← All solutions · E.T.I. Physics · Statics supplement · AD2026
Textbook problems hand you numbers. This one hands you the formula. Derive it once from ΣF = 0 and every two-cable problem becomes a substitution.
Drag the angles. The arrows scale with the real tensions, and the two balance equations under them stay at zero — that is what equilibrium means.
An object of weight W hangs from a point P. Two cables hold P still: one up-left at θl from the horizontal, one up-right at θr. Both angles are between 0° and 90°, and they need not be equal.
Find Tl and Tr in terms of W, θl and θr only — no numbers anywhere.
| Force | Direction | x-component | y-component |
|---|---|---|---|
| Tl | up and left | −Tl cos θl | +Tl sin θl |
| Tr | up and right | +Tr cos θr | +Tr sin θr |
| W | straight down | 0 | −W |
One minus sign per arrow that points the wrong way. The left cable pulls leftward, so its x-component is negative. Gravity pulls down, so its y-component is negative. Gravity has no x-component at all — that is the only reason this problem is solvable in two lines.
P does not move, so the three forces add to zero. A vector equation is two scalar equations.
Equation (1) says the two horizontal pulls cancel. Equation (2) says the two vertical pulls together hold up the weight. Two equations, two unknowns — the physics is finished. Everything from here is algebra.
We eliminate Tl and solve for Tr. Read the grey line under each step: it says what was done, not just what came out.
Divide equation (1) by cos θl. Legal because θl < 90°, so the cosine is never zero.
Every Tl is gone. One unknown left, and it appears twice.
Now the unknown multiplies one bracket. Whatever is inside, it is just a number once the angles are fixed.
Write sin θr as cos θl sin θr / cos θl and add the two fractions. Look hard at that numerator before going on.
The angle-addition formula, exactly as it appears in the trig table. This single line is why the answer is clean instead of a mess of products. Everything collapses into the sum of the two angles.
Multiply both sides by the reciprocal. Done — and not one number was used along the way.
Do not repeat the six steps. The problem is symmetric under swapping left and right: relabel l ↔ r everywhere and the same derivation runs unchanged, because sin(θl + θr) does not care about the order of its arguments. So swap the labels in the boxed result and you have Tl.
Both tensions share the same denominator. What differs is the numerator — and notice it is swapped: the right tension carries the cosine of the left angle. That is not a typo. It comes straight from equation (1), which ties the two cables together: you cannot change one cable without the other feeling it. Divide the two formulas and the denominator cancels:
The steeper cable has the smaller cosine in its own angle, which puts the larger cosine on top of its own tension. The steeper cable always carries more load. Push the sliders and watch the ratio obey this.
A derived formula is worth nothing until it reproduces the cases you can check by common sense. Each button up top sets the sliders to the case — run it while you read.
The cosines cancel and out drops the formula from the first day of statics. ✓
A horizontal cable has no vertical component, so it can hold nothing up. The vertical one takes everything. ✓
Why a clothesline sags, why a tightrope cannot be pulled straight, and why the tension blows up long before the cable is level. ✓
W = 200 N, θl = 55°, θr = 40°, so θl + θr = 95° and sin 95° = 0.9962.
Check it against the equations you started from, never against the formula you just used:
The left cable is steeper, so it carries more: 153.8 against 115.2, a ratio of 1.336 — exactly cos 40° / cos 55°. The formula knew that before any number was substituted.
The formula answers more than it was asked. Differentiate it and it tells you how the tension responds when a cable is tilted — which is what you actually need to know before you build anything.
Take Tr and differentiate with respect to the left angle, using the quotient rule. The numerator that comes out is −[sin θl sin(θl+θr) + cos θl cos(θl+θr)], and the cosine-difference identity collapses that bracket to a single cos θr. All four derivatives come out the same way:
Real suspensions move. Let the hanging point drift sideways by a small angle δ, so that one angle opens as the other closes:
One angle gains exactly what the other loses, so the sum never changes. And the sum is the only thing in the denominator. That single observation does most of the work: sin(θl+θr) is frozen, and each tension depends on just one moving angle.
It is also real geometry, not a convenience. The angle between the two cables at P is 180° − (θl+θr), and the points that see a fixed segment under a fixed angle form a circular arc through its endpoints. Constant sum means P is travelling along the circle that passes through both anchors. Perpendicular cables put P on the circle whose diameter is the anchor-to-anchor span — Thales, from secondary-school geometry, arriving uninvited in a statics problem.
With the denominator frozen, differentiate each formula with respect to δ alone:
Section 06 found a minimum at θl+θr = 90°, where ∂Tl/∂θl = 0. Tempting conclusion: build it perpendicular and the left cable will not feel the swing. It is wrong, and it is worth seeing why.
That partial derivative holds θr fixed while θl moves — the sum changes, and P leaves the circle. The swing does the opposite: both angles move, the sum is frozen. Same point in the plane, different direction of travel, different rate of change. A partial derivative only answers for the one direction it was taken in.
Run the numbers at θl = 50°, θr = 40°, W = 200 N: a swing of 2° moves Tl by +4.4 N, not zero. The perpendicular configuration is a genuine static optimum. It promises nothing about the dynamics.
E.T.I. Physics · Statics supplement · Anuar Kafuri Zarzosa · Tecnológico de Monterrey · AD2026