Equilibrium on Inclined Planes

AS · M1 · 12 min

A particle on a slope is the setting for more Paper 4 questions than any other: blocks held on ramps, crates about to slide, particles pushed by horizontal forces. Everything in this note uses tools you already have (resolving and the friction law), applied in the directions that suit a slope: parallel to it and perpendicular to it. The skill to master is handling friction, which can act up or down the plane depending on which way the particle is about to move.

Setting up: the two directions

A line of greatest slope is a line down the plane in the steepest direction, the way a ball would roll. In Paper 4 all forces act in the vertical plane containing a line of greatest slope, so the problem is two-dimensional.

On a plane inclined at $\alpha$ to the horizontal, resolve:

parallel to the plane (along the line of greatest slope), where friction and any applied force along the plane act;
perpendicular to the plane, where the normal reaction acts.

With this choice, $R$ appears only in the perpendicular equation and $F$ only in the parallel one. The weight is the only force that always needs splitting:

Components of weight on a plane inclined at α

mg\sin\alpha \text{ down the plane}, \qquad mg\cos\alpha \text{ into the plane}

If you are unsure which is sin and which is cos, use the check from Resolving forces: on a flat plane ( $\alpha = 0$ ) nothing pulls the particle along, so the parallel component must be $mg\sin\alpha$ .

Smooth planes

On a smooth plane there is no friction, so a particle can only stay at rest if some other force balances $mg\sin\alpha$ .

Force parallel to a smooth plane

A particle of mass $3\ \text{kg}$ is held at rest on a smooth plane inclined at $30^\circ$ to the horizontal by a force $P\ \text{N}$ acting up the line of greatest slope. Find $P$ and the normal reaction.

Solution

Parallel to the plane: $P = 30\sin 30^\circ = 15\ \text{N}$ .

Perpendicular to the plane: $R = 30\cos 30^\circ = 26.0\ \text{N}$ .

A horizontal force on a slope must itself be resolved. A horizontal force $P$ pushing towards the slope makes angle $\alpha$ with the plane, so it contributes $P\cos\alpha$ up the plane and $P\sin\alpha$ into the plane. That second component increases the normal reaction.

A particle of weight 20 N on a smooth plane inclined at 20°, held in equilibrium by a horizontal force P.

Horizontal force on a smooth plane

A particle of weight $20\ \text{N}$ rests on a smooth plane inclined at $20^\circ$ . It is held in equilibrium by a horizontal force $P\ \text{N}$ acting in the vertical plane containing a line of greatest slope. Find $P$ and the normal reaction.

Solution

Parallel to the plane:

P\cos 20^\circ = 20\sin 20^\circ \quad\Rightarrow\quad P = 20\tan 20^\circ = 7.28

Perpendicular to the plane (the horizontal force pushes into the plane):

R = 20\cos 20^\circ + P\sin 20^\circ = 18.794 + 2.490 = 21.3\ \text{N}

Check: resolving horizontally and vertically instead, $R\cos 20^\circ = 20$ , so $R = 21.3\ \text{N}$ , which agrees.

Rough planes with no applied force

A particle at rest on a rough plane inclined at α. Friction acts up the plane.

A particle resting on a rough slope with no other forces has three forces on it: weight, normal reaction and friction up the plane.

R = mg\cos\alpha, \qquad F = mg\sin\alpha

It stays at rest provided $F \le \mu R$ :

mg\sin\alpha \le \mu mg\cos\alpha \quad\Rightarrow\quad \tan\alpha \le \mu

Key result

A particle placed on a rough plane inclined at $\alpha$ , with no other forces, stays at rest if $\tan\alpha \le \mu$ and slides down if $\tan\alpha > \mu$ . If it is about to slide, $\mu = \tan\alpha$ . The mass does not matter.

This gives a simple experiment for measuring $\mu$ : tilt a plane until the block just starts to slide and measure the angle.

On the point of slipping

A particle rests on a rough plane inclined at angle $\alpha$ , where $\tan\alpha = \tfrac34$ .

(a) The particle is on the point of slipping. Find $\mu$ .

(b) Instead $\mu = 0.8$ and the particle has mass $4\ \text{kg}$ . Show that it remains at rest and find the frictional force.

Solution

(a) Limiting equilibrium with no other forces: $\mu = \tan\alpha = 0.75$ .

