Inclined Surfaces and Resolving Forces

Spec mapping: OCR H556 Module 3.2 — Forces in action (free-body diagrams, resolving forces into perpendicular components, equilibrium and Newton's second law on inclined surfaces; introduction to friction as a contact force). Refer to the official OCR H556 specification document for exact wording.

Many A-Level mechanics problems involve objects on slopes — a trolley on a ramp, a skier on a hill, a book on a tilted desk, a stack of stationery on a sloping shelf. On a slope, gravity is no longer aligned with a convenient axis, and the normal reaction is no longer simply equal to the weight. To analyse these situations you must resolve forces — express a vector as the sum of two perpendicular components — and then apply Newton's second law along each direction separately.

This lesson develops the resolving techniques used throughout Module 3.2 and is a skill that reappears in circular motion (Module 5.4), projectile motion (the previous lesson in this course), electric and gravitational fields (Module 5.5), and even the Young modulus practical (PAG 4). It also introduces friction at the level of qualitative understanding (the formal $\mu_s$ vs $\mu_k$ distinction is more an undergraduate topic, but the idea of a friction coefficient bounding the friction force is examinable.)

Key Definition: A vector $\vec{V}$ in a plane can be uniquely resolved into two perpendicular components $V_x = V\cos\alpha$ and $V_y = V\sin\alpha$ , where $\alpha$ is the angle between $\vec{V}$ and the $x$ -axis. On an incline, the smart axis choice is to align $x$ along the slope and $y$ perpendicular to the slope — making the motion one-dimensional along $x$ .

The Idea of Components

A single vector can always be written as the vector sum of two perpendicular vectors called its components. On a horizontal floor, gravity has components $(0, -mg)$ in the natural $(x, y)$ Cartesian axes — simple. On a slope of angle $\theta$ to the horizontal, it is far more convenient to choose axes along and perpendicular to the slope, because:

The motion is along the slope (one-dimensional along the new $x$ -axis).
The normal contact force from the slope is purely perpendicular (it has only a $y$ -component).
Only the weight needs to be decomposed.

Consider a block of mass $m$ on a smooth slope at angle $\theta$ to the horizontal. The weight $\vec{W} = m\vec{g}$ points straight down. With the slope-aligned axes (positive $x$ down-slope, positive $y$ away from slope), the weight decomposes as:

Component parallel to slope (positive down-slope): $W_{\parallel} = mg \sin\theta$ .
Component perpendicular to slope (negative — pointing into the slope): $W_{\perp} = mg \cos\theta$ .

Exam Tip: Students routinely muddle $\sin$ and $\cos$ . The mnemonic is: as $\theta$ increases from $0^{\circ}$ (flat) to $90^{\circ}$ (vertical wall), the parallel component grows from $0$ to $mg$ — that is the behaviour of $\sin\theta$ . The perpendicular component shrinks from $mg$ to $0$ — that is $\cos\theta$ . Check the limits whenever in doubt.

A quick sanity check: at $\theta = 0^{\circ}$ , $W_{\parallel} = 0$ (nothing slides on a flat floor) and $W_{\perp} = mg$ (full weight on the floor). At $\theta = 90^{\circ}$ , $W_{\parallel} = mg$ (free fall — no contact with the "wall") and $W_{\perp} = 0$ (no normal force from a vertical wall on a free-hanging body). These limiting cases are always worth verifying.

Normal Reaction on a Slope

Because the block does not accelerate perpendicular to the slope (it stays on the surface — neither lifts off nor sinks into it), the forces perpendicular to the slope must sum to zero. So the normal contact force balances the perpendicular weight component:

$N = mg \cos\theta$

Note that $N < mg$ for any $\theta > 0$ — the slope does not bear the full weight. This has a striking consequence for friction: since static friction satisfies $F_{\text{friction}} \leq \mu_s N$ , the maximum friction available on a steep slope is less than on a flat surface. This is the underlying reason why steep slopes are slippery; the slope cannot "grip" the body as firmly as a horizontal surface can.

