Two masses over one pulley
Hang a light string over a smooth, light pulley fixed to the ceiling. Tie a block of mass \(m_1\) to one end and a block of mass \(m_2\) to the other, with \(m_1 > m_2\). This is the Atwood machine. Let it go. The heavier block falls, the lighter block rises, and both have the same size of acceleration \(a\), because the string cannot stretch.
Take the direction of motion as positive for each block: down for \(m_1\), up for \(m_2\). The string pulls each block upward with the same tension \(T\).
\[ m_1 g - T = m_1 a, \qquad T - m_2 g = m_2 a \]
\[ m_1 g - T = m_1 a, \qquad T - m_2 g = m_2 a \]
Why. the second law for each block, each along its own direction of motion.
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Your turn. which line follows? Look for the step that removes the unknown tension.
\[ a = \dfrac{(m_1 - m_2)\,g}{m_1 + m_2} \]
Read it. divide by the total mass. The difference of the weights drives the total mass.
Put the acceleration back into either equation to find the tension.
\[ T = m_2(g + a) = \dfrac{2 m_1 m_2\, g}{m_1 + m_2} \]
The tension lies between the two weights. It is more than \(m_2 g\), because it must accelerate \(m_2\) upward. It is less than \(m_1 g\), because \(m_1\) is accelerating downward.
Use the sim to test the formula at its limits. With equal masses, \(a = 0\) and \(T = mg\): the blocks balance, like a pair of scales. With \(m_2\) very small, \(a\) approaches \(g\) and \(T\) approaches \(2m_2 g\): the heavy block almost falls freely. Every acceleration of an Atwood machine lies between \(0\) and \(g\).
George Atwood taught at Trinity College, Cambridge. In 1784 he described a machine to test Newton's laws of motion in the lecture room. A falling stone moves too fast to time with a pendulum clock. Atwood hung two nearly equal masses over a light, carefully made pulley. Only the small difference of their weights drove the pair, so the motion was slow enough to follow by eye and to time. By adding small riders to one side he could change the driving force and watch the acceleration change in step. The machine did for Newton's second law what Galileo's slope had done for free fall.
Blocks of \(6\,\text{kg}\) and \(4\,\text{kg}\) hang over a smooth light pulley. Take \(g = 10\,\text{m/s}^2\). Find the tension in the string in newtons.
In an Atwood machine with unequal masses, the tension in the string is
The string constraint
So far the claim "both blocks have the same size of acceleration" came from common sense. Now make it exact. A string that cannot stretch is an inextensible string. Its total length \(L\) never changes. Write \(L\) in terms of the positions of the bodies it is tied to. Then any motion of one body must be matched by the others, so that \(L\) stays the same. This relation is a constraint.
For the Atwood machine, measure the depth of each block below the pulley: \(y_1\) and \(y_2\). The string runs from one block up to the pulley, over it, and down to the other block.
\[ y_1 + y_2 + (\text{the part on the pulley}) = L \]
The part on the pulley does not change. So when \(y_1\) grows, \(y_2\) must shrink by the same amount. Differentiate twice with respect to time: \(a_1 + a_2 = 0\). One block goes down exactly as fast as the other goes up.
Measure the position of each body from a fixed point, along the string. Write the length of each straight part of the string in terms of these positions, and set their sum equal to a constant. Differentiating twice with respect to time gives the relation between the accelerations. It holds for any light, inextensible string that stays taut.
In an Atwood machine, the heavier block moves down \(30\,\text{cm}\). What does the lighter block do?
A movable pulley
A movable pulley hangs in a loop of the string instead of being fixed to the ceiling. In the arrangement below, one end of the string is tied to the ceiling. The string runs down, under the movable pulley, up over a fixed pulley, and down to a block \(B\). A block \(A\) hangs from the movable pulley.
Measure the depth of the movable pulley below the ceiling, \(y_A\), and the depth of block \(B\) below the fixed pulley, \(y_B\). Two strands of the string run from the ceiling level down to the movable pulley. Each has length \(y_A\), apart from constant bits. The third strand has length \(y_B\).
\[ 2y_A + y_B = \text{constant} \quad\Rightarrow\quad 2a_A + a_B = 0 \]
If \(B\) falls by \(x\), each of the two strands holding \(A\) must shorten by \(x/2\), so \(A\) rises by \(x/2\). Block \(B\) moves twice as far as \(A\), with twice its velocity and twice its acceleration.
The forces change too. The light movable pulley hangs on two strands, each pulling with \(T\). So the string lifts the pulley, and block \(A\) with it, with a force \(2T\). A movable pulley doubles the force and halves the motion.
In the arrangement of the figure, \(m_A = 4\,\text{kg}\) and \(m_B = 3\,\text{kg}\). The pulleys and the string are light and smooth. Take \(g = 10\,\text{m/s}^2\). Find the accelerations of both blocks and the tension.
Guess that \(B\) falls. Then \(A\) rises, with \(a_B = 2a\) and \(a_A = a\). \(B\) is pulled up by \(T\); \(A\) and its pulley are pulled up by \(2T\).
\(B\): \(30 - T = 3(2a)\). \(A\): \(2T - 40 = 4a\).
From \(B\), \(T = 30 - 6a\). Then \(60 - 12a - 40 = 4a\), so \(16a = 20\) and \(a = 1.25\,\text{m/s}^2\). So \(a_A = 1.25\,\text{m/s}^2\) up, \(a_B = 2.5\,\text{m/s}^2\) down, and \(T = 30 - 7.5 = 22.5\,\text{N}\).
\(B\) falls at \(2.5\,\text{m/s}^2\) and \(A\) rises at \(1.25\,\text{m/s}^2\), with a tension of \(22.5\,\text{N}\). The lighter block wins, because the pulley doubles its pull.
In a movable-pulley system like the one in the figure, the block on the free end moves down \(60\,\text{cm}\). How far does the block on the movable pulley rise, in centimetres?
In the movable-pulley arrangement, \(m_A\) hangs from the movable pulley and \(m_B\) from the free end. For which masses does nothing move?
Why can the acceleration of an Atwood machine never exceed \(g\)?
The driving force is the difference of the weights, \((m_1 - m_2)g\), and it must accelerate both masses, \(m_1 + m_2\). The difference of two positive masses is always smaller than their sum, so \(a = g(m_1 - m_2)/(m_1 + m_2)\) is always less than \(g\). It approaches \(g\) only when \(m_2\) is negligible and the heavy block falls almost freely.
- Taking the tension in an Atwood machine to be the weight of one of the blocks.The tension lies strictly between the two weights: \(T = 2m_1m_2g/(m_1 + m_2)\).
- Giving both blocks the same acceleration when one hangs on a movable pulley.Write the string length, \(2y_A + y_B = \text{constant}\). The block on the free end has twice the acceleration.
- Forgetting that a movable pulley is held by two strands.A light movable pulley feels \(2T\) upward, one \(T\) from each strand.