K323 / 2027
Dynamics overview

Lesson 6 of 7

Three-force equilibrium

  • I can identify and draw the forces on one body in up to two dimensions.Syllabus

    Syllabus K323, 3(h). Identify and draw the forces on one body in up to two dimensions.

  • I can solve two-dimensional, three-force problems for a static point mass by a graphical method.Syllabus

    Syllabus K323, 3(i). Solve two-dimensional, three-force problems for a static point mass by a graphical method.

Make a guess. You are not marked.

A picture hangs from a nail by two strings. Longer strings are fitted, so each string hangs closer to vertical. What happens to the tension in each string?

  1. It decreases, because each string now pulls more nearly upwards.
  2. It increases, because longer strings stretch more and pull harder.
  3. It stays the same, because the weight of the picture is unchanged.
Show the answer

It decreases, because each string now pulls more nearly upwards.

The upward parts of the two tensions must add up to the weight. When a string is closer to vertical, more of its pull is upwards, so a smaller tension is enough.

For a stationary point mass, the forces have zero resultant. Three force vectors drawn head to tail therefore form a closed triangle.

Use graphical vector addition: preserve each arrow's direction, choose a scale, then measure lengths. If needed, review W = mg to find the known weight first.

A point-mass model represents a small body as a single point. Here we use it to find the tensions in two supporting strings. Each string pulls away from the mass along its own direction.

Draw the forces before drawing the triangle

A 0.50 kg mass hangs stationary from two strings. The left string is 30° above the horizontal to the left; the right string is 45° above the horizontal to the right. Take g = 10 N/kg.

Three forces on a stationary point mass

Selected body: 0.50 kg point mass

Forces on a mass held by two stringsA stationary 0.50 kg point mass experiences three forces. Its 5.0 N weight acts vertically downwards. Left-string tension TL acts upwards and left, 30 degrees above the left horizontal. Right-string tension TR acts upwards and right, 45 degrees above the right horizontal. Both tensions pull away from the mass. The dashed horizontal line is an angle reference, not a force.TLLeft stringTRRight string30°45°5.0 NWeightby Earth

The dashed horizontal line is an angle reference. TL and TR act along the strings, away from the mass.

Forces on the point mass: a known 5.0 N weight downwards, an unknown left tension TL, and an unknown right tension TR. The strings supply the two tension directions.

The weight is W = mg = 0.50 x 10 = 5.0 N downwards. There are three forces on the mass. Their directions are known, but the two tensions' magnitudes are not.

Worked construction

Find both tensions with a scale drawing

  1. Choose a paper scale: let 1 cm represent 1 N.
  2. Draw the known force: draw AB 5.0 cm vertically downwards to represent the weight.
  3. Draw the left-tension direction: from B, draw a ray upwards and left, at 30° above the left horizontal.
  4. Draw a line parallel to the right tension: through A, draw a line at 45° to the horizontal. Extend it backwards, downwards and left, until it meets the first ray at C.
  5. Check the arrow directions: the closed sequence is A to B, B to C, then C to A. BC points along the left tension; CA points along the right tension.
  6. Measure and convert: BC is about 3.7 cm and CA about 4.5 cm. At this scale, TL is about 3.7 N and TR about 4.5 N.

A closed triangle of forces

Follow the arrowheads: A to B to C to A.

Head-to-tail construction for three-force equilibriumThe closed force triangle follows A to B to C to A. AB is 5.0 N vertically downwards. BC is TL, about 3.7 N, directed upwards and left at 30 degrees above the left horizontal. CA is TR, about 4.5 N, directed upwards and right at 45 degrees above the right horizontal. The three arrows return to A, so their resultant is zero. All three sides use the same force scale. On paper, a scale of 1 centimetre per newton gives AB 5.0 centimetres, BC about 3.7 centimetres and CA about 4.5 centimetres.ABC5.0 NTRabout 4.5 NTLabout 3.7 N

For your paper drawing, use 1 cm per N. Use the labelled values on screen; its size changes with your device.

The force triangle closes: AB is the weight, BC is TL and CA is TR. On paper at 1 cm per N, the supplied approximate lengths are 5.0 cm, 3.7 cm and 4.5 cm respectively.

The free-body diagram shows the forces acting at the mass. The separate triangle moves those arrows head to tail to add them; it does not show the physical positions of the strings. Returning to A means that the resultant vector is zero.

Use a paper construction for ruler measurements. A screen diagram changes size with the display and need not preserve physical centimetres. Use the stated scale and angles, draw carefully, then measure. Small differences in graphical answers are expected.

Do not add the two tension magnitudes and set that sum equal to the weight. Their directions differ, so they must be added as vectors. Drawing C to B instead of B to C would also reverse the left tension and break the intended head-to-tail sequence.

Try a different arrangement

This optional paper task uses a 6.0 N downward weight, supported by two strings that each make 60° with the horizontal on their respective sides. Draw the free-body diagram, choose a scale and construct the closed force triangle. There is no answer to submit.

Show the construction result

At 1 cm per N, draw the 6.0 cm downward weight. Construct the left-tension direction from its head, then the line parallel to the right tension through its tail, just as in the worked method.

Each tension side measures about 3.5 cm, so both tensions are approximately 3.5 N. The two sides have equal lengths because the arrangement is symmetric. Their vector sum balances the weight; the tension is not 6.0 / 2 = 3.0 N because both strings pull at an angle.