Forces and Equilibrium

In mechanics, understanding forces is essential. Before you can apply Newton’s laws to solve a problem, you must be able to identify every force acting on an object, draw an accurate free-body diagram, and determine whether the object is in equilibrium. This lesson covers the main types of force and the conditions for equilibrium, with particular attention to inclined plane problems and friction.

Types of Force

Weight (W)

Weight is the gravitational force acting on an object. It always acts vertically downward towards the centre of the Earth.

W = mg

where m is the mass in kg and g = 9.81 m/s² (near Earth’s surface). Weight is measured in newtons (N).

Normal Contact Force (N or R)

When two surfaces are in contact, they exert a force on each other perpendicular to the contact surface. This is the normal contact force (sometimes called the normal reaction).

On a horizontal surface, the normal force acts vertically upward. On a slope, it acts perpendicular to the slope surface — not vertically.

Friction (F)

Friction is a force that opposes the relative motion (or attempted motion) between two surfaces in contact. It acts along the surface, in the opposite direction to the motion or the tendency to move.

The maximum friction force is given by:

F_max = mu R

where mu is the coefficient of friction (dimensionless) and R is the normal contact force. Important points:

mu is a property of the two surfaces in contact.
The friction force can be anywhere from 0 up to mu R. It is only equal to mu R when the object is on the point of sliding or is already sliding.
Static friction (no sliding) can be less than mu R — it matches the applied force up to the maximum.

Reference Table: Typical Coefficients of Friction

Surface Pair	Approximate mu
Rubber on dry concrete	0.6 – 0.8
Rubber on wet concrete	0.4 – 0.5
Steel on steel (dry)	0.6
Steel on steel (lubricated)	0.05 – 0.1
Wood on wood	0.25 – 0.5
Ice on ice	0.03
Teflon on steel	0.04
Tyres on ice	0.1 – 0.2

Tension (T)

Tension acts in a string, rope, or cable when it is pulled taut. It acts along the string, pulling the connected object towards the string. For a light (massless) string, the tension is the same throughout its length.

Drag and Air Resistance

Drag forces oppose motion through a fluid (air, water). They increase with speed — typically proportional to v for low speeds and v² for high speeds. The direction of drag is always opposite to the velocity of the object.

Conditions for Equilibrium

An object is in equilibrium when:

The resultant force is zero (translational equilibrium).
The resultant moment about any point is zero (rotational equilibrium).

For this lesson, we focus on translational equilibrium. This means:

Sum of forces in the horizontal direction = 0
Sum of forces in the vertical direction = 0

Or equivalently, when three forces act on a body in equilibrium, they can be drawn as a closed triangle (the triangle of forces).

flowchart TD
    A["Is the object accelerating?"]
    A -->|"No (constant velocity or rest)"| B["Object is in EQUILIBRIUM\nResultant force = 0"]
    A -->|Yes| C["Object is NOT in equilibrium\nApply F = ma"]
    B --> D["Sum of horizontal forces = 0\nSum of vertical forces = 0"]
    C --> E["Find resultant force\nand use F = ma"]

Resolving Forces on an Inclined Plane

Inclined plane problems are among the most common in A-Level mechanics. The key is to resolve forces into components parallel and perpendicular to the slope, rather than horizontal and vertical.

Consider an object of mass m on a smooth slope at angle theta to the horizontal:

Perpendicular to slope: R = mg cos(theta) Parallel to slope (down the slope): Component of weight = mg sin(theta)

If the slope is smooth (no friction), the object accelerates down the slope with:

a = g sin(theta)

Quick Reference: Forces on Slopes

Slope Angle	mg sin(theta) (down slope)	mg cos(theta) (into slope)	a = g sin(theta)
10°	0.174 mg	0.985 mg	1.70 m/s²
20°	0.342 mg	0.940 mg	3.36 m/s²
30°	0.500 mg	0.866 mg	4.91 m/s²
45°	0.707 mg	0.707 mg	6.94 m/s²
60°	0.866 mg	0.500 mg	8.50 m/s²

Worked Example 1: Object on a Smooth Slope

A 4.0 kg box is placed on a smooth slope at 25° to the horizontal. Find the acceleration and the normal reaction force.

Component of weight parallel to slope = mg sin(25°) = 4.0 x 9.81 x sin(25°) = 4.0 x 9.81 x 0.4226 = 16.6 N

This is the net force down the slope (smooth slope, no friction): a = g sin(25°) = 9.81 x 0.4226 = 4.15 m/s²

Normal reaction: R = mg cos(25°) = 4.0 x 9.81 x cos(25°) = 4.0 x 9.81 x 0.9063 = 35.6 N

Note that R is less than the full weight (39.2 N) — on a slope, the normal force is always less than the weight.

Friction on an Inclined Plane

flowchart TD
    A["Object on rough slope at angle θ"]
    A --> B["Calculate mg sin θ\n(force down slope)"]
    A --> C["Calculate R = mg cos θ\n(normal reaction)"]
    C --> D["Calculate F_max = μ R"]
    B --> E{"mg sin θ > F_max?"}
    D --> E
    E -->|"Yes"| F["Block SLIDES\nFriction = F_max\nNet force = mg sin θ - F_max\na = Net force / m"]
    E -->|"No"| G["Block stays STATIONARY\nFriction = mg sin θ\n(less than F_max)"]

Worked Example 2: Object on a Rough Slope

A 10 kg block rests on a rough slope at 30° to the horizontal. The coefficient of friction is mu = 0.40. Determine whether the block slides and find the friction force.

Component of weight down the slope: mg sin(30°) = 10 x 9.81 x 0.500 = 49.1 N

Normal reaction: R = mg cos(30°) = 10 x 9.81 x 0.866 = 85.0 N

Maximum friction: F_max = mu R = 0.40 x 85.0 = 34.0 N

Since the component of weight down the slope (49.1 N) exceeds the maximum friction (34.0 N), the block will slide.

Net force down the slope = 49.1 - 34.0 = 15.1 N Acceleration = 15.1 / 10 = 1.5 m/s² down the slope

Worked Example 3: Object in Equilibrium on a Slope

If instead mu = 0.70 in the example above:

Maximum friction: F_max = 0.70 x 85.0 = 59.5 N

Since the component of weight down the slope (49.1 N) is less than F_max (59.5 N), the block does not slide. Friction = 49.1 N (matching the gravitational component, not the maximum).

Worked Example 4: Finding the Critical Angle

At what angle does a block just begin to slide on a surface with mu = 0.40?

At the point of sliding: mg sin(theta) = mu mg cos(theta) tan(theta) = mu = 0.40 theta = arctan(0.40) = 21.8°

This is the angle of friction — the steepest angle at which the block remains stationary. This result is independent of mass.

Equilibrium with Multiple Forces

Worked Example 5

A 5.0 kg crate is held stationary on a 20° slope by a horizontal force P. The slope is smooth. Find P and the normal reaction.

Resolving parallel to slope (up the slope positive): P cos(20°) = mg sin(20°) P cos(20°) = 5.0 x 9.81 x sin(20°) P x 0.940 = 5.0 x 9.81 x 0.342 = 16.8 P = 16.8 / 0.940 = 17.8 N

Resolving perpendicular to slope: R = mg cos(20°) + P sin(20°) R = 5.0 x 9.81 x 0.940 + 17.8 x 0.342 R = 46.1 + 6.09 = 52.2 N

Worked Example 6: Three Forces in Equilibrium

A 15 N weight hangs from a point where two strings meet. One string goes to the left wall at 30° above horizontal; the other goes to the right wall at 45° above horizontal. Find the tension in each string.

Vertical: T₁ sin(30°) + T₂ sin(45°) = 15 Horizontal: T₁ cos(30°) = T₂ cos(45°)

From horizontal: T₁ = T₂ cos(45°) / cos(30°) = T₂ x 0.707 / 0.866 = 0.816 T₂

Substituting: 0.816 T₂ x 0.500 + T₂ x 0.707 = 15 0.408 T₂ + 0.707 T₂ = 15 1.115 T₂ = 15 T₂ = 13.5 N

T₁ = 0.816 x 13.5 = 11.0 N

Common Mistakes and Misconceptions

Normal force equals weight: Only true on a horizontal surface with no other vertical forces. On a slope, R = mg cos(theta), which is less than mg.
Friction always equals mu R: Friction can be anywhere from 0 to mu R. It only equals mu R when the object is sliding or on the verge of sliding.
Forgetting to check whether the object slides: Before calculating acceleration, always compare mg sin(theta) with mu R to determine if the object actually moves.
Resolving in the wrong directions on slopes: On an inclined plane, resolve parallel and perpendicular to the slope, not horizontal and vertical.
Applying an incorrect horizontal force on a slope: If a horizontal force acts on an object on a slope, it has components both parallel and perpendicular to the slope. Both must be included.

Exam Tips for Edexcel Physics

Always draw a labelled free-body diagram for slope problems, showing weight (vertically down), normal reaction (perpendicular to slope), and friction (along slope opposing motion).
When a question says "smooth," it means frictionless (mu = 0). When it says "rough," friction is present.
For "show that" problems on slopes, give the full resolution of forces into components, not just the final formula.
In questions asking "does the object move?", compare the gravitational component along the surface with the maximum friction force. State clearly which is larger.
Remember that the angle of friction (arctan mu) is a useful shortcut for determining whether an object slides on a given slope.

Key Points

Five main forces: weight, normal reaction, friction, tension, drag.
Weight always acts vertically downward; normal force is perpendicular to the surface.
Friction opposes motion: F_max = mu R; actual friction can be less than this.
On an inclined plane, resolve forces parallel and perpendicular to the slope.
Equilibrium requires: resultant force = 0 and resultant moment = 0.
On slopes: the normal force = mg cos(theta), the gravitational pull down the slope = mg sin(theta).
Critical angle for sliding: theta = arctan(mu), independent of mass.

