A knot that is not symmetric
In the last lesson the two strings made equal angles, and symmetry gave equal tensions. Most real knots are not symmetric. A load hangs from a knot, one string makes \(30°\) with the horizontal and the other makes \(60°\). The two tensions are now different, and symmetry cannot find them. Two general methods can, and this lesson compares them on the same knot.
The method of components
Choose \(x\) horizontal and \(y\) vertical. Split every slanting force into its two components. Write the sum of the \(x\) components equal to zero and the sum of the \(y\) components equal to zero. The two equations are then solved together. The method never fails, and it works in the same way for three forces or for ten.
A load of mass \(10\,\text{kg}\) hangs from a knot. The left string makes \(30°\) with the horizontal and the right string makes \(60°\) with the horizontal. Take \(g = 10\,\text{m/s}^2\). Find the tensions \(T_1\) in the left string and \(T_2\) in the right string.
The body is the knot. The forces are \(W = 100\,\text{N}\) down, \(T_1\) at \(30°\) above the horizontal to the left, and \(T_2\) at \(60°\) above the horizontal to the right. Take \(x\) to the right and \(y\) upward.
Horizontal: \(T_2\cos 60° - T_1\cos 30° = 0\). Vertical: \(T_1\sin 30° + T_2\sin 60° - 100 = 0\).
The first equation gives \(0.5\,T_2 = 0.866\,T_1\), so \(T_2 = 1.732\,T_1\). The second becomes \(0.5\,T_1 + 1.732 \times 0.866\,T_1 = 100\), that is \(2\,T_1 = 100\). So \(T_1 = 50\,\text{N}\) and \(T_2 = 86.6\,\text{N}\).
The steeper string carries more: \(86.6\,\text{N}\) against \(50\,\text{N}\). Check: \(50^2 + 86.6^2 = 10\,000 = 100^2\), as it must be when the two strings are perpendicular.
A load hangs from a knot held by two strings of different steepness. Which string carries the larger tension?
The triangle of forces
For three forces the force polygon is a triangle. The sides of the triangle are the three forces, drawn to scale, and the angles of the triangle come from the directions of the forces. So every tool for a triangle can be used on the forces: right-angle trigonometry, the sine rule, even a careful drawing with a ruler. For the knot above, the strings are perpendicular, so the triangle is right-angled with the weight as its hypotenuse.
Let three forces act at a point and be in equilibrium. Then the three forces can be represented in size and direction by the sides of a triangle taken in order. Conversely, three forces that form a closed triangle taken in order are in equilibrium.
For the knot, the angle of the triangle between \(W\) and \(T_2\) is \(30°\), because the right string makes \(30°\) with the vertical. The side opposite this angle is \(T_1\), and the side next to it is \(T_2\). So \(T_1 = W\sin 30° = 50\,\text{N}\) and \(T_2 = W\cos 30° = 86.6\,\text{N}\). Two lines of arithmetic replace two simultaneous equations.
A block rests on a smooth incline. It is held by a string along the slope, and a hand also pushes it horizontally. Which method is the better choice?
Lami's theorem
When the force triangle is not right-angled, the sine rule is the tool. The sine rule needs the interior angles of the triangle. But a free-body diagram shows the angles between the forces, measured with all three arrows starting at the knot. Lami's theorem joins the two. It is stated directly in terms of the angles in the free-body diagram, so the triangle never has to be drawn.
Let three concurrent forces \(P\), \(Q\) and \(R\) keep a body in equilibrium. Then each force is proportional to the sine of the angle between the other two: \[ \dfrac{P}{\sin\alpha} = \dfrac{Q}{\sin\beta} = \dfrac{R}{\sin\gamma} \] where \(\alpha\) is the angle between \(Q\) and \(R\), \(\beta\) the angle between \(R\) and \(P\), and \(\gamma\) the angle between \(P\) and \(Q\).
\[ \vec P + \vec Q + \vec R = 0 \]
Why. the three forces are in equilibrium, so their vector sum is zero and, drawn head to tail, they close a triangle.
