Difference between revisions of "Sliding motion along an inclined plane"

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This article discusses a scenario/arrangement whose statics/dynamics/kinematics can be understood using the ideas of classical mechanics.
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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).

We assume that the block undergoes no rotational motion, i.e., it does not roll or topple. Although the diagram here shows a block with square cross section, this shape assumption is not necessary as long as we assume that the block undergoes no rotational motion.

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.

Scenario	Key addition/complication	Picture
sliding motion along a frictionless inclined plane	$\mu_k = \mu_s = 0$
toppling motion along an inclined plane	The block can topple
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
sliding motion along a frictionless circular incline	Incline is frictionless, but circular instead of linear (planar)
sliding motion along a circular incline	Incline is circular instead of linear (planar)
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

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).

The four candidate forces

Assuming no external forces are applied, there are four candidate forces on the block:

Force (letter)	Nature of force	Condition for existence	Magnitude	Direction
$mg$	gravitational force	unconditional	$mg$ where $m$ is the mass and $g$ is the acceleration due to gravity	vertically downward, hence an angle $\theta$ with the normal to the incline and an angle $(\pi/2) - \theta$ with the incline
$N$	normal force	unconditional	adjusts so that there is no net acceleration perpendicular to the plane of the incline	outward normal to the incline
$f_s$	static friction	no sliding	adjusts so that there is no net acceleration along the incline	up the incline (note that this could change if external forces were applied to push the block up the incline).
$f_k$	kinetic friction	sliding	$\mu_kN$ where $N$ is the normal force	opposite direction of motion -- hence down the incline if sliding up, up the incline if sliding down. Assuming the block was initially at rest, it can only slide down, hence up the incline is the only possibility.

Here is the force diagram (without components) in the no sliding case:

Here is the force diagram (without taking components) in the sliding down case:

Here is the force diagram (without taking components) in the sliding up case:

Component-taking for the gravitational force.

Taking components of the gravitational force

KEY FORCE CONCEPT (GRAVITY): For an object of mass $m$ close to the surface of the earth, gravity acts in a downward direction (toward the center of the earth) and has magnitude $mg$ , regardless of the presence or absence of other forces and regardless of the object's motion, velocity, or acceleration. Gravity acts on the block with the same magnitude and direction regardless of whether the block is sliding up, sliding down, or remaining where it is. Thus, the analysis of the gravitational force and its components is completely universal.

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

Normal component

mg\cos \theta

of gravitational force

mg

should cancel out normal force.

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.

KEY FORCE CONCEPT (ACTION-REACTION): Just because two forces are equal in magnitude and opposite in direction does not imply that they form an action-reaction pair in the sense of Newton's third law of motion. An action-reaction pair is a pair occurs between two bodies that exert forces on each other, not for a pair of forces both acting on the same body. In this case, $N$ and $mg\cos \theta$ do not form an action-reaction pair. The reason that they are equal is different, it is because the net acceleration in the normal direction is zero.

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, the case of no sliding, where an upward static friction $f_s$ acts (subject to the constraint $f_s \le \mu_sN$ ), 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.

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

What about the inclined plane?

A natural question that might occur at this stage is: what is the force diagram of the inclined plane? And, what is the effect on the inclined plane of the normal force and friction force exerted by the block on it?

The key thing to note is that by assumption, the inclined plane is fixed. This means that whatever mechanism is being used to fix the inclined plane is generating the necessary forces to balance the forces exerted by the block, so that by Newton's first law of motion, the inclined plane does not move. For instance, if the inclined plane is fixed to the floor with screws, then the contact force exerted from the floor (normal force or friction) exactly balances all the other forces on the inclined plane (including its own gravitational force and the contact force from the block).

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

Energy changes