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Old Oct 7th 2009, 11:14 AM   #1
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Particle on a plane

Question:
A force of magnitude P is applied to the particle in a horizontal direction, towards the plane, as shown in the graphic.

Assuming that $\displaystyle \mu < \tan\theta$, find the minimum value of P which is necessary to ensure that the particle remains at rest on the plane.

Answer so far:

$\displaystyle \textbf{F} = -\textsl{F} \textbf{i} $
$\displaystyle \textbf{N} = \textsl{N} \textbf{j}$
$\displaystyle \textbf{W} = -mg \sin\theta \textbf{i} - mg \cos\theta \textbf{j}$
$\displaystyle \textbf{P} = P \cos\theta \textbf{i} - P \sin\theta \textbf{j}$

Since the particle remains at rest:
$\displaystyle \textbf{F}+\textbf{N}+\textbf{W}+\textbf{P} = 0$

Solving in i- and j-directions:
$\displaystyle 0=-F - mg \sin\theta + P \cos\theta \quad \Rightarrow F = - P \cos\theta + mg \sin\theta$

$\displaystyle 0 = N - mg \cos\theta - P \sin\theta \quad \Rightarrow N = mg \cos\theta + P \sin\theta$

...

Thank you very much in advance!

Honey $\displaystyle \pi$
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Old Oct 8th 2009, 02:52 AM   #2
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I think you have pretty much solved it. Just introduce the coeff of friction $\displaystyle \mu$ by writing $\displaystyle F\ =\ \mu\ N$
Then plug in the value of N from the other eqn.

If needed, use the fact that for a minimum the first order derivative is zero and the second order derivative > 0
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