At a vegetable stall, the seller hooks a bag of potatoes on a spring balance. The balance reads \(2\,\text{kg}\). Now hook a second balance below the first, and hang the bag from the second. Many students expect each balance to read \(1\,\text{kg}\), sharing the load. Both balances read \(2\,\text{kg}\). Where does the extra reading come from?
One force all along a light string
Cut a light string at any point and look at a short piece. The piece has no mass, so its weight is zero. Only two forces act on the piece: the pull of the string above and the pull of the string below. The piece is in equilibrium, so the two pulls are equal. Every cut gives the same result, so the tension is the same at every point of a light string.
A light spring does the same
A light spring behaves in the same way. The spring is pulled at both ends by equal forces, and the size of either force is the tension in the spring. A spring balance is a light spring with a scale, and it reads that tension. Hook a balance to a wall and pull the free end with \(10\,\text{N}\): the wall pulls back with \(10\,\text{N}\), and the balance reads \(10\,\text{N}\). The balance reads one end's force, not the sum of the two.
The trick in Fig 6.1 is the choice of the cut. Each cut draws a line across one spring, and the body below the line is then a free body. The only upward force on that body is the pull of the spring just above the cut. The weight below the line is the same \(20\,\text{N}\) each time, because light balances add no weight. So every balance in the chain reads the full load.
Two students pull the two ends of a light spring balance in opposite directions, each with a force of \(10\,\text{N}\). The balance is at rest. What does it read?
Springs in series stretch more
Now let the springs have different spring constants \(k_1\) and \(k_2\), joined end to end, which is called springs in series. The same force \(F\) runs through both springs. Each spring stretches by its own amount: \(x_1 = F/k_1\) and \(x_2 = F/k_2\). The total stretch is the sum of the two. So a chain of springs is softer than either spring alone.
\[ F = k_1 x_1 = k_2 x_2 \]
Why. both springs are light and in series, so the same force \(F\) acts through each one. Each spring obeys Hooke's law with its own constant.
\[ x = x_1 + x_2 = \dfrac{F}{k_1} + \dfrac{F}{k_2} \]
Why. the lower end moves by the stretch of the upper spring plus the stretch of the lower spring.
\[ \dfrac{F}{k} = \dfrac{F}{k_1} + \dfrac{F}{k_2} \]
Why. the single spring that would replace the chain has constant \(k\), and it must stretch by the same \(x\) under the same \(F\), so \(x = F/k\).
\[ \dfrac{1}{k} = \dfrac{1}{k_1} + \dfrac{1}{k_2} \]
Read it. divide every term by \(F\). The reciprocals of the spring constants add.
Two light springs with constants \(k_1 = 200\,\text{N/m}\) and \(k_2 = 300\,\text{N/m}\) hang in series from a hook. A block of mass \(2\,\text{kg}\) hangs at rest from the lower spring. Take \(g = 10\,\text{m/s}^2\). Find the stretch of each spring and the constant of the equivalent single spring.
Cut above the block: the lower spring pulls up with \(F = 20\,\text{N}\). Cut above the lower spring: the upper spring also pulls with \(20\,\text{N}\), because the lower spring is light.
The stretches are \(x_1 = 20/200 = 0.10\,\text{m}\) and \(x_2 = 20/300 = 0.067\,\text{m}\). The total is \(0.167\,\text{m}\).
The equivalent constant is \(1/k = 1/200 + 1/300 = 5/600\), so \(k = 120\,\text{N/m}\). Check: \(20/120 = 0.167\,\text{m}\), the same total stretch.
Two identical light springs, each with \(k = 100\,\text{N/m}\), are joined in series and hung from a hook. A load of weight \(10\,\text{N}\) hangs at rest from the lower end. Find the total stretch of the chain, in centimetres.
Robert Hooke was curator of experiments at the Royal Society in London. He found that the stretch of a spring grows in step with the load on it. In 1676 he hid the discovery in an anagram, ceiiinosssttuv, so that no one could claim it before he was ready. Two years later he printed the answer: "ut tensio, sic vis", as the extension, so the force. Hooke used the law to design springs for watches. Every spring balance at a vegetable stall still works on his rule.
Three errors that come back
Mark enough answer sheets and the same three errors appear again and again. The first is a force that belongs to another body, such as the push of the block on the incline drawn on the block. The second is a missing normal force, usually where a body leans on a wall or sits on a slope. The third is a tension drawn in the wrong direction: pointing into the body, or not along the string. Each error changes the answer, and each can be caught before any arithmetic.
How each error is caught
Every error leaves a trace. A force that belongs to another body is exerted by the chosen body. So the question "who exerts this force on my body?" has no answer. A missing force leaves the force polygon open, because the remaining arrows cannot close. A tension in the wrong direction fails the rule that a string pulls away from the body, towards the point where the string is tied. Ask the three questions for every diagram, until the questions become a habit.
A string is like a hand that can only grip and pull, never push. Wherever a string touches a body, picture the hand pulling the body towards the far end of the string. If your arrow points the other way, the hand would be pushing, and a string cannot push.
A bucket hangs at rest from a rope tied to a hook in a beam. A student draws three forces on the bucket: its weight, the tension of the rope, and the pull of the hook on the rope. Which is the error?
A ball hangs from a string that is tied to a vertical wall above it, and the ball rests against the smooth wall. The string makes an angle with the wall. A student draws only the weight and the tension. How does the force polygon reveal the error?
A block rests on a smooth incline, held by a string along the slope. Which of these arrows in a student's free-body diagram of the block are errors?
Explain how drawing the force polygon can show that a force is missing from a free-body diagram, even before any calculation.
For a body in equilibrium the forces, drawn head to tail, must close. If one force is missing, the remaining arrows leave a gap. The gap is the arrow that the missing force should fill, in size and in direction. So an open polygon signals the error and also shows where to look for the missing contact, for example a wall or a slope.
The hook of a construction crane often carries a load cell, an electronic spring balance, between the hook and the load. The load cell reads the same force just above the load or higher up the cable. The reason is that the cable is light compared with the load. Crane operators rely on this: one reading anywhere in the chain gives the whole load, and a warning sounds before the crane is overloaded.
- Splitting the load between two balances hung in series.Cut the chain just above each balance. The body below always has the full weight, so each light balance reads the full load.
- Reading twice the force on a balance pulled from both ends.A balance reads the tension in its spring, the force at one end. Equal pulls at both ends are what equilibrium requires.
- Drawing a tension pointing into the body.A string can only pull. The tension on the body points along the string, away from the body.