(b) $\sin\alpha = 0.6$ , $\cos\alpha = 0.8$ . $R = 40 \times 0.8 = 32\ \text{N}$ , so $\mu R = 0.8 \times 32 = 25.6\ \text{N}$ . The friction needed for equilibrium is $F = 40 \times 0.6 = 24\ \text{N}$ . Since $24 < 25.6$ , the particle remains at rest, and the frictional force is $24\ \text{N}$ up the plane.

Rough planes with an applied force: two cases

Now add a force $P$ up the plane. If $P$ is small, the particle tends to slide down, so friction acts up the plane to help $P$ . If $P$ is large, the particle tends to slide up, so friction acts down the plane, against $P$ . Between these, the particle is at rest with friction less than its maximum.

A particle on a rough plane inclined at 25°, held by a force P up the line of greatest slope. Friction is shown up the plane: the case where the particle is about to slide down.

Least and greatest force for equilibrium on a rough plane

Resolve perpendicular to the plane to find $R$ (in terms of $P$ if $P$ has a perpendicular component).
About to slide down (least $P$ ): friction $\mu R$ acts up the plane. Resolve parallel: $P + \mu R = mg\sin\alpha$ (adapt for components of $P$ ).
About to slide up (greatest $P$ ): friction $\mu R$ acts down the plane. Resolve parallel: $P = mg\sin\alpha + \mu R$ .
The particle is in equilibrium for $P_{\text{least}} \le P \le P_{\text{greatest}}$ .
If step 2 gives a negative value, friction alone can hold the particle ( $\tan\alpha \le \mu$ ), so the least value is $0$ .

Range of values for a force parallel to the plane

A particle of mass $5\ \text{kg}$ is on a rough plane inclined at $25^\circ$ to the horizontal, with coefficient of friction $0.3$ . A force of magnitude $P\ \text{N}$ acts on the particle up a line of greatest slope. Find the set of values of $P$ for which the particle remains at rest.

Solution

Perpendicular: $R = 50\cos 25^\circ = 45.315\ \text{N}$ , so $\mu R = 0.3 \times 45.315 = 13.595\ \text{N}$ .

The weight component down the plane is $50\sin 25^\circ = 21.131\ \text{N}$ .

About to slide down (friction up the plane):

P + 13.595 = 21.131 \quad\Rightarrow\quad P = 7.536

About to slide up (friction down the plane):

P = 21.131 + 13.595 = 34.726

So $7.54 \le P \le 34.7$ .

Forces at an angle to the plane

A force acting at an angle $\theta$ above the line of greatest slope has a component $P\cos\theta$ up the plane and a component $P\sin\theta$ away from the plane, which reduces the normal reaction. (A force angled into the plane increases $R$ instead.) Always find $R$ from the perpendicular equation; never assume $R = mg\cos\alpha$ when an applied force has a perpendicular component.

A particle of weight 40 N on a rough plane inclined at 35°. A force of 30 N acts at 20° above the line of greatest slope; the particle is about to move up the plane, so friction acts down it.

Finding the coefficient of friction

A particle of mass $4\ \text{kg}$ is on a rough plane inclined at $35^\circ$ . A force of $30\ \text{N}$ acts on it at $20^\circ$ above a line of greatest slope, in the vertical plane containing that line. The particle is about to move up the plane. Find the coefficient of friction.

Solution

Perpendicular to the plane (the force pulls partly away from the plane):

R + 30\sin 20^\circ = 40\cos 35^\circ \quad\Rightarrow\quad R = 32.766 - 10.261 = 22.505\ \text{N}

Parallel to the plane. The particle is about to move up, so friction acts down the plane:

30\cos 20^\circ = 40\sin 35^\circ + F \quad\Rightarrow\quad F = 28.191 - 22.943 = 5.248\ \text{N}

Limiting, so $F = \mu R$ :

\mu = \frac{5.248}{22.505} = 0.233

Horizontal force on a rough plane (exam standard)

A particle of mass $2\ \text{kg}$ is on a rough plane inclined at $30^\circ$ to the horizontal. The coefficient of friction is $0.4$ . A horizontal force of magnitude $P\ \text{N}$ , acting in the vertical plane containing a line of greatest slope, pushes the particle towards the plane. Find the least and greatest values of $P$ for which the particle remains in equilibrium.

