Resolving Forces and Inclined Planes

SPEC 8.2, 8.4 • FORCES AND NEWTON'S LAWS

Breaking forces into components and solving equilibrium problems on slopes

Resolving Forces

Key Concept

Any force can be split into two perpendicular components. We form a right-angled triangle with the force as the hypotenuse.

When a force acts at an angle, we resolve it into horizontal and vertical components:

The force F is the hypotenuse of a right-angled triangle

Memory Aid

"Cos goes Along, Sin goes Skyward" - cos gives horizontal (along), sin gives vertical (skyward).

Component Formulas

For a force F at angle θ to the horizontal:

Formulas

F_{x} = F cos θ

F_{y} = F sin θ

Horizontal Component

F_{x} = F cos θ

Use cos for the component adjacent to the angle

Vertical Component

F_{y} = F sin θ

Use sin for the component opposite to the angle

Finding Resultant

R = F_{x}^{2} + F_{y}^{2}

tan α = \frac{F _{y}}{F _{x}}

01Worked Example

A force of 50 N acts at 30° to the horizontal. Find the horizontal and vertical components.

1

Identify: F = 50 N, θ = 30°

2

Horizontal:

F_{x} = 50 cos 30° = 50 \times 0.866 = 43.3

N

3

Vertical:

F_{y} = 50 sin 30° = 50 \times 0.5 = 25

N

Answer: Horizontal = 43.3 N, Vertical = 25 N

02Worked Example

Two forces act at a point: 40 N at 60° to the horizontal and 30 N horizontally to the right. Find the resultant.

1

Force 1 (40 N at 60°):

F_{1 x} = 40 cos 60° = 20

N,

F_{1 y} = 40 sin 60° = 34.64

N

2

Force 2 (30 N horizontal):

F_{2 x} = 30

N,

F_{2 y} = 0

3

Total:

Σ F_{x} = 50

N,

Σ F_{y} = 34.64

N

4

Resultant:

R = 5 0^{2} + 34.6 4^{2} = 60.8

N

5

Direction:

tan α = \frac{34.64}{50}

, so

α = 34.7°

Answer: 60.8 N at 34.7° to horizontal

03Worked Example

Three forces act at a point: 10 N right, 8 N up, and 15 N at 45° below horizontal to the left. Find the resultant.

1

Force 1:

F_{1 x} = + 10

N,

F_{1 y} = 0

2

Force 2:

F_{2 x} = 0

,

F_{2 y} = + 8

N

3

Force 3:

F_{3 x} = - 15 cos 45° = - 10.61

N,

F_{3 y} = - 15 sin 45° = - 10.61

N

4

Totals:

Σ F_{x} = - 0.61

N,

Σ F_{y} = - 2.61

N

5

Resultant:

R = 0.6 1^{2} + 2.6 1^{2} = 2.68

N at 76.9° below horizontal (left)

Answer: 2.68 N at 76.9° below horizontal to the left

Forces on an Inclined Plane

When an object rests on a smooth slope, two main forces act:

Weight (W = mg)

Always vertically downward

Normal Reaction (R)

Perpendicular to the surface

Note: This lesson covers smooth (frictionless) slopes only.

Resolving on Inclined Planes

Key Strategy

For inclined plane problems, ALWAYS resolve ALL forces into components parallel and perpendicular to the slope.

Why? Because R already acts perpendicular to the slope, this makes the perpendicular equation simpler - R appears on its own!

Method

Draw a clear force diagram with the object ON the slope

Resolve ALL forces parallel to the slope

Resolve ALL forces perpendicular to the slope

Apply equilibrium conditions in each direction

Remember

The weight mg needs to be resolved. R does NOT need resolving (it's already perpendicular to slope).

Resolving Weight on a Slope

The weight mg resolves into two components:

Weight Components

Parallel (down slope): m g sin θ

Perpendicular (into slope): m g cos θ

Key Point

The angle θ between mg and mg cos θ is the same as the angle of the incline. This is why we use the incline angle when resolving weight.

Memory Aid

"Sin Slides" - the sin component is what makes objects slide down!

04Worked Example

A 5 kg box rests on a smooth plane at 25° to the horizontal. Find: (a) the component of weight parallel to the slope, (b) the normal reaction.

1

Weight:

W = m g = 5 \times 9.8 = 49

N

2

(a) Parallel to slope:

m g sin 25° = 49 \times 0.4226 = 20.7

N

3

(b) Perpendicular:

m g cos 25° = 49 \times 0.9063 = 44.4

N

4

Equilibrium perpendicular:

R = m g cos 25° = 44.4

N

Answer: (a) 20.7 N, (b) R = 44.4 N

Equilibrium on Inclined Planes

For an object in equilibrium on a slope, forces must balance in both directions:

Equilibrium Conditions

Parallel to slope (↗): Forces up slope = Forces down slope

Perpendicular to slope (↖): Forces into slope = Forces away from slope

Method

Resolving parallel to the slope:

R (↗), F = ma : [forces] = 0

Resolving perpendicular to the slope:

R (↖), F = ma : [forces] = 0

An object on a smooth slope can be held in equilibrium by:

A force acting parallel up the slope equal to mg sin θ

A horizontal force (which must be resolved)

A force at some other angle (resolve into ↗ and ↖ components)

05Worked Example

A 10 kg block is held on a smooth slope of 30° by force P acting parallel to and up the slope. Find P.

1

Weight:

W = m g = 10 \times 9.8 = 98

N

2

Resolving parallel to the slope:

•

R (↗), F = ma :

•

P - m g sin 30° = 0

•

P = 98 \times 0.5 = 49

N

Answer: P = 49 N

06Worked Example

A 4 kg block is held on a smooth slope of 40° by a horizontal force H. Find H and R.

1

Weight:

W = 4 \times 9.8 = 39.2

N

2

H has components: parallel =

H cos 40°

, perpendicular =

H sin 40°

3

Resolving parallel to the slope:

•

R (↗), F = ma :

•

H cos 40° - m g sin 40° = 0

•

H = m g tan 40° = 39.2 \times 0.839 = 32.9

N

4

Resolving perpendicular to the slope:

•

R (↖), F = ma :

•

R - m g cos 40° - H sin 40° = 0

•

R = 30.0 + 21.2 = 51.2

N

Answer: H = 32.9 N, R = 51.2 N

07Worked Example

A 5 kg block rests on a smooth plane inclined at 30° to the horizontal. A force P acts at 20° above the plane. Find P and the normal reaction R.

1

Weight:

W = 5 \times 9.8 = 49

N

2

P has components: parallel =

P cos 20°

, perpendicular =

P sin 20°

3

Resolving parallel to the slope:

•

R (↗), F = ma :

•

P cos 20° - m g sin 30° = 0

•

P = \frac{49 \times 0.5}{0.940} = 26.1

N

4

Resolving perpendicular to the slope:

•

R (↖), F = ma :

•

R + P sin 20° - m g cos 30° = 0

•

R = 49 \times 0.866 - 26.1 \times 0.342 = 33.5

N

Answer: P = 26.1 N, R = 33.5 N

Practice

Question 1: A force of 80 N acts at 55° to the horizontal. Calculate the horizontal and vertical components.

1

Resolving horizontally:

•

R (\to), F = ma :

•

F_{x} = 80 cos 55° = 45.9

N

2

Resolving vertically:

•

R (↑), F = ma :

•

F_{y} = 80 sin 55° = 65.5

N

Answer: Horizontal = 45.9 N, Vertical = 65.5 N

Practice

Question 2: A 3 kg object rests on a plane at 35° to the horizontal. Calculate the component of weight down the slope and perpendicular to the slope.

1

Weight:

W = 3 \times 9.8 = 29.4

N

2

Resolving parallel to the slope:

•

R (↗), F = ma :

•

W_{∥} = 29.4 sin 35° = 16.9

N

3

Resolving perpendicular to the slope:

•

R (↖), F = ma :

•

W_{⊥} = 29.4 cos 35° = 24.1

N

Answer: Down slope = 16.9 N, Perpendicular = 24.1 N

Practice

Question 3: A 12 kg crate is held on a smooth slope of 20° by a force P parallel to the slope. Find P and R.

1

Weight:

W = 12 \times 9.8 = 117.6

N

2

Resolving parallel to the slope:

•

R (↗), F = ma :

•

P - m g sin 20° = 0

•

P = 117.6 \times 0.342 = 40.2

N

3

Resolving perpendicular to the slope:

•

R (↖), F = ma :

•

R - m g cos 20° = 0

•

R = 117.6 \times 0.940 = 110.5

N

Answer: P = 40.2 N, R = 110.5 N

Summary

Key Formulas

ConceptFormula

Horizontal component

F_{x} = F cos θ

Vertical component

F_{y} = F sin θ

Weight parallel to slope

m g sin θ

Weight perpendicular to slope

m g cos θ

Resultant magnitude

R = F_{x}^{2} + F_{y}^{2}

Key Exam Skills

Always resolve parallel and perpendicular to the slope

Draw clear force diagrams with box ON the slope

"Sin Slides" - sin component goes down the slope

For equilibrium: forces balance in all directions