CivilSolve

Statics & Mechanics of Materials

Particle Equilibrium (Cable Tensions)

Equilibrium of concurrent forces at a point: known loads (e.g. a hung weight, direction 270°) plus one or two members/cables at known directions — solves ΣFx = 0 and ΣFy = 0 for the unknown forces. Positive result = tension (pulling away from the point); a negative result means compression, which a cable cannot carry (warned). Use for: 'find the tension in each cable', ring/hook/pulley junction problems, two-bar bracket forces. Not for: rigid bodies with moments (beam-reactions) or full trusses.

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

Load a sample problem:
known forcesThe known applied forces at the point (weights, applied pulls)
known forces 1
unknown directionsThe 1 or 2 members/cables whose force is unknown
unknown directions 1

Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

symmetric cables at 30° and 150° holding 100 N

Given

known forces
  1. 1.magnitude 100 Nangle deg 270
unknown directions
  1. 1.angle deg 30
  2. 2.angle deg 150

Assumptions

  • All forces are concurrent and coplanar; members are two-force (axial) so their force acts along their direction; positive = tension.

Solution steps

  1. Sum the known forces

    Everything already known collapses into one net force the unknowns must balance.

    ΣFx = 0 and ΣFy = 0

    Known net: Fx = -1.837e-17 kN, Fy = -0.1 kN

  2. Solve for member 1

    The member's force component balance along x and y fixes its magnitude.

    each T acts along its member direction

    0.1 kN at 30° (tension)

    = 0.1 kN

  3. Solve for member 2

    The member's force component balance along x and y fixes its magnitude.

    each T acts along its member direction

    0.1 kN at 150° (tension)

    = 0.1 kN

Results

Force in member 1 (positive = tension)

0.1kN

Force in member 2 (positive = tension)

0.1kN

Where this answer was checked
source
ΣFx = ΣFy = 0, symmetric case — equal tensions of 100 N
verified by
hand-recomputed
derivation
Symmetry gives T1 = T2 = T. Vertical: 2·T·sin30° = 100 → T = 100 N.

Example 2

asymmetric cables at 45° and 120° holding 200 N

Given

known forces
  1. 1.magnitude 200 Nangle deg 270
unknown directions
  1. 1.angle deg 45label cable AB
  2. 2.angle deg 120label cable AC

Assumptions

  • All forces are concurrent and coplanar; members are two-force (axial) so their force acts along their direction; positive = tension.

Solution steps

  1. Sum the known forces

    Everything already known collapses into one net force the unknowns must balance.

    ΣFx = 0 and ΣFy = 0

    Known net: Fx = -3.674e-17 kN, Fy = -0.2 kN

  2. Solve for cable AB

    The member's force component balance along x and y fixes its magnitude.

    each T acts along its member direction

    0.1035 kN at 45° (tension)

    = 0.1035 kN

  3. Solve for cable AC

    The member's force component balance along x and y fixes its magnitude.

    each T acts along its member direction

    0.1464 kN at 120° (tension)

    = 0.1464 kN

Results

Force in cable AB (positive = tension)

0.1035kN

Force in cable AC (positive = tension)

0.1464kN

Where this answer was checked
source
2×2 equilibrium solved by hand — unequal tensions
verified by
hand-recomputed
derivation
x: T1·cos45 + T2·cos120 = 0 → 0.70711·T1 = 0.5·T2 → T2 = 1.41421·T1. y: T1·sin45 + T2·sin120 = 200 → 0.70711·T1 + 1.22474·T1 = 200 → T1 = 200/1.93185 = 103.53 N; T2 = 146.41 N.

Example 3

US 3-4-5 cables at 53.13° and 126.87° holding 100 lbf

Given

known forces
  1. 1.magnitude 100 lbfangle deg 270
unknown directions
  1. 1.angle deg 53.13
  2. 2.angle deg 126.87

Assumptions

  • All forces are concurrent and coplanar; members are two-force (axial) so their force acts along their direction; positive = tension.

Solution steps

  1. Sum the known forces

    Everything already known collapses into one net force the unknowns must balance.

    ΣFx = 0 and ΣFy = 0

    Known net: Fx = -1.837e-14 lbf, Fy = -100 lbf

  2. Solve for member 1

    The member's force component balance along x and y fixes its magnitude.

    each T acts along its member direction

    62.5 lbf at 53.13° (tension)

    = 62.5 lbf

  3. Solve for member 2

    The member's force component balance along x and y fixes its magnitude.

    each T acts along its member direction

    62.5 lbf at 126.9° (tension)

    = 62.5 lbf

Results

Force in member 1 (positive = tension)

0.0625kip

Force in member 2 (positive = tension)

0.0625kip

Where this answer was checked
source
Symmetric 3-4-5 geometry — T = 62.5 lbf each
verified by
hand-recomputed
derivation
sin53.13° = 0.8. Vertical: 2·T·0.8 = 100 → T = 62.5 lbf.