A child sits on a garden swing. A friend pushes the seat sideways and holds it there, waiting to let go. The further the swing is held from the vertical, the harder the friend must push. The ropes also feel tighter. Why should a sideways push make the ropes pull harder?
Three forces on the bob
Replace the swing by a small bob on a string. A horizontal push \(F\) holds the bob at rest, with the string at an angle \(\theta\) to the vertical. Three forces act on the bob: the weight \(W\) down, the push \(F\) horizontal, and the tension \(T\) along the string towards the support. The weight and the push are perpendicular, so the force triangle is right-angled, and the tension is its hypotenuse.
What the triangle gives
The vertical component of the tension balances the weight, \(T\cos\theta = W\). The horizontal component balances the push, \(T\sin\theta = F\). Divide the second equation by the first to get \(\tan\theta = F/W\). Square and add the two equations to get \(T^2 = W^2 + F^2\). The tension is always larger than the weight, because the string now carries the push as well as the weight.
A bob of mass \(4\,\text{kg}\) hangs from a light string. A horizontal force holds the bob at rest with the string at \(37°\) to the vertical. Take \(g = 10\,\text{m/s}^2\), \(\sin 37° = 0.6\) and \(\cos 37° = 0.8\). Find the horizontal force and the tension.
The body is the bob. The forces are \(W = 40\,\text{N}\) down, the horizontal push \(F\), and the tension \(T\) along the string at \(37°\) to the vertical.
Vertical: \(T\cos 37° - 40 = 0\). Horizontal: \(F - T\sin 37° = 0\).
\(T = 40 / 0.8 = 50\,\text{N}\). Then \(F = 50 \times 0.6 = 30\,\text{N}\).
The push is \(30\,\text{N}\) and the tension is \(50\,\text{N}\), a quarter more than the weight. Check with the triangle: \(30^2 + 40^2 = 2500 = 50^2\), and \(\tan 37° = 30/40 = 0.75\).
The formula \(\tan\theta = F/W\) also answers a question students often ask: can a horizontal push hold the string exactly horizontal? At \(\theta = 90°\), \(\tan\theta\) has no finite value. So no finite push can do it. However hard the friend pushes, the ropes of the swing stay a little below the horizontal. Some part of the tension must always be vertical to carry the weight.
A bob of mass \(2\,\text{kg}\) hangs on a string and is held aside by a horizontal force of \(15\,\text{N}\). Take \(g = 10\,\text{m/s}^2\). Find the tension in the string, in newtons.
The horizontal push on a hanging bob is slowly increased, and the bob stays at rest at each new position. What happens to the angle of the string and to the tension?
What a light pulley does
A string that runs over a pulley pulls in one direction on one side and in another direction on the other side. A light pulley has negligible mass and turns without friction. Such a pulley cannot change the tension in the string. The tension is the same on both sides, and the pulley only turns the direction of the pull round a corner. This is why a single pulley at the top of a flagpole lets you raise a flag by pulling down.
When a light string passes over a light, frictionless pulley, the tension has the same value on both sides of the pulley. The pulley changes the direction of the string, never the size of the tension.
Solving two bodies joined by one string
When two bodies hang from the two ends of one string, draw a free-body diagram for each body. The same letter \(T\) appears in both diagrams, because the pulley keeps the tension the same. Each diagram gives its own equations. The shared \(T\) links them: find \(T\) from one body and use it in the other.
Block A of mass \(4\,\text{kg}\) rests on a smooth incline of \(30°\). A light string runs from A, parallel to the slope, over a light pulley at the top, to block B, which hangs freely. Both blocks are at rest. Take \(g = 10\,\text{m/s}^2\). Find the mass of B, the tension and the normal force on A.
Two bodies. On A: \(W_A = 40\,\text{N}\) down, \(N\) perpendicular to the slope, \(T\) up the slope. On B: \(W_B = m_B g\) down and \(T\) up. The light pulley makes the tension the same on both.
For A, along the slope: \(T - 40\sin 30° = 0\); across the slope: \(N - 40\cos 30° = 0\). For B, vertically: \(T - m_B \times 10 = 0\).
From A: \(T = 20\,\text{N}\) and \(N = 34.6\,\text{N}\). From B: \(m_B = 20/10 = 2\,\text{kg}\).
A hanging block of \(2\,\text{kg}\) holds a \(4\,\text{kg}\) block on a \(30°\) incline, because only \(mg\sin 30°\), half the weight of A, acts along the slope. The string carries \(20\,\text{N}\) on both sides of the pulley.
The force on the pulley
The string presses on the pulley on both sides, so the axle of the pulley must be held by its support. The two segments of the string pull the pulley along their own directions, each with a force \(T\). The support feels the vector sum of these two pulls. If the two segments make an angle \(\phi\) with each other, the sum has size \(2T\cos(\phi/2)\). When both segments hang straight down, \(\phi = 0\) and the support carries \(2T\).
A light string passes over a light pulley. One segment of the string hangs vertically and the other runs horizontally, so the segments make \(90°\) with each other. The tension is \(50\,\text{N}\). Find the size of the force the pulley's support must provide, in newtons.
A block of weight \(30\,\text{N}\) hangs from a light string that passes over a light, frictionless pulley. A student holds the other end of the string at an angle of \(45°\) below the horizontal, and the block is at rest. With what force does the student pull?
Why must the pulley be light and frictionless for the tension to be the same on both sides?
Suppose the pulley had friction at its axle. Part of the pull on one side would then turn the pulley against the friction, and the two tensions would differ. If the pulley had mass, turning it would also need a difference in the two tensions. A light, frictionless pulley needs no such difference, so the same tension runs round it. Pulleys with mass come in Rotational Mechanics.
A villager draws water from a well with a rope over a pulley. The full bucket has a mass of about \(12\,\text{kg}\) and is held still halfway up. Roughly what force must the villager pull with, in newtons? Take \(g = 10\,\text{m/s}^2\).
On the morning of the fifteenth of August, the national flag rises to the top of a tall mast. The person hoisting it stands on the ground. A small pulley at the top of the mast makes this possible. The rope runs up the mast, over the pulley and down to the hands. Pulling down on one side raises the flag on the other, with the same tension in both parts of the rope. The pulley does not reduce the effort; it only turns an upward job into a downward pull.
- Writing a tension equal to the weight for a string held aside by a push.A leaning string carries the push as well: \(T = \sqrt{W^2 + F^2}\), always more than \(W\).
- Giving different tensions to the two sides of a light pulley.A light, frictionless pulley keeps one tension all round. Use the same letter \(T\) on both sides.
- Adding the two tensions on a pulley as numbers to find the force on its support.Add them as vectors. For segments at angle \(\phi\), the support feels \(2T\cos(\phi/2)\), which is \(2T\) only when the segments are parallel.