How do I remember whether or goes down the slope?

Test at = 0. A flat surface should have no force down the slope, and 0 = 0 while 0 = 1 — so is the one down the slope.

R4–R6MechanicsStretch

Forces on Inclined Planes

A slope makes every force problem harder for one reason: the weight no longer points along or across the direction of motion. Choosing axes $\textbf{along and perpendicular to the slope}$ fixes that, and turns an awkward two-dimensional problem back into two one-dimensional ones.

40 min Forces and Newton’s Laws

The big picture

The choice of axes is the whole technique. With horizontal and vertical axes, the normal reaction and any friction both need resolving and three forces become six components. Tilt the axes to match the slope and only the weight needs resolving — the normal reaction lies entirely on one axis, friction entirely on the other, and the problem becomes routine. Recognising that the angle between the weight and the perpendicular to the slope equals the slope angle is the geometric fact that makes it work, and it is worth deriving once so the $\sin$ and $\cos$ never have to be guessed.

What you'll be able to do

Resolve weight into components along and perpendicular to a slope
Analyse motion on smooth and rough inclined planes
Determine whether a body slides or remains in equilibrium on a slope
Find the angle of friction and the coefficient from a limiting case

Why $\sin$ goes down the slope

Consider a slope inclined at $\theta$ to the horizontal. The weight acts vertically downwards, and the perpendicular to the slope is tilted from the vertical by exactly $\theta$ — the same angle, because both are rotations from the horizontal and vertical respectively.

So the angle between the weight and the inward perpendicular is $\theta$ . The component $\textbf{along}$ that perpendicular is therefore the adjacent one, $mg\cos\theta$ , and the component down the slope is the opposite one, $mg\sin\theta$ .

A quick check confirms the assignment. At $\theta = 0$ the surface is flat, and the formulas give no force down the slope and the full weight pressing in — correct. At $\theta = 90^{\circ}$ the surface is vertical, giving the full weight down the slope and nothing pressing in — also correct.

W_{\parallel} = mg\sin\theta

Down the slope.

W_{\perp} = mg\cos\theta

Into the slope.

R = mg\cos\theta

For a body on the slope with no other perpendicular forces.

Why this works

That end-point check takes five seconds and settles the $\sin$ / $\cos$ question permanently. If a formula does not give the right answer at $0^{\circ}$ and $90^{\circ}$ , it is the wrong way round.

Smooth slopes

On a smooth slope the only force along it is the weight component $mg\sin\theta$ , so applying $F = ma$ gives $mg\sin\theta = ma$ and the mass cancels.

The acceleration is therefore $g\sin\theta$ , independent of mass — the same reason all objects fall at $g$ , since the driving force and the inertia both scale with $m$ .

Perpendicular to the slope there is no acceleration, so the normal reaction balances the perpendicular weight component: $R = mg\cos\theta$ .

a = g\sin\theta

Smooth slope — mass cancels.

Example 1A 4 kg block slides down a smooth slope inclined at

30^{\circ}

. Find the acceleration and the normal reaction. Take

g = 9.8

1Along the slope:

a = g\sin30^{\circ} = 9.8 \times 0.5

2Perpendicular:

R = mg\cos30^{\circ} = 4(9.8)(0.8660)

Answer

a = 4.9\ \mathrm{m\,s^{-2}}

;

R = 33.9\ \mathrm{N}

(3 s.f.).

Tip — The normal reaction on a slope is $mg\cos\theta$ , always less than $mg$ . Writing $R = mg$ on an incline is a routine lost mark.

Rough slopes

With friction present, the maximum friction is $\mu R = \mu mg\cos\theta$ — note that it depends on the angle, because the normal reaction does.

For a body sliding down, friction acts up the slope and the equation along the slope is $mg\sin\theta - \mu mg\cos\theta = ma$ . The mass cancels again, giving $a = g\left(\sin\theta - \mu\cos\theta\right)$ .

For a body being pushed or pulled up, friction acts down the slope and adds to the weight component instead — the direction of friction always opposes the motion, so it must be decided before writing the equation.

a = g\left(\sin\theta - \mu\cos\theta\right)

Sliding down a rough slope.

Example 1A block slides down a slope at

35^{\circ}

with

\mu = 0.2

. Find its acceleration. Take

g = 9.8

\sin35^{\circ} = 0.5736

\cos35^{\circ} = 0.8192

a = 9.8\left(0.5736 - 0.2 \times 0.8192\right)

= 9.8\left(0.5736 - 0.1638\right) = 9.8 \times 0.4098

Answer

a = 4.02\ \mathrm{m\,s^{-2}}

(3 s.f.)

Why this works

If $\mu\cos\theta$ exceeded $\sin\theta$ the formula would give a negative acceleration, which is impossible for a body released from rest — it simply would not move. That comparison is the test in the next section.

Will it slide? The angle of friction

A body on a rough slope stays put if the weight component down the slope does not exceed the maximum friction: $mg\sin\theta \leq \mu mg\cos\theta$ .

