Forces on a Slope | CAIE 9702 Paper 1 Notes

Paper 1

Physics9702

Forces on a Slope

On a slope, gravity still pulls straight down, but the surface is tilted. Resolving along and perpendicular to the slope gives the useful force directions.

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. If no other force has a component perpendicular to the slope and the object does not accelerate away from it, 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. — The weight acts vertically. Its component along the slide drives the downhill motion; the perpendicular component presses the child into the surface.

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.

Your turnExam-style

A 12 kg block slides down a smooth slope at 25° to the horizontal. Calculate its acceleration down the slope.

Show worked answer

Smooth means there is no friction, so the only force component along the slope is $mg\sin 25^\circ$ .
$a = F/m = g\sin 25^\circ = 4.1 \text{ m s}^{-2}$ .

Answer: 4.1 m s⁻² down the slope.

Ways the exam changes it: a block in equilibrium on a slope, where friction equals $W\sin\theta$ and the three forces form a closed triangle; finding the contact force $N = W\cos\theta$ ; or adding a rope at a different angle and resolving again.

Forces on a Slope

On a slope, gravity still pulls straight down, but the surface is tilted. Resolving along and perpendicular to the slope gives the useful force directions.

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. If no other force has a component perpendicular to the slope and the object does not accelerate away from it, 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.

Your turnExam-style

A 12 kg block slides down a smooth slope at 25° to the horizontal. Calculate its acceleration down the slope.

Show worked answer

Smooth means there is no friction, so the only force component along the slope is $mg\sin 25^\circ$ .
$a = F/m = g\sin 25^\circ = 4.1 \text{ m s}^{-2}$ .

Answer: 4.1 m s⁻² down the slope.

Ways the exam changes it: a block in equilibrium on a slope, where friction equals $W\sin\theta$ and the three forces form a closed triangle; finding the contact force $N = W\cos\theta$ ; or adding a rope at a different angle and resolving again.