On a village road a tractor tows a loaded trolley with a rope. As the tractor speeds up, the rope is pulled tight. The tractor pulls one end of the rope forward. The rope pulls the trolley forward at the other end. How hard is the rope pulling at each end, and is it the same pull as the one the engine gives the tractor?
What a string does
A string, a rope, a cable or a chain can only pull. It pulls along its own length, away from the body it is tied to. The size of that pull is the tension \(T\). A string that is slack has no tension at all. You met tension as one of the everyday forces in Forces and Newton's Laws. Now two bodies share one string, and the question is whether the pull is the same at both ends.
Why a light string has one tension
Take a short piece of string that is accelerating with the bodies, with acceleration \(a\). Let its mass be \(m_s\). The part of the string ahead of it pulls it forward with \(T_1\). The part behind pulls it back with \(T_2\). Apply the second law to the piece alone.
\[ T_1 - T_2 = m_s\,a \]
Why. the only horizontal forces on the piece are the pulls of its neighbours. Their difference accelerates the piece's own mass.
\[ m_s \to 0 \]
Why. a light string is one whose mass is negligible compared with the bodies it joins.
\[ T_1 - T_2 = 0, \quad T_1 = T_2 \]
Read it. with no mass, the piece needs no net force to accelerate. The two pulls on it must be equal. This holds for every piece, so the tension is the same all along the string.
The result needs a light string, not a string at rest. A heavy rope that accelerates has a larger tension at its front end than at its back end. The rope's own mass must be accelerated too. The problems in this chapter use light strings, so one string carries one tension.
A smooth, light pulley turns a string without changing its tension. The pulley has no mass and no friction at its axle, so the string pulls with the same \(T\) on both sides. A pulley changes the direction of the pull, not its size.
A light string joins two blocks that are speeding up together. Which statement about the tension is correct?
Two blocks pulled across a table
Two blocks lie on a smooth table, joined by a light string. The rear block has mass \(m_1\) and the front block \(m_2\). A force \(F\) pulls the front block forward. The string is taut and cannot stretch, so the two blocks move together with one acceleration \(a\).
There are two ways to find \(a\). The first treats the two blocks and the string as one body of mass \(m_1 + m_2\). The tension is then an internal force. The string pulls the rear block forward and the front block back with the same \(T\), and the two pulls cancel in the sum. Only \(F\) is left.
\[ F = (m_1 + m_2)\,a \quad\Rightarrow\quad a = \dfrac{F}{m_1 + m_2} \]
The second way finds the tension. Isolate the rear block. The only horizontal force on it is the tension, pulling forward. That tension must give the rear block the common acceleration.
\[ T = m_1 a = \dfrac{m_1 F}{m_1 + m_2} \]
Check the front block too. It has \(F\) forward and \(T\) backward, so \(F - T = m_2 a\). With the numbers of the figure, \(30 - 10 = 4 \times 5\). The equation holds. The front block feels a smaller net force than \(F\), because part of \(F\) is passed back through the string to the rear block.
This answers the tractor's question. The rope pulls the trolley forward with just enough force to give the trolley the common acceleration. The engine's push, passed through the road to the tractor, must accelerate both the tractor and the trolley. So the rope's pull is smaller than the driving force.
Blocks of \(3\,\text{kg}\) and \(7\,\text{kg}\) lie on a smooth table, joined by a light string. A force of \(50\,\text{N}\) pulls the \(7\,\text{kg}\) block. Find the tension in the string in newtons.
The same force now pulls the \(3\,\text{kg}\) block instead, and the \(7\,\text{kg}\) block trails behind. What happens to the tension?
A block on a table and a hanging mass
A block of mass \(M\) lies on a smooth table. A light string runs from it to a smooth pulley at the edge of the table and down to a hanging block of mass \(m\). When the system is let go, the hanging block falls and drags the table block along. The string is taut, so both blocks have the same size of acceleration \(a\). One moves horizontally and the other vertically; the pulley has only turned the direction.
Write the second law for each block along its own direction of motion. For the table block, forward is towards the pulley. For the hanging block, forward is downward.
\[ \text{table block: } T = Ma, \qquad \text{hanging block: } mg - T = ma \]
Add the two equations. The tension cancels, and the acceleration follows.
\[ mg = (M + m)\,a \quad\Rightarrow\quad a = \dfrac{mg}{M + m}, \qquad T = Ma = \dfrac{Mmg}{M + m} \]
The tension is less than the weight of the hanging block. It must be. If the string pulled up with the full \(mg\), the hanging block would be in balance and would not fall. The hanging weight has two jobs. Part of it accelerates the hanging block itself, and the rest passes through the string to the table block.
A \(3\,\text{kg}\) block on a smooth table is joined over a smooth pulley to a hanging \(2\,\text{kg}\) block. The system is released from rest. Take \(g = 10\,\text{m/s}^2\). Find the acceleration, the tension, and the speed after the hanging block has fallen \(0.5\,\text{m}\).
Both blocks move with the same size of acceleration \(a\). Forward for the table block is towards the pulley; forward for the hanging block is down.
Table block: \(T = 3a\). Hanging block: \(20 - T = 2a\).
Adding: \(20 = 5a\), so \(a = 4\,\text{m/s}^2\), and \(T = 3 \times 4 = 12\,\text{N}\). Then \(v^2 = 2as = 2 \times 4 \times 0.5 = 4\), so \(v = 2\,\text{m/s}\).
The blocks accelerate at \(4\,\text{m/s}^2\) with a tension of \(12\,\text{N}\), and move at \(2\,\text{m/s}\) after \(0.5\,\text{m}\).
A block on a smooth table is pulled by a hanging mass \(m\) over a pulley at the edge. While the system accelerates, the tension in the string is
A \(4\,\text{kg}\) block on a smooth table is pulled by a hanging block over an edge pulley. Rank the hanging masses by the tension they produce, smallest first. Take \(g = 10\,\text{m/s}^2\).
When two blocks joined by a string are treated as one body, why does the tension not appear in the equation?
The tension is an internal force of the combined body. The string pulls the rear block forward with \(T\) and pulls the front block backward with the same \(T\). When the forces on the whole system are added, these two pulls cancel. Only forces from outside the system, such as the applied force, can change the motion of the system as a whole. To find \(T\), a single block must be isolated, so that the tension becomes an external force on it.
A goods train has one locomotive in front and fifty or more wagons behind it, linked by couplers. When the train starts, the coupler just behind the locomotive must accelerate every wagon, so it carries the largest pull. The last coupler pulls only the last wagon and carries the smallest. This is the two-block result with many blocks. Very long goods trains are often given a second locomotive in the middle, so that no single coupler has to pull the whole train.
- Taking the tension to be equal to the hanging weight \(mg\).The hanging block accelerates, so \(mg - T = ma\) and \(T < mg\). \(T = mg\) only when the system is at rest or moves steadily.
- Using the applied force \(F\) as the force on the rear block.The rear block feels only the tension. Find \(a\) for the whole system, then \(T = m_1 a\).
- Putting \(T\) into the equation for the combined system.For the system as one body the tension is internal and cancels. It appears only when a single block is isolated.