Worked exercise: inclined plane with kinetic friction

Problem

A block slides down an inclined plane of $\theta = 40^\circ$ with a coefficient of kinetic friction $\mu_d = 0{,}30$. Calculate the block’s acceleration along the plane.

Block sliding down the inclined plane: weight $\vec P$, normal reaction $\vec N$ and kinetic friction $\vec F_a$ opposing the motion.

Solution

Method. We choose axes oriented along the plane ($x$, positive downward) and perpendicular to it ($y$). The weight $\vec P$ resolves into $P_x = mg\sin\theta$ (driving) and $P_y = mg\cos\theta$ (pressing). The constraint reaction balances $P_y$: $N = mg\cos\theta$. The kinetic friction is $F_a = \mu_d N = \mu_d mg\cos\theta$, opposing the motion.

Newton’s second law along $x$ gives: $ma = mg\sin\theta - \mu_d mg\cos\theta \;\Longrightarrow\; a = g(\sin\theta - \mu_d\cos\theta)$

With the numbers: $a = 9{,}81\cdot(0{,}643 - 0{,}30\cdot 0{,}766) \approx 9{,}81\cdot 0{,}413 \approx 4{,}05\ \mathrm{m/s}^2$

$\ev{a = g(\sin\theta - \mu_d\cos\theta) \approx 4{,}05\ \mathrm{m/s}^2}$

Without friction one would have $g\sin 40^\circ \approx 6{,}30\ \mathrm{m/s}^2$: friction removes about a third of the acceleration.

Links

Topics: Dinamica · Attrito Concepts: Attrito dinamico · Forza normale · Forza peso · Seconda legge di Newton Skills: Scomposizione sul piano inclinato Methods: Metodo del diagramma di corpo libero Objects: Piano inclinato · Blocco