Particle equilibrium in two dimensions
With a good free-body diagram, two equilibrium equations solve for two unknown forces. Shallow cable angles hold some surprises.
The two equations
For concurrent forces in a plane, equilibrium requires \(\sum F_x = 0\) and \(\sum F_y = 0\). With at most two unknowns, these two equations always suffice. Write each force's components with signs from the FBD, then solve.
Two cables at different angles
For the 20 kg traffic light, cables at 30° and 45°: \(\sum F_x = 0\) gives \(T_{AC}\cos 45° = T_{AB}\cos 30°\), and \(\sum F_y = 0\) gives \(T_{AB}\sin 30° + T_{AC}\sin 45° = 196.2\). Solving, \(T_{AB} \approx 143.6\) N and \(T_{AC} \approx 175.9\) N.
Checking the answer
Substitute back: 143.6 x 0.5 + 175.9 x 0.7071 = 71.8 + 124.4 = 196.2 N, and 143.6 x 0.866 = 124.4 = 175.9 x 0.7071. The steeper cable, AC, carries more of the load, which matches intuition.
Shallow angles, huge tensions
A weight W hung from the middle of a cable whose two halves each make angle \(\theta\) with the horizontal needs tension \(T = \frac{W}{2\sin\theta}\). At 30°, T = W; at 10°, T is about 2.9W; at 2°, T is about 14.3W. A washing line or power cable can never be pulled perfectly straight under load.
Lami's theorem
For exactly three concurrent forces in equilibrium, each force is proportional to the sine of the angle between the other two: \(\frac{F_1}{\sin\alpha_1} = \frac{F_2}{\sin\alpha_2} = \frac{F_3}{\sin\alpha_3}\). It gives a quick check, but the component method works in every case.
Inclined planes
A block on a smooth incline held by a rope parallel to the slope: resolving along the slope, the rope tension equals W sin theta, and perpendicular to the slope, the normal force equals W cos theta. On a 30-degree smooth slope, holding a 50 kg block takes 50 x 9.81 x 0.5, about 245 N.
Choosing axes wisely
Aligning one axis with an unknown force makes that force appear in only one equation, so each equation can be solved in turn rather than simultaneously. On an incline, axes along and perpendicular to the slope usually achieve this.
Worked example
A 75 kg climber hangs from the middle of a rope stretched between two rocks, each half sagging at 15 degrees below the horizontal. The tension in each half is 75 x 9.81 / (2 sin 15) = 735.75 / 0.5176, about 1,421 N, nearly twice the climber's weight. Rope and anchors must be rated well above body weight.
Key takeaways
A particle in planar equilibrium satisfies sum Fx = 0 and sum Fy = 0, enough to find two unknowns. Write components with signs from the free-body diagram and check by substitution. A weight hung at the middle of a cable needs tension W over (2 sin theta), which becomes huge at shallow angles. Lami's theorem offers a three-force shortcut. On smooth inclines, rope tension is W sin theta and the normal force W cos theta. Align an axis with an unknown to simplify the algebra.
Flashcard
Sum Fx = 0, sum Fy = 0
Check your understanding
A weight W hangs from the middle of a cable whose halves are each 30 degrees below horizontal. The tension in each half is: