Physics 9702 · Paper 2

Forces on a Slope

On a slope, gravity still pulls straight down, but the surface is tilted. One resolving step handles every slope question.
Split the weight along the slope and into the slope, instead of horizontally and vertically.

Splitting the weight

On a slope, split the weight: W sin 30° pulls the block along the slope, W cos 30° presses it into the slope. The contact force N is at 90° to the surface, so N has no part along the slope.

For a slope at angle $\theta$ to the horizontal, the weight $W = mg$ splits into:
$W\sin\theta$ along the slope, pulling the object downhill;
$W\cos\theta$ into the slope, pressing the object onto the surface. The contact force balances this part: $N = W\cos\theta$ .
The contact force N is at 90° to the surface, so N has no part along the slope. Only $W\sin\theta$ and friction act along it.
Steeper slope, bigger $\sin\theta$ , bigger pull downhill. At $\theta = 0$ the pull is zero; at $90^\circ$ it is the whole weight.

Why the angle inside the split is θ

The weight is vertical. The angle between the weight and the slope surface is $90^\circ - \theta$ , so the part of W along the slope is $W\cos(90^\circ - \theta) = W\sin\theta$ . The slope's tilt angle reappears between the weight and the normal: a useful check is that the two parts must obey $(W\sin\theta)^2 + (W\cos\theta)^2 = W^2$ .

Worked example

Smallest case: the downhill pull

A crate of weight 500 N rests on a slope at 30° to the horizontal. Find the part of its weight that acts down the slope.

$W\sin\theta = 500 \times \sin 30^\circ = 250 \text{ N}$ .

Answer

250 \text{ N down the slope}

From the split to the motion

A child sliding down a long straight waterslide at about thirty degrees, seen from the side. — On a slope, only the part of the weight along the slope drives the motion. The weight acts straight down. Splitting it into a part along the slope and a part into the slope shows the along-slope part is what makes the slider go faster.

Worked example

Exam version: down a waterslide

A child of mass 40 kg slides down a waterslide at 30° to the horizontal. Friction of 120 N acts up the slope. Find the child's acceleration (9702-style, from the coursebook).

Weight: $W = 40 \times 9.81 = 392 \text{ N}$ . Downhill part: $392\sin 30^\circ = 196 \text{ N}$ .
Resultant along the slope: $F = 196 - 120 = 76 \text{ N}$ .
Newton's second law (lesson 3.02): $a = \dfrac{F}{m} = \dfrac{76}{40} = 1.9 \text{ m s}^{-2}$ down the slope.

Answer

a = 1.9 \text{ m s}^{-2}

Common mistake

Mixing the two parts:

\sin

goes with the motion along the slope,

\cos

goes with the contact force. If your slope gets easier as it gets steeper, the two are switched.

Ways the exam changes it: a block in equilibrium on a slope, where friction equals $W\sin\theta$ exactly and the three forces close a triangle (9702/13/M/J/25 Q14); finding the contact force $N = W\cos\theta$ ; a rope holding the object at a further angle (resolve along the slope again).

PreviousThree Forces: Resolve or Close the Triangle NextDensity

Forces on a Slope

On a slope, gravity still pulls straight down, but the surface is tilted. One resolving step handles every slope question.
Split the weight along the slope and into the slope, instead of horizontally and vertically.

Splitting the weight

On a slope, split the weight: W sin 30° pulls the block along the slope, W cos 30° presses it into the slope. The contact force N is at 90° to the surface, so N has no part along the slope.

For a slope at angle $\theta$ to the horizontal, the weight $W = mg$ splits into:
$W\sin\theta$ along the slope, pulling the object downhill;
$W\cos\theta$ into the slope, pressing the object onto the surface. The contact force balances this part: $N = W\cos\theta$ .
The contact force N is at 90° to the surface, so N has no part along the slope. Only $W\sin\theta$ and friction act along it.
Steeper slope, bigger $\sin\theta$ , bigger pull downhill. At $\theta = 0$ the pull is zero; at $90^\circ$ it is the whole weight.

Why the angle inside the split is θ

Worked example

Smallest case: the downhill pull

A crate of weight 500 N rests on a slope at 30° to the horizontal. Find the part of its weight that acts down the slope.

$W\sin\theta = 500 \times \sin 30^\circ = 250 \text{ N}$ .

Answer

250 \text{ N down the slope}

From the split to the motion

Worked example

Exam version: down a waterslide

A child of mass 40 kg slides down a waterslide at 30° to the horizontal. Friction of 120 N acts up the slope. Find the child's acceleration (9702-style, from the coursebook).

Weight: $W = 40 \times 9.81 = 392 \text{ N}$ . Downhill part: $392\sin 30^\circ = 196 \text{ N}$ .
Resultant along the slope: $F = 196 - 120 = 76 \text{ N}$ .
Newton's second law (lesson 3.02): $a = \dfrac{F}{m} = \dfrac{76}{40} = 1.9 \text{ m s}^{-2}$ down the slope.

Answer

a = 1.9 \text{ m s}^{-2}

Common mistake

Mixing the two parts:

\sin

goes with the motion along the slope,

\cos

goes with the contact force. If your slope gets easier as it gets steeper, the two are switched.

Ways the exam changes it: a block in equilibrium on a slope, where friction equals $W\sin\theta$ exactly and the three forces close a triangle (9702/13/M/J/25 Q14); finding the contact force $N = W\cos\theta$ ; a rope holding the object at a further angle (resolve along the slope again).