Resolution of Forces: Inclined Planes and Lami's Theorem

When Forces Act at Angles

A box dragged by a rope pulled at an angle, a car parked on a hill, a picture hanging from two strings, a boat pulled by two tugs: in each case the forces do not all act along one line. To find their combined effect, physicists use two complementary skills: adding forces to find a single resultant, and resolving a force into perpendicular components. This tutorial covers the resolution of forces as tested in WAEC, NECO and JAMB physics and in PHY 101: resultants by the parallelogram law and the cosine rule, components of a force, objects on smooth and rough inclined planes, the equilibrium of three forces, the triangle of forces and Lami's theorem, with worked examples throughout. It builds on scalars and vectors in the measurement tutorial and on Newton's laws.

Finding the Resultant of Two Forces

The resultant of several forces is the single force that has the same effect as all of them together. For two forces along the same line, simply add them (same direction) or subtract them (opposite directions). For forces at an angle, use the parallelogram law: draw the two forces from one point as adjacent sides of a parallelogram, and the diagonal from that point represents the resultant in size and direction.

Forces at right angles

When two forces P and Q are perpendicular, the parallelogram is a rectangle, so:

R = √(P² + Q²), at an angle θ to P where tan θ = Q ÷ P

Worked example 1

Forces of 3 N and 4 N act on a body at right angles. Find their resultant.

R = √(9 + 16) = 5 N, at tan⁻¹(4/3) ≈ 53.1° to the 3 N force.

Forces at any angle: the cosine rule

For two forces P and Q with an angle θ between them:

R² = P² + Q² + 2PQ cos θ

Note the plus sign. It differs from the usual cosine rule because θ here is the angle between the forces, not the angle inside the triangle.

Worked example 2

Two forces of 10 N and 6 N act at a point with an angle of 60° between them. Find the size and direction of their resultant.

R² = 100 + 36 + 2 × 10 × 6 × cos 60° = 136 + 60 = 196, so R = 14 N.

Direction, using the sine rule: sin α = 6 sin 60° ÷ 14 ≈ 0.371, so α ≈ 21.8° from the 10 N force.

Check the extremes: at θ = 0° the formula gives R = P + Q, and at θ = 180° it gives R = |P − Q|, as expected.

The equilibrant

The equilibrant is the single force that would balance the others and hold the body in equilibrium. It is equal in size to the resultant and acts in the opposite direction.

Resolving a Force into Components

Any force F can be replaced by two perpendicular forces, called its components, that together have exactly the same effect. If F makes an angle θ with the horizontal:

Horizontal component: F cos θ
Vertical component: F sin θ

The component adjacent to the angle uses cos, and the component opposite the angle uses sin.

Worked example 3

A trolley is pulled by a rope with a force of 50 N at 30° above the horizontal. Find the horizontal and vertical components.

Horizontal: 50 cos 30° ≈ 43.3 N, which moves the trolley forwards.

Vertical: 50 sin 30° = 25 N, which partly lifts the trolley and so reduces the force it exerts on the ground.

Adding many forces by components

To find the resultant of several forces at different angles, resolve each one into x and y components, add all the x components (ΣFx) and all the y components (ΣFy), then combine them: R = √(ΣFx² + ΣFy²), with tan θ = ΣFy ÷ ΣFx. This method works for any number of forces, and it is far easier than drawing several parallelograms.

Objects on an Inclined Plane

For an object on a slope, it is easiest to resolve forces parallel and perpendicular to the surface instead of horizontally and vertically. The weight mg, acting straight down, splits into:

Down the slope: mg sin θ, which tends to make the object slide.
Into the slope: mg cos θ, which is balanced by the normal reaction R = mg cos θ.

Here θ is the angle of the slope to the horizontal. A quick check: on a flat surface (θ = 0) there is no force down the slope, and on a vertical wall (θ = 90°) the whole weight acts down it.

Worked example 4: a smooth slope

A 10 kg block rests on a smooth (frictionless) plane inclined at 30°. Find the force down the slope, the normal reaction and the acceleration. (g = 10 m/s²)

Weight = 100 N. Down the slope: 100 sin 30° = 50 N. Normal reaction: 100 cos 30° ≈ 86.6 N.

Acceleration = 50 ÷ 10 = 5 m/s². In general, on a smooth slope a = g sin θ, whatever the mass.

Worked example 5: a rough slope

The same block is on a rough plane at 30° with a coefficient of friction μ = 0.2. Find its acceleration down the slope, and the force parallel to the slope needed to push it up at constant speed.