Acceleration Down a Smooth Slope

If the slope is smooth (frictionless), the only force along the slope is the parallel component of weight:

$F_{\parallel} = mg \sin\theta$

Applying Newton II along the slope ( $x$ -direction, positive down-slope):

$ma = mg \sin\theta \quad \Longrightarrow \quad a = g \sin\theta$

The mass cancels — every body, irrespective of mass, accelerates down a smooth slope at the same rate $g \sin\theta$ . This is a beautiful generalisation of the universal-free-fall result: just as all bodies fall vertically at $g$ , all bodies slide down a smooth slope at $g \sin\theta$ .

Limiting cases: $\theta = 0^{\circ}$ , $a = 0$ (nothing slides on a flat floor); $\theta = 90^{\circ}$ , $a = g$ (recovers free fall, no surface to constrain the body).

Worked Example 1 — Smooth Ramp

A 2.5 kg trolley is released from rest on a smooth ramp inclined at $25^{\circ}$ to the horizontal. The ramp is 1.60 m long. Find (a) the acceleration down the ramp, (b) the speed at the bottom, (c) the time taken to slide down.

(a) $a = g \sin\theta = 9.81 \times \sin 25^{\circ} = 9.81 \times 0.4226 = 4.15$ m s $^{-2}$ .

(b) Using $v^2 = u^2 + 2as$ : $v^2 = 0 + 2 \times 4.15 \times 1.60 = 13.27$ , so $v = 3.64$ m s $^{-1}$ .

Cross-check via energy: vertical drop $h = 1.60 \sin 25^{\circ} = 0.676$ m. Conservation of energy: $\frac{1}{2}mv^2 = mgh$ , so $v^2 = 2gh = 2 \times 9.81 \times 0.676 = 13.27$ , giving $v = 3.64$ m s $^{-1}$ . ✓

Energy methods and force methods always agree — for a smooth surface.

Friction on a Slope

If the slope is rough, friction acts opposite to the direction of motion (or opposite to the tendency of motion if the body is stationary). Let $\mu$ be the coefficient of friction. Although OCR A-Level does not require numerical computation of $\mu_s$ vs $\mu_k$ , you should know:

Static friction $\mu_s$ — applies to a stationary body. Static friction takes whatever value (up to a maximum $\mu_s N$ ) is needed to keep the body stationary. Once $F_{\text{applied}}$ exceeds $\mu_s N$ , the body starts to slide.
Kinetic friction $\mu_k$ — applies once sliding has begun. Kinetic friction has a fixed magnitude $\mu_k N$ regardless of velocity.
Typically $\mu_s > \mu_k$ — it takes more force to start motion than to keep it going. This is why pushing a stationary fridge across a kitchen floor is hardest at the moment of breakaway.
$\mu$ is dimensionless (it is a ratio of two forces). Typical values: $\mu \approx 0.5$ for wood-on-wood, $\mu \approx 0.7$ for rubber tyre on dry tarmac, $\mu \approx 0.05$ for ice on ice, $\mu \approx 0.001$ for ball-bearings.

The Angle of Repose

The condition for a body to be on the verge of slipping on a rough slope is that the maximum static friction equals the parallel component of weight:

$mg \sin\theta = \mu_s mg \cos\theta \quad \Longrightarrow \quad \tan\theta = \mu_s$

This defines the angle of repose $\theta_R$ — the maximum slope angle at which the body remains stationary. For wood on wood ( $\mu_s \approx 0.5$ ), $\theta_R = \arctan(0.5) \approx 27^{\circ}$ . For granular materials (sand, gravel, scree), $\mu_s$ ranges from $0.5$ to $0.8$ , giving angles of repose of about $30^{\circ}$ to $40^{\circ}$ — which is why mountain scree slopes have such a remarkably constant pitch globally.

