Free Pearson Edexcel International A Level Physics lesson

Vectors, components and equilibrium

Choose axes once, then let directions stay visible.

What you’ll learn

  • Distinguish scalar and vector quantities and resolve vectors along chosen axes.
  • Construct resultants by scale drawing and calculate perpendicular resultants.
  • Build a free-body diagram and test equilibrium in two directions.

One vector, two perpendicular descriptions

A scalar is described by magnitude and unit; a vector additionally needs direction. Mass, time, distance and energy are scalars. Displacement, velocity, acceleration, force and momentum are vectors. A negative scalar value in a chosen convention is not what makes a quantity a vector. Bold symbols or an arrow over a symbol identify vectors; the corresponding plain symbol often gives magnitude. Draw arrows with direction and a labelled scale rather than relying on an unlabelled length.

Resolve a vector by projecting it onto two perpendicular axes. If an arrow of magnitude F makes angle θ above positive horizontal, its components are F cos θ horizontally and F sin θ vertically. Sketch the right triangle and identify which side is adjacent to the stated angle before using a calculator in the correct angle mode. Components can be negative when arrows point opposite the positive axes. Changing axes changes the component numbers, while the physical vector remains the same.

Components replace the original vector in a calculation; they are not two extra physical forces. On a slope it is often simpler to choose axes parallel and perpendicular to the surface. Weight then has a downslope component mg sin α and a component into the slope mg cos α, where α is the slope angle to horizontal. These expressions follow from the geometry, not from a rule that sine always means horizontal. Use a scaled drawing to check the relative sizes and signs.

Fx = F cos θ Fy = F sin θ θ measured from the positive x-axis

Symbols and units

  • F is force magnitude in N; Fx and Fy are signed perpendicular components in N.
  • θ is the angle from positive x; use a consistent angle mode when evaluating sin and cos.

How the equation explains the idea

  1. In the right-triangle resolution, the adjacent side is F cos θ and the opposite side is F sin θ.
  2. At θ = 0°, Fx = F and Fy = 0; at θ = 90°, Fx = 0 and Fy = F. These limiting checks catch swapped components.
  3. Components outside the first quadrant may be negative; their signs preserve direction.
Rotate a fixed force and its components exchange size

Conditions. Computed components of one force with fixed magnitude F = 10 N, rotating from 0° to 90°. Fx = F cos θ and Fy = F sin θ; angles are converted to radians for calculation.

Angle above positive x / degrees; Force component / N.

Fx decreases from 10 N to zero while Fy increases from zero to 10 N. At 45° both are about 7.07 N. Their squares add to 100 N², so the total magnitude remains 10 N, not Fx + Fy.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

A cable inclined above a floor

  1. A cable pulls with 50.0 N at 30.0° above the horizontal.
  2. The horizontal component is adjacent: Fx = 50.0 cos 30.0° = 43.3 N.
  3. The vertical component is opposite: Fy = 50.0 sin 30.0° = 25.0 N.

The 50.0 N cable force is represented by +43.3 N horizontally and +25.0 N vertically.

Keep this idea. Draw the angle and axes before choosing sine or cosine; components describe the same vector.

Add arrows without losing their directions

To add two coplanar vectors by drawing, choose a scale and draw the first arrow with the correct direction and length. Move the second arrow without rotating it so its tail meets the first head. The resultant runs from the original starting point to the final arrowhead. Measure its length and direction with a ruler and protractor. A parallelogram construction gives the same resultant from the shared tail to the opposite corner. State the scale and likely drawing uncertainty.

For perpendicular components Rx and Ry, the magnitude is √(Rx² + Ry²). Find a reference angle using tan θ = |Ry|/|Rx|, then use the signs to identify the quadrant. A calculator’s inverse tangent alone may not name the physical direction correctly. For an arbitrary pair of coplanar vectors, a head-to-tail scaled drawing is sufficient for this specification; resolving each into the same axes and adding signed components is also a useful extension.

The resultant depends on the angle between arrows. Two equal perpendicular forces do not cancel, and adding their magnitudes gives too large a resultant. Subtracting a vector means adding its reversed arrow. Keep vectors of different physical quantities separate: a velocity cannot be added to a force. A result should include both magnitude and direction, together with units; a numerical magnitude alone does not fully specify the resultant vector.

Rx = ΣFx Ry = ΣFy R = √(Rx² + Ry²)

Symbols and units

  • Rx and Ry are resultant force components in N; R is resultant magnitude in N.
  • ΣFx and ΣFy mean separate sums of signed components along perpendicular axes.

How the equation explains the idea

  1. Resolve and add all x components, then do the same for y. The resulting components belong to one resultant vector.
  2. Apply Pythagoras: R² = Rx² + Ry², so R = √(Rx² + Ry²). Components 3 N and 4 N give a 5 N magnitude, not 7 N.
  3. Recover direction from the component signs as well as the ratio Ry/Rx; an inverse tangent by itself can choose the wrong quadrant.

Worked example

A perpendicular displacement

  1. A survey robot moves 12 m east then 5 m north.
  2. Displacement magnitude = √(12² + 5²) = 13 m.
  3. Direction = tan⁻¹(5/12) = 22.6° north of east. Its travelled distance is 17 m.

The resultant displacement is 13 m at 22.6° north of east, not 17 m.

Keep this idea. A resultant joins the beginning to the end of a head-to-tail construction; check its quadrant as well as its size.

Balance one body along both axes

A free-body diagram isolates one selected body and shows only forces acting on it. Label each force with its agent: weight from Earth, normal contact from a surface, tension from a cable and friction along a contact. For an extended rigid body, put forces at their points or lines of application and represent its weight at the centre of gravity. Do not draw its velocity as if it were a force, or include a reaction force acting on a different body.

For a particle in translational equilibrium, the resultant is zero along every axis. Thus ΣFx = 0 and ΣFy = 0. This allows rest or constant velocity; it does not require each individual force to vanish. When a cable is inclined, use its components to balance the relevant directions. Normal force is perpendicular to the surface and is not automatically mg: another vertical force, a slope or vertical acceleration can change it.

An extended object also needs zero resultant moment to remain in rotational equilibrium; balancing horizontal and vertical forces alone is insufficient. For a particle supported by two strings, a closed head-to-tail force triangle can show equilibrium without introducing a spurious extra balancing force. A proposed negative normal contact in a situation with an unrestrained surface means the assumed contact has failed, because that surface cannot pull the object toward itself.

For translational equilibrium: ΣFx = 0 and ΣFy = 0

Symbols and units

  • ΣFx and ΣFy are signed resultant force components in N.
  • Translational equilibrium means the acceleration of the centre of mass is zero in each direction.

How the equation explains the idea

  1. Newton’s second law applies independently to perpendicular axes. Zero acceleration requires both sums to vanish.
  2. For a stationary load held by two identical ropes angled θ above horizontal, horizontal components cancel and 2T sin θ = mg. Thus T = mg/(2 sin θ), so shallower ropes require larger tension.

Worked example

An inclined pull changes the contact force

  1. A 4.00 kg box moves at steady speed on a horizontal surface, pulled by 20.0 N at 30.0° above horizontal. Use g = 9.81 m s⁻².
  2. Horizontal balance gives resistance = 20.0 cos 30.0° = 17.3 N.
  3. Vertical balance gives N + 20.0 sin 30.0° = 4.00 × 9.81, so N = 29.24 N.

The contact force is 29.2 N, smaller than the box’s 39.2 N weight.

Keep this idea. Choose one body, name its external forces and test each axis; an extended body also requires a moment balance.

Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.

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