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 Equilibrium and Elasticity Equilibrium and Elasticity Physics Help Forum 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  Tags particle, plane Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post yungman Advanced Waves and Sound 0 May 30th 2013 10:05 PM RAz Periodic and Circular Motion 1 Oct 25th 2010 11:25 AM Sundae Kinematics and Dynamics 1 May 13th 2009 08:16 AM Apprentice123 Kinematics and Dynamics 2 Apr 3rd 2009 06:13 PM joker1 Nuclear and Particle Physics 0 Mar 12th 2009 01:10 PM 