Solution

Weight $20\ \text{N}$ : components $20\sin 30^\circ = 10$ down the plane and $20\cos 30^\circ = 17.321$ into it. The horizontal force has components $P\cos 30^\circ$ up the plane and $P\sin 30^\circ = 0.5P$ into it.

Perpendicular to the plane:

R = 17.321 + 0.5P

Least $P$ (about to slide down, friction up the plane):

P\cos 30^\circ + 0.4(17.321 + 0.5P) = 10

P(0.8660 + 0.2) = 10 - 6.928 \quad\Rightarrow\quad 1.0660P = 3.072 \quad\Rightarrow\quad P = 2.88

Greatest $P$ (about to slide up, friction down the plane):

P\cos 30^\circ = 10 + 0.4(17.321 + 0.5P)

P(0.8660 - 0.2) = 16.928 \quad\Rightarrow\quad 0.6660P = 16.928 \quad\Rightarrow\quad P = 25.4

The particle is in equilibrium for $2.88 \le P \le 25.4$ .

Notice how the normal reaction depends on $P$ here, so the friction term changes between the two cases. Forgetting the $0.5P$ in $R$ is the commonest error in this type of question.

Two conditions, two unknowns

Some questions give both limiting situations and ask you to find the mass or the coefficient of friction. Write one equation for each case and solve simultaneously; often adding and subtracting the equations is quickest. Practice question 7 is of this type.

Common mistakes

Slope errors

$mg\cos\alpha$ down the plane. The component down the plane is $mg\sin\alpha$ .
$R = mg\cos\alpha$ when another force has a perpendicular component. A horizontal push or a pull at an angle to the plane changes $R$ .
Friction always up the plane. Friction opposes the likely motion. If the particle is about to move up, friction acts down.
Using $F = \mu R$ when the particle is just "at rest". Only use it at the limit; otherwise find $F$ from equilibrium and check.
Resolving the horizontal force as $P\sin\alpha$ along the plane. A horizontal force makes angle $\alpha$ with the plane, so its component along the plane is $P\cos\alpha$ .

Exam technique

Exam tip

State the direction you are resolving in each time: "Resolving parallel to the plane", "Resolving perpendicular to the plane". It makes your working easy to mark.
For "find the set of values" or "find the least and greatest" questions, label your two cases clearly ("about to slide down", "about to slide up") and draw friction the right way in each.
If the angle is given as $\sin\alpha = 0.6$ or $\tan\alpha = \tfrac{3}{4}$ , use exact sin and cos; the answers are usually designed to come out neatly.
Mark schemes award the perpendicular equation, the parallel equation and the use of $F = \mu R$ separately. Even if you mix up sin and cos, a three-term equation with all the right forces usually earns the method mark.

Summary

Resolve parallel and perpendicular to the plane. Weight gives $mg\sin\alpha$ down the plane and $mg\cos\alpha$ into it.
With no applied force, a particle on a rough plane is in equilibrium if $\tan\alpha \le \mu$ ; about to slip means $\mu = \tan\alpha$ .
A force up the plane gives a range of equilibrium: least value when about to slide down (friction up), greatest when about to slide up (friction down).
A horizontal force $P$ contributes $P\cos\alpha$ along the plane and $P\sin\alpha$ into it; a force at $\theta$ above the plane contributes $P\cos\theta$ along and $P\sin\theta$ away from it.
Always find $R$ from the perpendicular equation before using $\mu R$ .