\[ \text{interior angle between the sides } Q \text{ and } R = 180° - \alpha \]
Why. in the free-body diagram \(Q\) and \(R\) start at the same point with angle \(\alpha\) between them. Head to tail, one arrow is carried to the end of the other, and the angle inside the triangle becomes the supplement, \(180° - \alpha\). The same holds for the other two corners.
\[ \dfrac{P}{\sin(180° - \alpha)} = \dfrac{Q}{\sin(180° - \beta)} = \dfrac{R}{\sin(180° - \gamma)} \]
Why. the sine rule for the triangle divides each side by the sine of the interior angle opposite it. The side \(P\) is opposite the corner where \(Q\) and \(R\) meet.
\[ \dfrac{P}{\sin\alpha} = \dfrac{Q}{\sin\beta} = \dfrac{R}{\sin\gamma} \]
Read it. \(\sin(180° - x) = \sin x\) for every angle \(x\). The supplements disappear, and each force is matched with the angle between the other two, as the free-body diagram shows it.
Each force goes with the angle between the other two, the angle it does not touch. Students often pair a force with the angle next to it. Before substituting, point to the force, then point to the angle between the other two arrows, and only then write the fraction.
A load of mass \(10\,\text{kg}\) hangs from a knot. A horizontal string ties the knot to a wall. A second string ties the knot to the ceiling at \(30°\) to the vertical. Take \(g = 10\,\text{m/s}^2\). Find the tension \(T_1\) in the ceiling string and \(T_2\) in the horizontal string.
The body is the knot, with three forces: \(W = 100\,\text{N}\) down, \(T_2\) horizontal towards the wall, and \(T_1\) up along the ceiling string. The angle between \(T_1\) and \(T_2\) is \(90° + 30° = 120°\). The angle between \(T_2\) and \(W\) is \(90°\). The angle between \(W\) and \(T_1\) is \(180° - 30° = 150°\).
Lami's theorem, each force with the angle between the other two: \(\dfrac{T_1}{\sin 90°} = \dfrac{T_2}{\sin 150°} = \dfrac{100}{\sin 120°}\).
\(\dfrac{100}{\sin 120°} = \dfrac{100}{0.866} = 115.5\,\text{N}\). So \(T_1 = 115.5 \times 1 = 115.5\,\text{N}\) and \(T_2 = 115.5 \times 0.5 = 57.7\,\text{N}\).
The ceiling string carries more than the whole weight, \(115.5\,\text{N}\). Check by components: vertical, \(115.5\cos 30° = 100\); horizontal, \(115.5\sin 30° = 57.7\). Both balance.
A load of weight \(50\,\text{N}\) hangs from a knot held by two strings. String A makes \(30°\) with the vertical on one side, and string B makes \(60°\) with the vertical on the other side. Use Lami's theorem to find the tension in string B, in newtons.
Which conditions must hold before Lami's theorem can be used?
Three strings pull on a ring, and the ring stays at rest. The angle between every pair of strings is \(120°\). What can you say about the three tensions?
The sine rule uses the interior angles of the force triangle, but Lami's theorem uses the angles between the forces in the free-body diagram. Explain why both give the same answer.
Move two forces head to tail to form the triangle. The angle between them inside the triangle is then \(180°\) minus their angle in the free-body diagram. The sine of an angle equals the sine of its supplement, \(\sin(180° - x) = \sin x\). So replacing each interior angle by the angle in the free-body diagram changes no sine, and the sine rule becomes Lami's theorem.
Bernard Lamy was a French priest who taught mathematics and philosophy in schools of the Oratorian order. Like many teachers, he looked for the shortest honest proof of each result for his students. In a small book on mechanics he stated the result for three forces in equilibrium. Each force is proportional to the sine of the angle between the other two. The result was not new in substance, since it follows from the triangle of forces. But Lamy wrote it in a form that a student can use straight from a diagram. The theorem still carries his name, usually spelt Lami.
- Pairing a force with an angle next to it in Lami's theorem.Each force goes with the angle between the other two, the only angle that force does not touch.
- Using Lami's theorem with four forces.Lami needs exactly three forces. For four or more, use components, or first combine two forces into one.
- Using interior angles of the force triangle in Lami's theorem.Lami uses the angles between the forces with their tails together, as in the free-body diagram. The interior angles are their supplements.