Worked Example 2 — Block on a Rough Slope

A $4.0$ kg block sits on a slope at $30^{\circ}$ . The static friction force currently acts at $15$ N up the slope. Find the resulting acceleration.

Parallel weight component: $W_{\parallel} = mg \sin\theta = 4.0 \times 9.81 \times 0.500 = 19.62$ N (down the slope).
Friction $= 15$ N (up the slope).
Net force: $F_{\text{net}} = 19.62 - 15 = 4.62$ N (down the slope).
Acceleration: $a = F_{\text{net}} / m = 4.62 / 4.0 = 1.16$ m s $^{-2}$ (down the slope).

The block is accelerating down the slope; the friction has been overcome but only partially. If the friction were larger ( $\geq 19.62$ N), the block would be stationary.

Worked Example 3 — Object Held by a Rope on a Slope

A $10$ kg crate is held stationary on a smooth slope of $20^{\circ}$ by a rope parallel to the slope. Find (a) the tension in the rope, (b) the normal reaction.

Resolve weight:

$W_{\parallel} = 10 \times 9.81 \times \sin 20^{\circ} = 98.1 \times 0.342 = 33.6$ N.
$W_{\perp} = 10 \times 9.81 \times \cos 20^{\circ} = 98.1 \times 0.940 = 92.2$ N.

(a) Equilibrium along the slope: $T = W_{\parallel} = 33.6$ N (rope holds the crate against gravity along the slope).

(b) Equilibrium perpendicular: $N = W_{\perp} = 92.2$ N.

Note: had the rope pulled at an angle to the slope, both the tension and the normal reaction would shift — see Worked Example 5 below.

Choosing Axes — A Crucial Skill

On a slope, the conventional (horizontal, vertical) axes lead to messy algebra because the motion is diagonal. The smart move is to rotate your axes so that one axis lies along the slope and the other is perpendicular to it. Then:

Motion is one-dimensional (along the slope $x$ -axis).
The normal reaction has only one non-zero component (perpendicular).
Only the weight needs to be decomposed.

Always state your choice of axes explicitly at the start of your working. Examiners reward clarity, and unclear axis conventions are the source of an enormous fraction of self-inflicted sign errors.

flowchart TD
  A[Slope problem starts] --> B[Choose axes: x along slope, y perpendicular]
  B --> C[State sign convention: positive x is down-slope]
  C --> D[Draw FBD: weight, normal N, friction f, applied force]
  D --> E[Resolve weight: W parallel = mg sin theta, W perp = mg cos theta]
  E --> F{Body in equilibrium?}
  F -->|Yes| G[Sum of forces along x = 0 and along y = 0]
  F -->|No| H[Newton II: sum F along x = ma, sum F along y = 0]
  G --> I[Solve for unknown forces]
  H --> J[Solve for a; then SUVAT if needed]

Pulleys and Inclined Planes — Compound Problems

A classic compound problem is a block on a slope connected by a string over a pulley to another block hanging vertically. This combines the slope decomposition with multi-body Newton II.

Worked Example 4 — Atwood-Style Pulley

A $3.0$ kg block on a smooth $30^{\circ}$ slope is connected by a light inextensible string (over a frictionless pulley at the top of the slope) to a $2.0$ kg mass hanging vertically. Find the common acceleration and the tension.

Strategy: write Newton II for each block in its own natural axis (slope-aligned for the slope block, vertical for the hanging block). The string tension $T$ is the same on both sides of the pulley (light string, frictionless pulley), and the magnitude of the acceleration is the same for both blocks (inextensible string).