Practice

Question

A particle of mass $1.5\ \text{kg}$ is held at rest on a smooth plane inclined at $40^\circ$ by a force acting up a line of greatest slope. Find the force and the normal reaction.
A particle rests on a rough plane inclined at $25^\circ$ , with no other forces acting. Find the least possible value of the coefficient of friction.
A particle of mass $3\ \text{kg}$ is placed on a rough plane inclined at $\alpha$ , where $\tan\alpha = \tfrac{5}{12}$ . The coefficient of friction is $0.5$ . Show that the particle remains at rest and find the frictional force.
A particle of mass $5\ \text{kg}$ is held at rest on a smooth inclined plane by a horizontal force of $20\ \text{N}$ . Find the angle of inclination of the plane and the normal reaction.
A particle of mass $8\ \text{kg}$ is on a rough plane inclined at $30^\circ$ with $\mu = 0.2$ . A force $P\ \text{N}$ acts up a line of greatest slope. Find the set of values of $P$ for which the particle is in equilibrium.
A particle of mass $6\ \text{kg}$ is on a rough plane inclined at $20^\circ$ with $\mu = 0.35$ . A force of magnitude $P\ \text{N}$ acts at $15^\circ$ above a line of greatest slope. The particle is about to move up the plane. Find $P$ .
A particle of mass $m\ \text{kg}$ is on a rough plane inclined at $30^\circ$ . When a force of $10\ \text{N}$ acts up a line of greatest slope, the particle is about to slide down. When the force is increased to $30\ \text{N}$ , the particle is about to slide up. Find $m$ and the coefficient of friction.
A particle of mass $0.5\ \text{kg}$ is on a rough plane inclined at $40^\circ$ with $\mu = 0.25$ . A horizontal force $P\ \text{N}$ acts on it, in the vertical plane containing a line of greatest slope, pushing it towards the plane. Find the least and greatest values of $P$ for equilibrium.

Answers

Parallel: $P = 15\sin 40^\circ = 9.64\ \text{N}$ . Perpendicular: $R = 15\cos 40^\circ = 11.5\ \text{N}$ .
About to slip gives $\mu = \tan 25^\circ = 0.466$ , so $\mu \ge 0.466$ .
$\sin\alpha = \tfrac{5}{13}$ , $\cos\alpha = \tfrac{12}{13}$ . $R = 30 \times \tfrac{12}{13} = 27.69\ \text{N}$ , $\mu R = 13.8\ \text{N}$ . Friction needed: $F = 30 \times \tfrac{5}{13} = 11.5\ \text{N}$ . Since $11.5 < 13.8$ , the particle stays at rest with friction $11.5\ \text{N}$ up the plane.
Parallel: $20\cos\alpha = 50\sin\alpha$ , so $\tan\alpha = 0.4$ and $\alpha = 21.8^\circ$ . Perpendicular: $R = 50\cos\alpha + 20\sin\alpha = \sqrt{50^2 + 20^2} = 53.9\ \text{N}$ (or $46.42 + 7.43$ ).
$R = 80\cos 30^\circ = 69.28$ , $\mu R = 13.86$ ; weight component $40$ . About to slide down: $P = 40 - 13.86 = 26.1$ . About to slide up: $P = 40 + 13.86 = 53.9$ . So $26.1 \le P \le 53.9$ .
Perpendicular: $R = 60\cos 20^\circ - P\sin 15^\circ$ . Parallel (friction down the plane): $P\cos 15^\circ = 60\sin 20^\circ + 0.35(60\cos 20^\circ - P\sin 15^\circ)$ . So $P(\cos 15^\circ + 0.35\sin 15^\circ) = 20.521 + 19.734 = 40.255$ , giving $P = \dfrac{40.255}{1.05651} = 38.1\ \text{N}$ .
About to slide down: $10 + \mu R = 10m\sin 30^\circ$ . About to slide up: $30 = 10m\sin 30^\circ + \mu R$ , where $R = 10m\cos 30^\circ$ in both. Adding: $40 = 20m\sin 30^\circ = 10m$ , so $m = 4$ . Subtracting: $20 = 2\mu R = 2\mu(40\cos 30^\circ)$ , so $\mu = \dfrac{10}{34.64} = 0.289$ .
Weight $5\ \text{N}$ : $5\sin 40^\circ = 3.214$ down the plane, $5\cos 40^\circ = 3.830$ into it. $R = 3.830 + P\sin 40^\circ$ . Least (friction up): $P\cos 40^\circ + 0.25(3.830 + P\sin 40^\circ) = 3.214$ , so $P(\cos 40^\circ + 0.25\sin 40^\circ) = 2.256$ and $P = 2.43$ . Greatest (friction down): $P\cos 40^\circ = 3.214 + 0.25(3.830 + P\sin 40^\circ)$ , so $P(\cos 40^\circ - 0.25\sin 40^\circ) = 4.171$ and $P = 6.89$ . So $2.43 \le P \le 6.89$ .

Setting up: the two directions

Smooth planes

Rough planes with no applied force

Rough planes with an applied force: two cases

Forces at an angle to the plane

Two conditions, two unknowns

Common mistakes

Exam technique

Summary

Practice

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