Watch a crate being loaded into a truck at a godown. The workers do not lift the crate. They lay a wooden plank from the ground to the truck and pull the crate up the plank with a rope. On a gentle plank one worker holds a heavy crate with ease. On a steep plank the same worker struggles. The crate has the same weight both times. What has changed?
Three forces on the incline
Take a block on a smooth incline, held at rest by a string that runs parallel to the slope. A smooth surface has no friction, so the incline can only push perpendicular to itself. Three forces act on the block. The weight \(W = mg\) acts vertically down. The normal force \(N\) acts perpendicular to the slope, away from it. The tension \(T\) acts along the string, up the slope.
Choose the axes to fit the slope
Two of the three forces already lie along the slope or across it. So choose \(x\) along the slope, upward, and \(y\) perpendicular to the slope, away from it. Only the weight must then be split into components. The angle between the weight and the \(y\) axis equals the angle of the incline \(\theta\). So the weight has a component \(W\sin\theta\) down the slope and a component \(W\cos\theta\) into the slope.
To check which component gets the sine, imagine the incline flat, with \(\theta = 0\). The whole weight then presses into the surface, and nothing pulls the block along it. \(\cos 0 = 1\) and \(\sin 0 = 0\), so the component into the slope must be \(W\cos\theta\). Now imagine a vertical wall, \(\theta = 90°\): the whole weight acts along the surface.
A block of mass \(2\,\text{kg}\) rests on a smooth incline of \(30°\). A string parallel to the slope holds the block at rest. Take \(g = 10\,\text{m/s}^2\). Find the tension in the string and the normal force on the block.
The body is the block. The forces are the weight \(W = 20\,\text{N}\) vertically down, the normal force \(N\) perpendicular to the slope, and the tension \(T\) up the slope. Take \(x\) up the slope and \(y\) perpendicular to the slope, away from it.
Along the slope: \(T - W\sin 30° = 0\). Perpendicular to the slope: \(N - W\cos 30° = 0\).
\(T = 20 \times 0.5 = 10\,\text{N}\). \(N = 20 \times 0.866 = 17.3\,\text{N}\).
The string pulls with \(10\,\text{N}\) and the incline pushes with \(17.3\,\text{N}\). Both are smaller than the weight. Check with the force triangle: \(10^2 + 17.3^2 = 400 = 20^2\).
The incline is made steeper while the string stays parallel to the slope. What happens to the normal force \(N\) and the tension \(T\)?
At \(15°\) the normal force is \(97\,\%\) of the weight and the string carries only \(5.2\,\text{N}\). At \(60°\) the normal force is half the weight and the string carries \(17.3\,\text{N}\). This is why the workers at the godown choose a long, gentle plank. The plank takes most of the weight of the crate, and the rope needs to supply only \(mg\sin\theta\).
A crate of mass \(5\,\text{kg}\) is held at rest on a smooth ramp of \(37°\) by a rope parallel to the ramp. Take \(g = 10\,\text{m/s}^2\), \(\sin 37° = 0.6\) and \(\cos 37° = 0.8\). Find the tension in the rope, in newtons.
A student writes \(N = mg\sin\theta\) for a block resting on a smooth incline of angle \(\theta\), held by a string along the slope. How can the student check this quickly?
Simon Stevin was a Flemish engineer who worked in Holland. He drew a closed chain of equal balls laid over a block with two unequal slopes. If the chain began to slide one way, Stevin argued, it would slide for ever, because the chain looks the same after every step. Perpetual motion is impossible, so the chain stays at rest. The loose part hanging below the block is symmetric and can be removed. So four balls on the long, gentle slope balance two balls on the short, steep slope. From this, Stevin showed that the pull of a weight along a slope is proportional to the height over the length, which is \(\sin\theta\). He printed the drawing on his title page with the words "Wonder en is gheen wonder": a wonder, and yet no wonder.
A load on two strings
Now hang a lamp from a knot, and tie the knot to the ceiling with two strings. The lamp hangs below the knot by a third, short string. Take the knot as the body, because all three strings meet there. The lamp string pulls the knot down with a force equal to the weight of the lamp \(W\). Each ceiling string pulls the knot along the string, up and outward.
The vertical components carry the weight
Let each ceiling string make an angle \(\theta\) with the vertical. The horizontal components of the two tensions point in opposite directions. For a symmetric hang they cancel when the two tensions are equal. The vertical components both point up, and together they must balance \(W\). Each vertical component is \(T\cos\theta\), so \(2T\cos\theta = W\).
A lamp of mass \(4\,\text{kg}\) hangs from a knot. Two strings run from the knot to the ceiling, each at \(60°\) to the vertical. Take \(g = 10\,\text{m/s}^2\). Find the tension in each string.
First the lamp: its string must pull it up with \(40\,\text{N}\), so that string pulls the knot down with \(40\,\text{N}\). Now the body is the knot, with three forces: \(40\,\text{N}\) down, \(T_1\) along the left string and \(T_2\) along the right string.
Horizontal: \(T_2\sin 60° - T_1\sin 60° = 0\). Vertical: \(T_1\cos 60° + T_2\cos 60° - 40 = 0\).
The first equation gives \(T_1 = T_2 = T\). The second gives \(2T \times 0.5 = 40\), so \(T = 40\,\text{N}\).
Each string pulls with \(40\,\text{N}\), as much as the whole weight of the lamp. Spreading the strings to \(60°\) from the vertical doubles the tension from the \(20\,\text{N}\) each string would carry if both hung straight down.
Move the slider in the simulation below. At small angles each string carries a little more than half the weight. At \(60°\) each string carries the whole weight. Beyond that the tension climbs fast, because \(\cos\theta\) falls towards zero. The force triangle shows why: its height stays equal to \(W\), while its two slanting sides lengthen without limit as the triangle flattens.
A load of mass \(6\,\text{kg}\) hangs from a knot held by two equal strings, each at \(30°\) to the vertical. Take \(g = 10\,\text{m/s}^2\). Find the tension in each string, in newtons.
A clothes line is tied tightly between two poles. A wet shirt is hung at the middle. Why can the line never be pulled perfectly straight, however hard it is tightened?
A picture frame hangs at rest from a nail by a wire. The two halves of the wire make an angle with each other at the nail. The frame is re-hung with a longer wire, so that the angle at the nail becomes smaller. What happens to the tension in the wire?
A student of mass \(60\,\text{kg}\) lies in a hammock. The two ropes each make about \(60°\) with the vertical. Roughly what is the tension in each rope, in newtons? Take \(g = 10\,\text{m/s}^2\).
The wires that carry electricity between two pylons always hang in a gentle curve. An engineer could pull a wire almost straight. But the tension would then be enormous, and the wire could snap on a cold night when it contracts. So the wires are hung with a planned sag. The sag gives each part of the wire a small upward component of tension. That component carries the weight of the wire at a safe tension.
- Writing \(N = mg\) for a block on an incline.On an incline the normal force balances only the component of the weight into the slope, \(N = mg\cos\theta\) when nothing else pushes across the slope.
- Swapping sine and cosine for the weight components on an incline.Test with \(\theta = 0\). On a level surface the whole weight presses into the surface, so the component into the slope is \(mg\cos\theta\).
- Assuming that each of two strings carries half the weight.That is true only when both strings hang vertically. Balance the vertical components: \(2T\cos\theta = W\).