A-Level Deep Dive: Forces and Equilibrium

Spec mapping

Edexcel 9PH0 specification, Topic 2 — Mechanics, sub-topic 2.4 (Forces) develops the conditions for equilibrium of a rigid body under coplanar forces, the resolution of weight on inclined planes, friction (smooth vs rough surfaces, coefficient of friction), and the principle of moments (refer to the official Pearson Edexcel specification document for exact wording). Equilibrium is treated more rigorously than at GCSE: both $\Sigma F = 0$ in every direction AND $\Sigma M = 0$ about every point are required — neither alone is sufficient. The techniques recur across Newton's laws (sub-topic 2.2) for non-zero acceleration, Topic 4 (Materials) for stress in wires and beams, Topic 6 (Further mechanics) for circular-motion equilibrium with centripetal force, and Topic 7 (Electric and magnetic fields) for charged-particle equilibrium. The Edexcel data, formulae and relationships booklet supplies trigonometric identities but does not state the moment formula $M = Fd$ or the friction relation $F = \mu R$ — both must be quoted in working.

Worked example with full mark scheme

Question (8 marks):

A uniform ladder of mass 18.0 kg and length 5.00 m rests against a smooth vertical wall, with its base on rough horizontal ground. The ladder makes an angle of 65° with the ground. Take $g = 9.81$ m s⁻².

(a) Draw a labelled free-body diagram showing all forces acting on the ladder. (2)

(b) By taking moments about the base of the ladder and resolving forces, determine the minimum coefficient of static friction $\mu$ between the ladder and the ground for the ladder to remain in equilibrium. (6)

Solution with mark scheme:

(a) B1 — four forces correctly drawn: weight $W = mg$ acting vertically downward at the midpoint of the ladder; normal reaction $N_W$ from the wall acting horizontally at the top of the ladder (no friction component because the wall is smooth); normal reaction $N_G$ from the ground acting vertically upward at the base; friction $f$ from the ground acting horizontally toward the wall at the base. B1 — angles labelled (65° between ladder and ground), point of application of weight at the geometric centre indicated.

(b) Step 1 — vertical equilibrium. Take upward as positive.

$\Sigma F_y = 0: \quad N_G - W = 0 \implies N_G = mg = 18.0 \times 9.81 = 176.6 \text{ N}$

M1 — applying $\Sigma F_y = 0$ correctly. The wall provides no vertical force because it is smooth.

Step 2 — horizontal equilibrium. Take rightward (toward wall) as positive.

$\Sigma F_x = 0: \quad f - N_W = 0 \implies f = N_W$

M1 — applying $\Sigma F_x = 0$ to relate friction at the base to the wall reaction.

Step 3 — moments about the base. Taking anticlockwise as positive. The weight acts at the midpoint, a perpendicular distance of $(L/2)\cos 65°$ from the line of the base; the wall reaction acts at the top, a perpendicular distance of $L\sin 65°$ from the base.

$\Sigma M_\text{base} = 0: \quad N_W (L\sin 65°) - W \left(\dfrac{L}{2}\cos 65°\right) = 0$

M1 — taking moments about the base; choosing this point eliminates $N_G$ and $f$ from the equation (their lines of action pass through the pivot, so their moment is zero). The $L$ cancels:

$N_W = \dfrac{W \cos 65°}{2 \sin 65°} = \dfrac{W}{2 \tan 65°}$

$N_W = \dfrac{176.6}{2 \times 2.1445} = \dfrac{176.6}{4.2891} = 41.2 \text{ N}$

A1 — $N_W = 41.2$ N (3 sf).

Step 4 — minimum coefficient of friction. At the verge of slipping, friction reaches its maximum value $f_\text{max} = \mu_\text{min} N_G$ . From step 2, $f = N_W = 41.2$ N, so:

$\mu_\text{min} = \dfrac{f}{N_G} = \dfrac{N_W}{W} = \dfrac{1}{2 \tan 65°} = \dfrac{41.2}{176.6} = 0.233$

M1 — recognising that minimum friction occurs when the ladder is on the verge of slipping. A1 — $\mu_\text{min} = 0.233$ (3 sf), notably independent of the ladder mass and length — only the geometry matters.

Total: 8 marks (B2 M4 A2).

Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A uniform horizontal beam of mass 12.0 kg and length 4.00 m is supported by two vertical cables: cable A is attached at one end of the beam, and cable B is attached 1.00 m from the other end. A small load of mass 8.00 kg is hung from the end of the beam furthest from cable A. Take $g = 9.81$ m s⁻².

(a) Draw a labelled free-body diagram showing the four forces acting on the beam. (1)

(b) By taking moments and applying force balance, calculate the tensions in cables A and B. (4)

(c) Explain what would happen to the tension in cable A if the load were moved to a position directly underneath cable B. (1)

Mark scheme decomposition by AO: