Source: Experimental Physics and Engineering Lab, Texas A&M University
Tags: multi-force equilibrium, four concurrent forces, cable geometry, tension limits, maximum weight, inverse trig, angle from geometry, particle statics, ENGR 214
These notes cover two more involved particle equilibrium problems: one with four concurrent forces at known angles where two force magnitudes are unknown, and one where cable geometry must be extracted from coordinate dimensions before solving for tensions and a maximum allowable weight. Both reinforce the same method (FBD, resolve into components, solve the system) but add complexity through more forces or a geometry step.
Concurrent forces
Forces whose lines of action all pass through a single point. In particle statics, every force problem is a concurrent-force problem because the particle is a point.
Inverse trigonometry for angle extraction
When a problem gives you side lengths rather than angles, you recover the angle with tan⁻¹(opposite/adjacent), sin⁻¹(opposite/hypotenuse), or cos⁻¹(adjacent/hypotenuse). This step comes before the equilibrium equations.
Maximum allowable load
The largest weight or force a system can support before any member reaches its tension (or compression) limit. You set the most-stressed member equal to its limit and solve backwards for the load.
Draw the free body diagram with every force labelled.
Extract angles from given geometry (inverse trig if needed).
Choose axes (usually horizontal x, vertical y).
Resolve each force into x and y components.
Write ΣFx = 0 and ΣFy = 0.
Solve the resulting system of equations (substitution or elimination).
This method works regardless of how many forces there are, as long as you have exactly two unknowns (for 2D).
Four forces meet at a single point: FA, FB, FC, FD.
Given: |FA| = 4.2 kN, |FB| = 2.4 kN.
Angles from the diagram: FA at 30° above horizontal (to the left), FB at 30° above horizontal (to the right), FC at 45° below horizontal (to the left), FD at 45° below horizontal (to the right).
Find: magnitudes of FC and FD.
Assumptions: static equilibrium, particle idealisation.
(FA cos 30 – FB cos 30) + (FC cos 45 – FD cos 45) = 0
Substituting known values:
(4.2 cos 30 – 2.4 cos 30) = –FC cos 45 + FD cos 45
3.64 – 2.08 = –0.707 FC + 0.707 FD
1.56 = 0.707(–FC + FD)
2.204 = –FC + FD
So: FD = 2.204 + FC ... (i)
FC sin 45 + FD sin 45 = FA sin 30 + FB sin 30
Substituting:
0.707(FC + FD) = 4.2 sin 30 + 2.4 sin 30 = 2.1 + 1.2 = 3.3
FC + FD = 4.667 ... (ii)
From (i) and (ii):
FC + (2.204 + FC) = 4.667
2FC = 2.463
FC ≈ 1.232 kN
FD = 2.204 + 1.232 ≈ 3.436 kN
With four forces but only two unknowns, you still only need two equations. The known forces move to the right-hand side and the unknowns stay on the left. The algebra is a straightforward pair of simultaneous equations.
A weight W hangs from point C, supported by two cables: AC and BC.
Cable AC connects to a wall at point A. Cable BC connects to a ceiling/wall at point B.
An additional horizontal force Fx = 150 lbs acts at point C.
Given: W = 205 lbs. Geometry from the diagram gives side lengths for the triangles (horizontal 12, vertical 16, horizontal 30 for the upper supports).
Find:
(a) Tension in cables AC and BC.
(b) Maximum W such that neither cable tension exceeds 220 lbs.
The problem gives coordinate dimensions, not angles, so the first step is trigonometry.
For the triangle to point B: θ = tan⁻¹(16/30) ≈ 28°
For the triangle to point A: θ = tan⁻¹(16/12) ≈ 53.15°
These angles are measured from the horizontal.
The angle for cable BC from horizontal: 28°. The angle for cable AC from horizontal: 53.15°.
