Physics Help Forum Particle on a plane
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 Oct 7th 2009, 11:14 AM #1 Junior Member   Join Date: Oct 2009 Posts: 1 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 $\displaystyle \pi$ Attached Images
 Oct 8th 2009, 02:52 AM #2 Physics Team   Join Date: Feb 2009 Posts: 1,425 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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