Sliding motion along an inclined plane

This article discusses a scenario/arrangement whose statics/dynamics/kinematics can be understood using the ideas of classical mechanics.
View other mechanics scenarios
The brown triangle is the fixed inclined plane and the black block is placed on it with a dry surface of contact.

The scenario here is a dry block (with a stable surface of contact) on a dry fixed inclined plane, with $\theta$ being the angle of inclination with the horizontal axis.

The extremes are $\theta = 0$ (whence, the plane is horizontal) and $\theta = \pi/2$ (whence, the plane is vertical).

Similar scenarios

The scenario discussed here can be generalized/complicated in a number of different ways. Some of these are provided in the table below.

sliding motion for adjacent blocks along an inclined plane Two blocks instead of one, with a normal force between them
sliding-cum-rotational motion along an inclined plane A sphere or cylinder on an inclined plane
pulley system on a double inclined plane Two sliding blocks, connected by a string over a pulley, forcing a relationship between their motion
block on free wedge on horizontal floor A block on the incline of a wedge that is free to move on a horizontal floor.
blocks on two inclines of a free wedge Blocks on two inclines of a wedge that is free to move on a horizontal floor.

Basic components of force diagram

Component-taking for the gravitational force.

A good way of understanding the force diagram is using the coordinate axes as the axis along the inclined plane and normal to the inclined plane. For simplicity, we will assume a two-dimensional situation, with no forces acting along the horizontal axis that is part of the inclined plane (in our pictorial representation, this no action axis is the axis perpendicular to the plane).

Taking components of the gravitational force

The most important thing for the force diagram is understanding how the gravitational force, which acts vertically downward on the block, splits into components along and perpendicular to the incline. The component along the incline is $mg\sin \theta$ and the component perpendicular to the incline is $mg\cos \theta$. The process of taking components is illustrated in the adjacent figure

Component perpendicular to the inclined plane

In this case, assuming a stable surface of contact, and that the inclined plane does not break under the weight of the block, and no other external forces, we get the following equation from Newton's first law of motion applied to the axis perpendicular to the inclined plane:

$\! N = mg \cos \theta$

where $N$ is the normal force between the block and the inclined plane, $m$ is the mass of the block, and $g$ is the acceleration due to gravity. $N$ acts inward on the inclined plane and outward on the block. Some observations:

Value/change in value of $\theta$ Value of $N$ Comments
$\theta = 0$ (horizontal plane) $N = mg$ The normal force exerted on a horizontal surface equals the mass times the acceleration due to gravity, which we customarily call the weight.
$\theta = \pi/2$ (vertical plane) $N = 0$ The block and the inclined plane are barely in contact and hardly pressed together.
$\theta$ increases from $0$ to $\pi/2$ $N$ reduces from $mg$ to $0$. The derivative $\frac{dN}{d\theta}$ is $-mg\sin \theta$ The force pressing the block and the inclined plane reduces as the slope of the incline increases.

For simplicity, we ignore the cases $\theta = 0$ and $\theta = \pi/2$ unless specifically dealing with them.

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:

Case Convention on the sign of acceleration $a$ Equation Simplified, after combining with $\! N = mg \cos \theta$
Block sliding down the inclined plane measured positive down the inclined plane, hence a positive number $\! ma = mg \sin \theta - \mu_kN$ $\! a = g (\sin \theta - \mu_k\cos \theta) = g \cos \theta (\tan \theta - \mu_k)$
Block sliding up the inclined plane measured positive down the inclined plane (note that acceleration opposes direction of motion, leading to retardation) $\! ma = mg \sin \theta + \mu_kN$ $\! a = g (\sin \theta + \mu_k \cos \theta) = g \cos \theta (\tan \theta + \mu_k)$
Block stationary no acceleration, so no sign convention $\! mg \sin \theta \le \mu_sN$ $\! \tan \theta \le \mu_s$

Value of the acceleration function

For a block sliding downward

As we saw above, if the block is sliding downward, the acceleration (downward positive) is given by:

$\! a = g(\sin \theta - \mu_k \cos \theta) = g\cos \theta(\tan \theta - \mu_k)$

The quotient $a/g$ where $a$ is downward acceleration and $g$ is the acceleration due to gravity, as a function of the angle (in radians). Note that each line corresponds to a particular value of friction coefficient $\mu_k$. The acceleration is measured in the downward direction. Note that when the value is negative, the block undergoes retardation and hence it will not start moving if it is initially placed at rest.

We see that $a/g$ is an increasing function of $\theta$ for fixed $\mu_k$. The point where it crosses the axis is $\tan^{-1}(\mu_k)$.

For a block sliding upward

As we saw above, if the block is sliding above, the acceleration (downward positive) is given by:

$\! a = g(\sin \theta + \mu_k \cos \theta) = g\cos \theta (\tan \theta + \mu_k)$

The quotient $a/g$ where $a$ is downward acceleration and $g$ is the acceleration due to gravity, as a function of the angle (in radians). Note that each line corresponds to a particular value of friction coefficient $\mu_k$. The acceleration is measured in the downward direction. Note that it is always negative, indicating that the block will undergo retardation.

