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55 ideas · 9 connections
Springs and Newton’s second law Springs and equilibrium Spring force and elastic energy Springs and graph interpretation Resolving forces in equilibrium Scalars and vectors2.3 Materials3.4 Dynamics3.2.1 Equilibrium3.2.3 Springs3.4.1 Elastic potential energy Force–extension graphs
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Springs and equilibrium

OCR A AS Physics · Two ideas, one explanation

A stationary load can deform a spring while the resultant force remains zero. In a multi-spring system, tension directions matter as much as their magnitudes.

ΣFx = 0 ΣFy = 0

Follow the reasoning

  1. Draw the forces on the load, including its weight.
  2. Resolve each tension along perpendicular axes.
  3. Apply zero resultant force in each direction.
  4. Use F = kx to connect each tension to that spring’s extension.
Specification outcomes (4)
  • 3.4.1(b) State and apply Hooke’s law.
  • 3.4.1(c) Define the force constant k of a spring or wire using F = kx.
  • 3.2.3(e) Describe equilibrium of an object under the action of forces and torques.
  • 3.2.3(f) Apply the condition for equilibrium of three coplanar forces using the triangle of forces.
Open the official OCR AS specification ↗

H156 v2.0 · May 2026

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