c) Calculate the acceleration assuming μk = 0.125.
d) What value of μk is needed to make the block slide down at constant velocity?
e) How great must μs be to prevent sliding?
f) In what way to the answers to d) and e) depend on the block's mass?

2. Homework Equations
Taking the incline as the x-axis,

Fwx = Fwsinθ
Fwy = Fwcosθ

∑Fx: Fwsin37° - FF = max
∑Fy: Fwcos37° + FN = may
tan θ = tan 37° = 0.75

3. The Attempt at a Solution

a)
Fwx = 20sin37° = 12 N
Fwy = 20cos37° = 16 N = FN

b) ∑Fx: 20sin37° = 2a, ∴ a = 6 m/s2

c) ∑Fx: 20sin37° - FF = 2a,
12 - ((μk)(FN)) = 2a
12 - ((0.125)(15.97) = 2a
12 - 2 = 2a
a = 10/2 = 5 m/s2

d) Constant velocity implies a = 0.
∑Fx: Fwsin37° - FF= 0, ∴FF= 12 N -- The force of friction balances the component of gravity pulling the box down the incline.
If FF = 12 N, μk = FF/FN = 12/16 = 0.75

e) I know that the coefficient of kinetic friction (μk) is usually lower than the coefficient of static friction (μs). If I were told what force was required to disrupt equilibrium (no motion), μs would be given by (Force required/FN). Without knowing this force I'm not sure how to proceed next.

f) My suspicion is that these values do not depend on mass at all, since acceleration is zero in each of the cases described in d) and e) (constant velocity and no motion until the friction force resisting motion is overcome.

Any and all help is greatly appreciated.
 

BvU

Science Advisor
Homework Helper
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Hello sleepy, and welcome to PF :)

In the e) case, you can consider a minute little force to start it sliding.
I don't agree with your force required/FN: there always is the 12 N from gravity that has to be compensated by friction.
For the answer, let the magnitude of this little force go to 0.

For f: you could repeat the calculations for a 4 kg block, but I think you already have seen that all forces that appear are proportional to m, so this m cancels
 

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