Friction = μR = 0.2 × 86.6 ≈ 17.3 N, acting against the motion.

Sliding down: net force = 50 − 17.3 = 32.7 N, so a = 32.7 ÷ 10 ≈ 3.27 m/s².

Pushing up at constant speed: the push must balance both the weight component and friction, which now acts down the slope: 50 + 17.3 ≈ 67.3 N.

The object stays at rest on its own if mg sin θ ≤ μ mg cos θ, that is when tan θ ≤ μ. The steepest angle at which it can stay put is the angle of friction, covered in the friction tutorial.

Equilibrium of Three Forces

A body acted on by forces is in equilibrium when there is no resultant force (and no resultant moment). For three non-parallel forces in equilibrium:

Their lines of action all pass through a single point.
They can be represented in size and direction by the three sides of a triangle, taken in order: the triangle of forces.
The sum of their components in any direction is zero. In particular, ΣFx = 0 and ΣFy = 0.

Lami's theorem

If three forces A, B and C keep a point in equilibrium, each force is proportional to the sine of the angle between the other two:

A ÷ sin α = B ÷ sin β = C ÷ sin γ

Here α is the angle between B and C, β the angle between A and C, and γ the angle between A and B.

Worked example 6: a weight on two strings

A 100 N weight hangs from two strings attached to a ceiling. One string makes 30° with the ceiling, and the other makes 60°. Find the tension T₁ in the 30° string and T₂ in the 60° string.

Method 1: components.

Horizontally: T₁ cos 30° = T₂ cos 60°, so T₂ = T₁ × 0.866 ÷ 0.5 ≈ 1.732 T₁.
Vertically: T₁ sin 30° + T₂ sin 60° = 100, so 0.5T₁ + 1.732T₁ × 0.866 = 100, which gives 2T₁ = 100.

So T₁ = 50 N and T₂ ≈ 86.6 N.

Method 2: Lami's theorem. The angle between the two strings is 180° − 30° − 60° = 90°. The angle between T₂ and the weight is 90° + 60° = 150°, and between T₁ and the weight it is 90° + 30° = 120°. So:

T₁ ÷ sin 150° = T₂ ÷ sin 120° = 100 ÷ sin 90°, giving T₁ = 100 × 0.5 = 50 N and T₂ = 100 × 0.866 ≈ 86.6 N, the same answers.

Notice that the steeper string carries more of the load. A string that is nearly horizontal needs a very large tension, which is why a washing line pulled perfectly taut would snap under a heavy load.

Common Mistakes with the Resolution of Forces

Swapping sin and cos. The component next to the angle uses cos, and the one opposite it uses sin.
Using the slope's angle with the wrong component: on an incline, the force down the slope is mg sin θ, and the normal reaction is mg cos θ.
Adding forces at an angle as if they were in a line. Forces of 3 N and 4 N at right angles give 5 N, not 7 N.
Using the minus-sign cosine rule with the angle between the forces. Use R² = P² + Q² + 2PQ cos θ.
Using the wrong angles in Lami's theorem. Each force goes with the angle between the other two forces.

Resolution of Forces: Practice Questions

Find the resultant of forces of 8 N and 6 N acting at right angles.
Find the horizontal and vertical components of a 20 N force acting at 40° to the horizontal.
Two forces of 5 N and 5 N act with an angle of 120° between them. Find their resultant.
A 5 kg box slides down a smooth slope inclined at 37°. Find its acceleration. (g = 10 m/s²)
A picture of weight 30 N hangs from a string over a nail, with each half of the string at 45° to the vertical. Find the tension in the string.
What is the equilibrant of a resultant force of 12 N acting due east?

Answers: 1. √(64 + 36) = 10 N. 2. Horizontal 20 cos 40° ≈ 15.3 N, and vertical 20 sin 40° ≈ 12.9 N. 3. R² = 25 + 25 + 50 cos 120° = 25, so R = 5 N. 4. a = g sin 37° ≈ 6.0 m/s². 5. 2T cos 45° = 30, so T ≈ 21.2 N. 6. 12 N acting due west.

Conclusion

Forces acting at angles are combined into a resultant using the parallelogram law: R = √(P² + Q²) for perpendicular forces, and R² = P² + Q² + 2PQ cos θ in general. They are resolved into components F cos θ and F sin θ, which makes it easy to add many forces and to analyse inclined planes, where the weight splits into mg sin θ down the slope and mg cos θ into it. Three forces in equilibrium form a closed triangle of forces and obey Lami's theorem, and resolving along two perpendicular directions solves any equilibrium problem.