X-direction (ΣFx = 0):
TBC cos 28 – TAC cos 53.15 – 150 cos 28 = 0
0.88 TBC = 0.6 TAC + 132.44
TBC = (0.6 TAC + 132.44) / 0.88 ... (i)
Y-direction (ΣFy = 0):
TBC sin 28 + TAC sin 53.15 – 150 sin 28 – W = 0
0.471 TBC + 0.8 TAC = 70.58 + 205 = 275.58 ... (ii)
Substitute (i) into (ii):
0.471 × [(0.6 TAC + 132.44) / 0.88] + 0.8 TAC = 275.58
0.535(0.6 TAC + 132.44) + 0.8 TAC = 275.58
0.321 TAC + 70.89 + 0.8 TAC = 275.58
1.121 TAC = 204.69
TAC ≈ 183 lbs
Back-substitute into (i):
0.88 TBC = 0.6(183) + 132.44 = 241.99
TBC ≈ 275 lbs
TBC is the larger tension (275 lbs when W = 205 lbs), so TBC will hit the 220 lb limit first.
Set TBC = 220 lbs in equation (i):
0.88(220) = 0.6 TAC + 132.44
193.6 = 0.6 TAC + 132.44
TAC = 61.16 / 0.6 ≈ 101.93 lbs
Now use equation (ii) with TBC = 220 and TAC = 101.93:
0.471(220) + 0.8(101.93) = 70.58 + W
103.62 + 81.54 = 70.58 + W
W ≈ 114 lbs
When you need the maximum load a system can carry, identify which member carries the highest tension for a unit load, set that member's tension equal to its limit, and solve backwards for the load. The critical member is the one whose tension-to-load ratio is largest.
Formula | Use |
|---|---|
θ = tan⁻¹(opposite / adjacent) | Extract angle from side lengths |
ΣFx = 0, ΣFy = 0 | Equilibrium conditions for a particle |
Fx = F cos θ, Fy = F sin θ | Force components (θ from horizontal) |
⚠️ Always extract angles from geometry before writing equilibrium equations. A common error is using the wrong angle (e.g. measuring from the wrong axis). Double-check with a quick sketch.
⚠️ When a problem asks for a "maximum weight" or "maximum load," that is a design-limit problem. Set the most-stressed member to its limit and solve. Make sure you identify the correct critical member first.
⚠️ Watch units. Question 3 uses kN, Question 4 uses lbs. Mixing units within a single equation is a fast route to a wrong answer.
⚠️ The substitution method (solve one equation for one unknown, plug into the other) is reliable but algebra-heavy. Keep your working neat and check intermediate results with a calculator.
⚠️ In Question 4, TBC > TAC even though they support the same weight. The cable closer to horizontal carries more load because a larger fraction of its tension goes to balancing the horizontal force, leaving less vertical capacity. This is a useful intuition for exam problems.
Q: A particle is in equilibrium under five concurrent forces. Three are known. How many equations can you write, and is that enough to find the remaining two forces?
A: In 2D, you get two equations (ΣFx = 0 and ΣFy = 0). Two unknown force magnitudes (with known directions) give two unknowns, which is solvable. If the directions are also unknown, you have four unknowns and the system is under-determined.
Q: A cable makes an angle with the horizontal. If you flatten the angle (make it more horizontal), what happens to the tension required to support the same vertical load?
A: The tension increases. As the cable becomes more horizontal, its vertical component (T sin θ) gets smaller per unit of tension, so more tension is needed to provide the same vertical support.
Q: In Question 4, why is TBC the critical cable rather than TAC?
A: TBC has a shallower angle (28° vs 53.15°), which means a larger share of its tension goes towards balancing horizontal forces. It ends up carrying more total tension for the same weight. When checking tension limits, the shallower cable hits its limit first.
Q: You are given a triangle with a horizontal side of 5 m and a vertical side of 12 m. What angle does the hypotenuse make with the horizontal?
A: θ = tan⁻¹(12/5) = tan⁻¹(2.4) ≈ 67.4°.
multi-force equilibrium, four concurrent forces, cable tension calculation, maximum weight problem, tension limit, inverse trigonometry in statics, angle from coordinates, particle idealisation, free body diagram, ENGR 214, PHYS 214, Texas A&M, force balance, simultaneous equations in statics, critical member, design load, cable geometry