# Changes

## Sliding motion along an inclined plane

, 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 $mg\sin \theta$. 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 $f_s$ acts, the case of ''sliding down'', where an upward kinetic friction $f_k = \mu_kN$ acts, and the case of ''sliding up'', where a downward kinetic friction $f_k = \mu_kN$ acts.:
{| class="sortable" border="1"
! Case !! Convention on the sign of acceleration $a$ !! Equation using $f_s, f_k$ !! Simplified, substituting $f_s,f_k$ in terms of $N,\mu_s,\mu_k$ !! Simplified, after combining with $\! N = mg \cos \theta$
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| Block sliding down the inclined plane stationary || measured positive down the inclined planeno acceleration, hence a positive number so no sign convention || $\! ma = mg \sin \theta - \mu_kN = f_s$ || $\! a = g (mg \sin \theta - \mu_kle \cos mu_sN$ || $\theta) = g ! \cos tan \theta (\tan \theta - le \mu_k)mu_s$
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| 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 || $\! ma = mg \sin \theta - f_k$ || $\! ma = mg \sin \theta + - \mu_kN$ || $\! a = g (\sin \theta + - \mu_k \cos \theta) = g \cos \theta (\tan \theta + - \mu_k)$
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| 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) || $\! ma = mg \sin \theta + f_k$ || $\le ! ma = mg \mu_sNsin \theta + \mu_kN$ || $\! a = g (\sin \theta + \mu_k \cos \theta) = g \cos \theta (\tan \theta + \le \mu_smu_k)$
|}