The mass cancels, leaving $\tan\theta \leq \mu$ . So whether an object slides depends only on the angle and the coefficient — not on how heavy it is, which is a genuinely surprising and examinable result.

The critical angle at which sliding just begins satisfies $\tan\theta = \mu$ , and is called the $\textbf{angle of friction}$ . Tilting a plank until an object starts to slide is a standard way of measuring $\mu$ experimentally.

\tan\theta \leq \mu \Rightarrow \text{remains at rest}

Independent of mass.

\tan\lambda = \mu

The angle of friction

\lambda

Example 1A block is on the point of sliding when a plank is tilted to

22^{\circ}

. Find the coefficient of friction.

1On the point of sliding means

\tan\theta = \mu

\mu = \tan22^{\circ}

Answer

\mu = 0.404

(3 s.f.)

Example 2Will a block with

\mu = 0.6

remain at rest on a slope of

28^{\circ}

1Compute

\tan28^{\circ} = 0.5317

2Compare with

\mu = 0.6

0.5317 \leq 0.6

AnswerYes — it remains in equilibrium, whatever its mass.

Tip — The mass cancelling is worth stating explicitly in an answer. "Since $\tan\theta \leq \mu$ , the block remains at rest regardless of its mass" is exactly what a mark scheme wants.

Think like an examiner

Take axes along and perpendicular to the slope and say that you have.
Use $R = mg\cos\theta$ , not $mg$ , and show the resolving step.
Decide the direction of friction from the direction of motion before writing equations.
Note when the mass cancels — it is often the point of the question.

Common misconceptions

Using

R = mg

on an inclined plane.

Only the perpendicular component presses on the surface:

R = mg\cos\theta

Swapping

\sin

and

\cos

for the slope components.

Down the slope is

mg\sin\theta

. Check at

\theta = 0

: no force down a flat surface.

Taking friction as acting down the slope for a descending body.

Friction opposes motion, so it acts up the slope when the body slides down.

Assuming a heavier block is more likely to slide.

Mass cancels: sliding depends only on

\tan\theta

against

\mu

Inclined planes

W_{\parallel} = mg\sin\theta

Down the slope.

R = mg\cos\theta

Normal reaction.

a = g\sin\theta

Smooth slope.

a = g(\sin\theta - \mu\cos\theta)

Rough, sliding down.

\tan\theta \leq \mu

Condition to stay at rest.

Stretch yourself

A 6 kg block rests on a rough slope inclined at $25^{\circ}$ with $\mu = 0.35$ . A force $P$ is applied up the slope, parallel to it. Find the range of values of $P$ for which the block remains in equilibrium. Take $g = 9.8\ \mathrm{m\,s^{-2}}$ .

Hint — Two limiting cases: on the point of sliding down (friction acts up) and on the point of sliding up (friction acts down).

Reveal solution

SolutionFirst find the components.

R = 6(9.8)\cos25^{\circ} = 58.8 \times 0.9063 = 53.29

N, so maximum friction is

0.35 \times 53.29 = 18.65

N. The weight component down the slope is

58.8\sin25^{\circ} = 58.8 \times 0.4226 = 24.85

N. On the point of sliding

\textbf{down}

P

is at its smallest and friction acts up the slope at its maximum:

P + 18.65 = 24.85

, giving

P = 6.20

N. On the point of sliding

\textbf{up}

P

is at its largest and friction acts down the slope:

P = 24.85 + 18.65 = 43.50

N. So equilibrium holds for

6.20\ \mathrm{N} \leq P \leq 43.5\ \mathrm{N}

(3 s.f.). The width of that range,

2\mu R = 37.3

N, is exactly twice the maximum friction — friction has to be overcome in both directions, which is why rough slopes tolerate such a wide band of applied forces.

\blacksquare

Questions students ask

Why choose axes along the slope?

Because then the normal reaction lies entirely on one axis and friction entirely on the other, so only the weight needs resolving. Horizontal and vertical axes would require resolving three forces instead of one.

How do I remember whether

\sin

\cos

goes down the slope?

Test at

\theta = 0

. A flat surface should have no force down the slope, and

\sin 0 = 0

while

\cos 0 = 1

— so

\sin

is the one down the slope.

Does a heavier object slide more easily?

No. The sliding condition

\tan\theta \leq \mu

contains no mass, because both the driving force and the maximum friction are proportional to the weight.

What is the angle of friction?

The angle

\lambda

at which a body is on the point of sliding, satisfying

\tan\lambda = \mu

. Tilting a surface until motion begins is a practical way to measure

\mu

Key takeaways

Take axes along and perpendicular to the slope so only the weight needs resolving.
Weight components are $mg\sin\theta$ down the slope and $mg\cos\theta$ into it.
The normal reaction on a slope is $mg\cos\theta$ , less than $mg$ .
On a smooth slope $a = g\sin\theta$ ; with friction, subtract $\mu g\cos\theta$ .
A body stays at rest when $\tan\theta \leq \mu$ , regardless of mass.