All subjects · Mathematics · WME02 Mechanics 2

Rods hinged to a wall: reactions at a hinge

Edexcel IAL Mathematics · WME02 Mechanics 2

A rod attached to a wall by a hinge can be held up by a string. The hinge can push in any direction, so its reaction has an unknown size and direction. You find it from its horizontal and vertical components.

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Forces on the rod

Weight at the centre of mass (midpoint if uniform), plus any loads.
Tension T along the string.
Hinge reaction with components X (horizontal) and Y (vertical). You can draw them in either direction; a negative answer means the other way.

Method

1. Moments about the hinge to find T (X and Y have no moment there).

2. Resolve horizontally to find X.

3. Resolve vertically to find Y.

4. Magnitude of the reaction = sqrt(X² + Y²); its direction is at arctan(Y/X) to the horizontal.

Exam points

This is equilibrium under non-parallel coplanar forces (spec 5.2). A question may say the reaction at the hinge acts at a given angle: then X and Y are R cos θ and R sin θ. If asked for the direction, give an angle with a clear reference (for example, above the horizontal).

Worked example

A uniform rod AB of length 2 m and weight 40 N is hinged at A to a vertical wall. It is held horizontal by a string from B to a point on the wall above A; the string makes 30° with the rod. Find the tension and the reaction at the hinge.

M(A): T sin 30° × 2 = 40 × 1, so T = 40 N.
Horizontally: X = T cos 30° = 34.6 N.
Vertically: Y + T sin 30° = 40, so Y = 20 N.
R = sqrt(1200 + 400) = 40 N at arctan(20/34.6) = 30° above the horizontal.

Answer: T = 40 N; hinge reaction 40 N at 30° above the horizontal

Key idea: Moments about the hinge for T, then resolve horizontally and vertically for X and Y. R = sqrt(X² + Y²).

Check yourself

The rod in the worked example also carries a 20 N load at B. Find the tension in the string.
Show answer
80 N. M(A): T sin 30° × 2 = 40 × 1 + 20 × 2 = 80, so T = 80 N.
With the 20 N load at B (T = 80 N), find the horizontal component of the hinge reaction, to 3 significant figures.
Show answer
69.3 N. X = 80 cos 30° = 69.3 N.
With the 20 N load at B (T = 80 N), find the vertical component of the hinge reaction.
Show answer
20 N. Y + 80 sin 30° = 60, so Y = 20 N.
A uniform rod AB of length 3 m and weight 30 N is hinged at A to a vertical wall and held horizontal by a light string from B to a point on the wall vertically above A. The string makes 60° with the rod. A 10 N load hangs 2 m from A. Find the tension, to 3 significant figures.
Show answer
25.0 N. T sin 60° × 3 = 30 × 1.5 + 10 × 2 = 65, so T = 25.0 N.
For the rod in the previous question, find the magnitude of the hinge reaction, to 3 significant figures.
Show answer
22.2 N. X = T cos 60° = 12.5 N, Y = 40 - T sin 60° = 40 - 65/3 = 18.3 N, R = 22.2 N.
Why do you usually take moments about the hinge first?
- The hinge carries no force
- Moments are always taken about the left end
- The unknown hinge reaction has no moment about it
- The tension acts through the hinge
Show answer
The unknown hinge reaction has no moment about it. Taking moments about the hinge removes X and Y, leaving T as the only unknown.

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Last updated 8 November 2026. Spotted a mistake? Tell us.