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Sliding motion along an inclined plane

524 bytes added, 21:34, 17 September 2010
Component along (down) the inclined plane
===Component along (down) the inclined plane===
For the axis ''down'' the inclined plane, the gravitational force component is <math>mg\sin \theta</math>. The magnitude and direction of the friction force are not clear. We have three cases, the case of ''no sliding'', where an upward static friction <math>f_s</math> acts, the case of ''sliding down'', where an upward kinetic friction <math>f_k = \mu_kN</math> acts, and the case of ''sliding up'', where a downward kinetic friction <math>f_k = \mu_kN</math> acts.:
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! Case !! Convention on the sign of acceleration <math>a</math> !! Equation using <math>f_s, f_k</math> !! Simplified, substituting <math>f_s,f_k</math> in terms of <math>N,\mu_s,\mu_k</math> !! Simplified, after combining with <math>\! N = mg \cos \theta</math>
| Block sliding down the inclined plane stationary || measured positive down the inclined planeno acceleration, hence a positive number so no sign convention || <math>\! ma = mg \sin \theta - \mu_kN = f_s</math> || <math>\! a = g (mg \sin \theta - \mu_kle \cos mu_sN</math> || <math>\theta) = g ! \cos tan \theta (\tan \theta - le \mu_k)mu_s</math>
| Block sliding up down the inclined plane || measured positive down the inclined plane (note that acceleration opposes direction of motion, leading to retardation) hence a positive number || <math>\! ma = mg \sin \theta - f_k</math> || <math>\! ma = mg \sin \theta + - \mu_kN</math> || <math>\! a = g (\sin \theta + - \mu_k \cos \theta) = g \cos \theta (\tan \theta + - \mu_k)</math>
| Block stationary sliding up the inclined plane || no measured positive down the inclined plane (note that accelerationopposes direction of motion, so no sign convention leading to retardation) || <math>\! ma = mg \sin \theta + f_k</math> || <math>\le ! ma = mg \mu_sNsin \theta + \mu_kN</math> || <math>\! a = g (\sin \theta + \mu_k \cos \theta) = g \cos \theta (\tan \theta + \le \mu_smu_k)</math>
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