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Constructing force triangles with unknown tensions

Key idea: For a static point mass held by two strings, the weight and tension directions fix a closed force triangle. Its side lengths represent the two tensions.

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A sign hangs from two strings. You know its weight and you can see which way each string runs, but you know neither tension. You cannot draw the triangle of forces in the usual way, because two of its sides have unknown lengths. You can still draw it: the two known directions fix where those sides must meet.

Treat the hanging object as a static point mass. Its weight acts vertically downwards, and each taut string pulls on it along the string, away from the object. Ignore any other forces. This model is about the balance of forces only, not about turning effects.

Find the missing corner

The weight arrow is the one side you can draw completely. Call its tail A and its head B. The two tension arrows must start at B and finish at A, meeting at an unknown corner C. Each tension direction gives a line that C must lie on:

  1. Choose a scale, such as 1 cm for 2 N. Draw the weight from A down to B.
  2. From B, draw a faint line in the direction of the left tension, up and to the left.
  3. From A, draw a faint line in the reverse of the right tension’s direction, down and to the left. It is reversed because the right-tension arrow must finish at A, so you are tracing it backwards from its end.
  4. Mark C where the two lines cross. The force arrows are A→B for the weight, B→C for the left tension and C→A for the right tension.
  5. Measure BC and CA with a ruler and convert both lengths with your scale.

Until you find C, the construction lines show only directions. Their lengths mean nothing, so do not measure a tension from a line you drew before finding C.

Constructing the missing corner of a force triangleThe free-body diagram has a 12 N weight downward, left tension 30 degrees above the leftward horizontal, and right tension 60 degrees above the rightward horizontal. In the triangle, weight runs from A down to B. The left-tension ray from B meets the reversed right-tension ray from A at C. The final arrows are A to B, B to C, and C to A.Forces on the point massFind C where the rays meet30°60°TₗTᵣW = 12 NTensions pull along their strings.ABC12 NTₗTᵣFrom B: actual direction of Tₗ.From A: reverse direction of Tᵣ.Final forces: A→B, B→C, C→A.
Constructing the missing corner of a force triangleThe free-body diagram has a 12 N weight downward, left tension 30 degrees above the leftward horizontal, and right tension 60 degrees above the rightward horizontal. In the triangle, weight runs from A down to B. The left-tension ray from B meets the reversed right-tension ray from A at C. The final arrows are A to B, B to C, and C to A.Forces on the point massTₗTᵣ30°60°W = 12 NTensions pull along their strings.Find C where the rays meetABC12 NTₗTᵣFrom B: actual direction of Tₗ.From A: reverse direction of Tᵣ.Final forces: A→B, B→C, C→A.
The dashed construction rays locate C without guessing a tension length. The final force arrows close A→B→C→A. Make your own drawing using a stated centimetre-to-newton scale; screen lengths are not centimetres.

The diagram on screen shows the construction, not a real centimetre scale. Make your own labelled scale drawing on paper.

Symmetric strings

When two strings make equal angles with the horizontal and support the weight symmetrically, the two tension sides of the triangle come out the same length, so the tensions are equal. It is the symmetry that makes them equal, not simply having two strings.

Worked example 1

Two strings at equal angles

Problem

A 10 N weight hangs symmetrically from two strings, each at 45° above the horizontal. Use a force triangle at 1 cm for 2 N to find each tension.

Show full solution

  1. Draw the known force to scale

    Method

    Draw the 10 N weight as a 5.0 cm arrow, A to B, vertically downwards.

    Reason

    At this scale each 2 N is 1 cm.

    Working

    Answer

    (10 N)/(2 N cm⁻¹) = 5.0 cm

  2. Find the missing corner

    Method

    From B, draw a line up and to the left at 45°. From A, draw the reversed right-tension line down and to the left at 45°. Mark C where they cross, and draw the arrows B→C and C→A.

    Reason

    The three forces are in equilibrium, so their arrows must close into a triangle.

    Working

    Answer

    The two tension sides are equal, each about 3.5 cm long.

  3. Convert the measured length

    Method

    Multiply each measured side by the scale.

    Reason

    The drawing gives lengths in centimetres; the answer must be a force in newtons.

    Working

    Answer

    T ≈ (3.5 cm)(2 N cm⁻¹) = 7.0 N

Check your understanding 1 Steeper strings

A 20 N object hangs symmetrically from two strings, each at 60° above the horizontal. Treat it as a static point mass. Choose a scale, draw a closed force triangle and estimate each tension. Label all three force arrows and their directions; record the scale, construction and measured lengths below.

Show answer

With 1 cm for 2 N, draw a 10 cm weight arrow downwards and find the missing corner from the two 60° directions. Each tension side measures about 5.8 cm, so each tension is about 12 N (more precisely, 11.5 N).

Self-review the construction: A→B is vertically down, B→C follows the left string up-left at 60°, and C→A follows the right string up-right at 60°. All lengths use one stated scale and the arrows close. Equivalent scales are valid; report a precision your drawing supports.

The tensions are equal because the angles are equal. Each is more than 10 N, half the weight, because only part of each tension acts upwards.

Check your understanding 2 Shallower strings

The same 20 N object now hangs symmetrically from two strings at 30° above the horizontal. Will each tension be larger or smaller than with the 60° strings? Predict first, then check with a labelled force triangle on paper using the same scale as before. Explain the change even though the weight is unchanged.

Show answer

Self-review that the 30° tension directions replace the 60° directions, while the weight arrow and scale stay unchanged. The arrows must still close, with the left tension up-left and right tension up-right.

Each tension is larger: about 20 N. At the same scale the weight side is still 10 cm, but each tension side is now about 10 cm.

A shallow string pulls mostly sideways, so it holds up less of the weight for each newton of tension. The tension has to grow until the upward parts of the two tensions add up to the weight again.

Strings at different angles

When the strings make different angles, the tensions are different, and the steeper string carries more of the weight. The construction is the same; only the two directions change. The figure above shows it for a 12 N weight on strings at 30° and 60°.

Guided practice 2

A sign on two strings at different angles

About 8 min

Problem

A sign of weight 12 N hangs from two strings. The left string is at 45° above the horizontal and the right string is at 60° above the horizontal. Treat the sign as a static point mass. Draw a scaled force triangle and find both tensions.

Construct the two-angle triangle on paper, then record your reasoning

First draw the sign’s three-force diagram, then a separate scaled triangle on paper. Use the question’s left angle of 45°, not the earlier figure’s 30°. Record where each construction ray starts and which one is reversed. The optional checks assess only numerical tensions.

Unit: N
Unit: N

Hints

Hint 1

start with a free-body diagram

Show the weight downwards and each tension along its own string, away from the sign.

Hint 2

draw the weight first

At 1 cm for 2 N, draw the weight as a 6.0 cm arrow downwards. Trace the left-tension direction from its head and the reversed right-tension direction from its tail.

Show solution step by step

  1. Construct the triangle

    Method

    Draw the 6.0 cm weight arrow, then the two construction lines, and mark C where they cross.

    Reason

    The sign is in equilibrium, so the three force arrows must close.

    Working

    Answer

    The force arrows are A→B (weight), B→C (left tension) and C→A (right tension).

  2. Measure and convert

    Method

    Measure BC and CA, and multiply each by 2 N cm⁻¹.

    Reason

    Unequal angles give unequal sides.

    Working

    Answer

    T_left ≈ (3.1)(2) = 6.2 N, T_right ≈ (4.4)(2) = 8.8 N

Check your understanding 3 Correct the rays, then construct a new support

A learner starts both actual tension directions at B, the head of the weight arrow, and calls B their intersection. Explain why this cannot locate the missing corner. Then independently construct a triangle for a static 18 N sign: the left string is at 60° above the horizontal and the right at 30°. Ignore other forces. Choose a sensible scale, label the forces and estimate both tensions.

Show answer

The right-tension arrow must finish at A, not start at B. Trace its direction backwards from A, down-left at 30°, while tracing the left tension forwards from B, up-left at 60°. Their intersection C fixes the missing corner; starting both rays at B creates no weight-balancing triangle.

For example, at 1 cm for 3 N draw A→B 6.0 cm downwards. B→C is about 5.2 cm up-left at 60°, and C→A is about 3.0 cm up-right at 30°. The tensions are about 15.6 N left and 9.0 N right. Check one stated scale, labelled force arrows, correct directions and closed head-to-tail arrows; any equivalent scale and order is valid. The left string is now steeper and has the larger tension. Its upward component is about 13.5 N, the right’s about 4.5 N, which add to the 18 N weight; the horizontal components cancel. These component values are a check on the drawing, not a required alternative solution.

Common mistakes

  • Turning an arrow when moving it from the free-body diagram to the triangle. Moving an arrow keeps its direction.
  • Drawing the right tension from A instead of towards A. Reverse a direction only for the construction line; the force itself is C→A.
  • Halving the weight to get each tension. Even with symmetric strings, each tension is more than half the weight when the strings slope.
  • Measuring a tension from a construction line before finding where the lines cross.
  • Making the drawing small, or giving many calculator figures. Report only the precision supported by your scale and measuring tools.

Spotted an error? Report a correction

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