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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.
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H156 v2.0 · May 2026

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PhysicsUK Solve10 marks

A sensor is suspended from a small ring by a light vertical cord. The ring is held by an inclined spring on its left and a light horizontal cord on its right, as shown in Figure 2. The spring slopes upwards to the left at 40.0° above the horizontal. The sensor's weight is 12.0 N. The ring and cords have negligible mass. All forces act in one plane through the centre of the ring, which is at rest. The spring constant is 80.0 N m⁻¹ and the spring obeys Hooke's law for extensions up to 0.300 m. (a) Use vertical equilibrium to calculate the tension in the inclined spring. [3] (b) Calculate the tension in the horizontal cord. [2] (c) Calculate the spring's extension and check whether using Hooke's law is consistent with the stated proportional range. [3] (d) A pupil claims that the spring tension is 12.0 N because the sensor is at rest. Explain why this claim is incorrect. [2]

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