The magnitude of acceleration (i.e., retardation) is maximum when $\theta = (\pi/2) - \tan^{-1}(\mu_k)$. In the extreme case that $\mu_k = 0$, retardation is maximum for $\theta = \pi/2$, and as $\mu_k$ increases, the angle for maximum retardation decreases. The magnitude of maximum possible retardation is:

$\! a = g\sqrt{1 + \mu_k^2}$

Behavior for a block initially placed at rest

Statics and dynamics

If the block is initially placed gently at rest, we have the following cases:

Case What happens
$\tan \theta < \mu_s$, or $\theta < \tan^{-1}(\mu_s)$ The block does not start sliding down, and instead stays stable where it is. The magnitude of static friction is $mg \sin \theta$, and its direction is upward along the inclined plane, precisely balancing and hence canceling the component of gravitational force along the inclined plane.
$\tan \theta > \mu_s$ or $\theta > \tan^{-1}(\mu_s)$ The block starts sliding down, and the downward acceleration is given by $a = g(\sin \theta - \mu_k \cos \theta) = g \cos \theta (\tan \theta - \mu_k)$.

The angle $\tan^{-1}(\mu_s)$ is termed the angle of repose, since this is the largest angle at which the block does not start sliding down.

Kinematics

The kinematic evolution in the second case is given as follows, if we set $t = 0$ as the time when the block is placed, we have the following (here, the row variable is written in terms of the column variable):

$\! v$ $\! t$ $\! s$ vertical displacement (call $\!h$) horizontal displacement (call $\! x$)
$\! v$ $\! v$ $\! gt\cos \theta(\tan \theta - \mu_k)$ $\! \sqrt{2sg\cos \theta (\tan \theta - \mu_k)}$ $\! \sqrt{2hg(1 - \mu_k \cot \theta)}$ $\! \sqrt{2xg(\tan \theta - \mu_k)}$
$\! s$ $\! v^2/(2g\cos \theta (\tan \theta - \mu_k))$ $\! (1/2)g \cos \theta (\tan \theta - \mu_k)t^2$ $\! s$ $\! h/\sin \theta$ $\! x/\cos \theta$
$\! h$ $\! v^2 \tan \theta/(2g(\tan \theta - \mu_k))$ $\! (1/2) g \cos \theta \sin \theta (\tan \theta - \mu_k)$ $\! s \sin \theta$ $\! h$ $\! x \tan \theta$
$\! x$ $\! v^2/(2g (\tan \theta - \mu_k))$ $\! (1/2)g \cos^2 \theta (\tan \theta - \mu_k)$ $\! s \cos \theta$ $\! h \cot \theta$ $\! x$

Given an initial speed downward

Statics and dynamics

If the block is given an initial downward speed, the acceleration is $g \cos \theta (\tan \theta - \mu_k)$ downward, and its long-term behavior is determined by whether $\tan \theta < \mu_k$ or $\tan \theta > \mu_k$:

• If $\tan \theta < \mu_k$, then the block's speed reduces with time. The magnitude of retardation is $g \cos \theta (\mu_k - \tan \theta)$. If the incline is long enough, this retardation continues until the block reaches a speed of zero, at which point it comes to rest and thence stays at rest.
• If $\tan \theta > \mu_k$, then the block's speed increases with time. The magnitude of acceleration is $g \cos \theta (\tan \theta - \mu_k)$. The block thus does not come to a stop and keeps going faster as it goes down the incline.

Kinematics

Fill this in later

Energy changes

Fill this in later

Given an initial speed upward

Statics and dynamics

If the block is given an initial upward speed, the acceleration is $g \cos \theta (\tan \theta + \mu_k)$ downward, i.e., the retardation is $g \cos \theta(\tan \theta + \mu_k)$. We again consider two cases:

Case What happens
$\! \tan \theta < \mu_k$ or $\! \theta < \tan^{-1}(\mu_k)$ The block's speed reduces with time as it rises. If the incline is long enough, the block eventually comes to a halt and stays still after that point.
$\! \tan \theta$ is between $\! \mu_k$ and $\! \mu_s$ The behavior is somewhat indeterminate, in the sense that whether the block starts sliding down after stopping its upward slide is unclear.
$\! \tan \theta > \mu_s$ The block's speed reduces with time as it rises, until it comes to a halt, after which it starts sliding downward, just as in the case of a block initially placed at rest.

Kinematics

Suppose the initial upward speed is $u$. Because of our downward positive convention, we will measure the velocity as $-u$. Then, we have:

Quantity Value
Time taken to reach highest point $\! t = \frac{u}{g \cos \theta (\tan \theta + \mu_k)}$
Displacement to highest point achieved (along incline) $\frac{u^2}{2g \cos \theta (\tan \theta + \mu_k)}$
Vertical height to highest point achieved $\frac{u^2 \tan \theta}{2g (\tan \theta + \mu_k)}$
Horizontal displacement to highest point achieved $\frac{u^2}{2g(\tan \theta + \mu_k)}$

In case that $\tan \theta > \mu_s$, the block is expected to slide back down and return to the original position. The kinematics of the downward motion are the same as those for a block initially placed at rest. Two important values are given below:

Quantity Value
Time taken on return journey $\! \frac{u}{g \cos \theta\sqrt{\tan^2 \theta - \mu_k^2}}$
Speed at instant of return $\! u \frac{\tan \theta - \mu_k}{\tan \theta + \mu_k}$

Energy changes

Fill this in later