Inclined plane with kinetic friction

Problem

A block of mass $m = 8{,}0\ \text{kg}$ rests on a $30^\circ$ inclined plane. The coefficient of kinetic friction between block and plane is $\mu_d = 0{,}25$. Calculate the acceleration of the block along the plane if it is left free to slide.

Solution

We choose the axes aligned with the plane: the $x$-axis along the plane, positive downward, and the $y$-axis perpendicular to the plane. We resolve the weight force $\vv{P} = m\vv{g}$ into its two components.

Along the plane (axis $x$) acts the component $P_x = mg\sin\theta$, while perpendicular to the plane the component is $P_y = mg\cos\theta$.

Along the $y$-axis the block does not accelerate, so the normal reaction balances the perpendicular component of the weight: $N = mg\cos\theta$

The kinetic friction force opposes the motion (the block slides downward, so friction is directed up the plane) and has magnitude: $F_a = \mu_d N = \mu_d\, mg\cos\theta$

We apply Newton’s second law along the plane: $mg\sin\theta - \mu_d\, mg\cos\theta = ma$

The mass cancels and we obtain: $a = g\,(\sin\theta - \mu_d\cos\theta)$

We substitute the numerical values with $g = 9{,}81\ \text{m/s}^2$, $\sin 30^\circ = 0{,}500$ and $\cos 30^\circ = 0{,}866$: $a = 9{,}81 \cdot (0{,}500 - 0{,}25 \cdot 0{,}866) = 9{,}81 \cdot 0{,}2835$

$\ev{a \approx 2{,}77\ \text{m/s}^2 \quad \text{directed down along the plane}}$

Since $\sin\theta > \mu_d\cos\theta$, the result is positive: the block does indeed slide downward, accelerating.

Links

Topics: Dynamics · Friction Concepts: Newton’s second law · Kinetic friction Skills: Resolution on an inclined plane Objects: Inclined plane · Block