Block on slope (along slope, positive down-slope):

$3.0\,g\sin 30^{\circ} - T = 3.0\,a$ $3.0 \times 9.81 \times 0.500 - T = 3.0\,a$ $14.715 - T = 3.0\,a \quad \text{(eq. 1)}$

Hanging block (vertical, positive upward — note the upward acceleration if the slope block descends, since the string is over the pulley):

$T - 2.0\,g = 2.0\,a$ $T - 19.62 = 2.0\,a \quad \text{(eq. 2)}$

Adding (1) and (2) to eliminate $T$ :

$14.715 - 19.62 = 5.0\,a \Rightarrow a = -\frac{4.905}{5.0} = -0.981 \text{ m s}^{-2}$

The negative sign means the assumed direction (3 kg block descending) is wrong: in fact the 3 kg block slides up the slope while the 2 kg hanging block descends.

Sanity check: the parallel weight of the slope block is $14.7$ N (trying to pull the slope block down-slope); the weight of the hanging block is $19.62$ N (trying to pull the hanging block down, which by string tension pulls the slope block up the slope). Since $19.62 > 14.7$ , the hanging block wins — the system accelerates with the hanging block descending. ✓

From (2): $T = 19.62 + 2.0 \times (-0.981) = 19.62 - 1.962 = 17.66$ N.

Worked Example 5 — Force Applied at an Angle on a Slope

A $5.0$ kg block on a smooth $15^{\circ}$ slope is pulled by a rope at $20^{\circ}$ above the slope with tension $30$ N. Find (a) the acceleration along the slope, (b) the normal reaction.

Choose axes along ( $x$ , positive up-slope) and perpendicular ( $y$ , positive away from slope) to the slope.

Decompose tension (rope makes angle $20^{\circ}$ with the slope):

$T_{\parallel} = 30 \cos 20^{\circ} = 28.19$ N (up the slope).
$T_{\perp} = 30 \sin 20^{\circ} = 10.26$ N (away from the slope).

Decompose weight:

$W_{\parallel} = 5.0 \times 9.81 \times \sin 15^{\circ} = 12.69$ N (down the slope, i.e. negative $x$ ).
$W_{\perp} = 5.0 \times 9.81 \times \cos 15^{\circ} = 47.38$ N (into the slope, i.e. negative $y$ ).

(a) Newton II along $x$ :

$T_{\parallel} - W_{\parallel} = ma \Rightarrow 28.19 - 12.69 = 5.0\,a \Rightarrow a = 3.10 \text{ m s}^{-2}$

up the slope.

(b) Equilibrium along $y$ (block stays on slope):

$N + T_{\perp} - W_{\perp} = 0 \Rightarrow N = W_{\perp} - T_{\perp} = 47.38 - 10.26 = 37.1 \text{ N}$

Notice the normal reaction is smaller than $mg \cos\theta$ (47.38 N) because the rope's perpendicular component lifts the block slightly off the slope. Had the rope been horizontal to the slope (parallel), $T_{\perp} = 0$ and $N = 47.38$ N. This subtlety matters for friction calculations where $F_{\text{friction}} = \mu N$ : angling the rope upward reduces $N$ and so reduces friction, making it easier to drag heavy objects — a classic ergonomics result.

Linking to Energy Methods

An alternative way to solve smooth-slope problems is via conservation of energy. A block sliding from rest through a vertical drop $h$ on a smooth slope arrives at the bottom with:

$\tfrac{1}{2}mv^2 = mgh \Rightarrow v^2 = 2gh$

Inclined Surfaces and Resolving Forces

Inclined Surfaces and Resolving Forces

The Idea of Components

Normal Reaction on a Slope

Acceleration Down a Smooth Slope

Worked Example 1 — Smooth Ramp

Friction on a Slope

The Angle of Repose

Worked Example 2 — Block on a Rough Slope

Worked Example 3 — Object Held by a Rope on a Slope

Choosing Axes — A Crucial Skill

Pulleys and Inclined Planes — Compound Problems

Worked Example 4 — Atwood-Style Pulley

Worked Example 5 — Force Applied at an Angle on a Slope

Linking to Energy Methods

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