Acceleration on an incline with friction

Problem

Sketch, as a function of the incline angle $\theta$, the qualitative behaviour of the acceleration of a block sliding down with kinetic friction $\mu_d = 0{,}30$, comparing it with the frictionless case. State the value of $\theta$ for which the acceleration is zero and the one for which it is maximum (on a $90^\circ$ incline).

Solution

Acceleration law. Along the incline, for a block sliding down, the component of the weight $mg\sin\theta$ pulls downwards and kinetic friction $\mu_d N = \mu_d\,mg\cos\theta$ opposes it. Newton’s second law gives: $a(\theta) = g\,(\sin\theta - \mu_d\cos\theta)$ In the frictionless case ($\mu_d = 0$) this reduces to $a_0(\theta) = g\sin\theta$.

Angle of zero acceleration. The block (already moving) slows down or stays at constant speed when $a = 0$: $\sin\theta = \mu_d\cos\theta \;\Rightarrow\; \tan\theta = \mu_d = 0{,}30 \;\Rightarrow\; \theta \approx 16{,}7^\circ$ For $\theta < 16{,}7^\circ$ the formula would give $a<0$: this means friction is able to stop the (moving) block and, from rest, static friction holds it still.

$\ev{a = 0 \text{ for } \theta \approx 16{,}7^\circ}$

Maximum. $a(\theta)$ increases with $\theta$ and reaches its maximum on the vertical plane, $\theta = 90^\circ$, where $\cos 90^\circ = 0$ (friction vanishes because $N\to 0$) and $\sin 90^\circ = 1$: $a(90^\circ) = g\,(1 - 0) = g \approx 9{,}8\;\mathrm{m/s^2}$

$\ev{a_\text{max} = g \text{ at } \theta = 90^\circ \text{ (free fall)}}$

Qualitative graph (comparison).

$\theta$	$a_0 = g\sin\theta$ (frictionless)	$a = g(\sin\theta-\mu_d\cos\theta)$ (with friction)
$0^\circ$	$0$	$\approx -2{,}9$ (stays still, $a=0$)
$16{,}7^\circ$	$\approx 2{,}8$	$0$
$30^\circ$	$4{,}9$	$\approx 2{,}4$
$45^\circ$	$\approx 6{,}9$	$\approx 4{,}8$
$90^\circ$	$9{,}8$	$9{,}8$

The frictionless curve $a_0 = g\sin\theta$ starts from $0$ and rises monotonically up to $g$. The curve with friction is shifted downwards and to the right: it starts from zero only at $\theta\approx 16{,}7^\circ$, always lies below the frictionless curve (by $\mu_d g\cos\theta$), but the two curves rejoin at $\theta = 90^\circ$, where both equal $g$ because the normal reaction — and hence friction — vanishes. The difference between the two curves is greatest at small $\theta$ and shrinks to zero towards $90^\circ$.

Links

Topics: Dynamics · Friction Concepts: Kinetic friction · Newton’s second law · Normal force · Weight force Skills: Resolution on an incline · Applying Newton’s laws · Use of trigonometry Methods: Free-body diagram method Objects: Inclined plane · Block