Strings, pulleys and constraint relations

One pull, two speeds

Pull a rope through a pulley that hangs from the ceiling, and a heavy box rises at the same speed as your hand. Hang the box from a second pulley that can move, and something changes. The box now rises at half the speed of your hand. Your pull on the rope is also half what you needed before.

Where does the number $2$ come from? Free-body diagrams alone do not tell us. We need one more idea: a string that does not stretch has a fixed length. Let us see how to use it.

Three steps for every connected system

1. Draw a free-body diagram for every body, and for every pulley that moves.
2. Write Newton's second law for each body along the direction in which it can move.
3. Write the constraint. This is the relation between the accelerations that the string forces.

Without step 3, we meet more unknowns than equations. For two blocks on a string over a fixed pulley, we have two accelerations $a_1$ and $a_2$ and one tension $T$ (three unknowns), but only two Newton equations. The constraint is the third equation. For the fixed pulley, it is simply that when one block goes down, the other goes up by the same amount.

The constraint from the length of the string

A string that stays tight and does not stretch has a constant length $L$. We measure the length of each straight part from a fixed point and add them up.

• Over a fixed pulley. The two hanging parts have lengths $x_1$ and $x_2$ below the pulley, so $x_1 + x_2 = L$. Differentiate: $v_1 + v_2 = 0$. Differentiate again: $a_1 = -a_2$. With "down" as positive for both blocks, one speeds up downward as the other speeds up upward.
• Round a movable pulley. Look at the picture. Block A is on a smooth table. Its string goes over a pulley at the edge and down round a movable pulley that carries the load B. Then it goes back up to a fixed point under the table. The length of string between A and the edge is $x$. Each of the two parts that hold the movable pulley has a length $y$.

The total length is $x + 2y = L$. Differentiate once: $v_A + 2v_B = 0$ with signs for the directions we chose. Differentiate again, and take sizes:

A moves twice as far, twice as fast and with twice the acceleration as B. This is the same number $2$ that we found for the pulley and the box: two parts of string hold the movable pulley, so the free end moves twice as far as the pulley. In general, suppose $k$ parts of the same string hold a movable pulley. Then the pulley moves $1/k$ as far as the free end.

Massless pulleys and the tension

In this lesson, every pulley is smooth and has no mass. Two facts follow.

• A string over a fixed smooth pulley has the same tension on both sides. The pulley turns freely, so nothing makes the tension change as the string passes over it.
• A movable pulley with no mass has a net force of zero on it. With no mass, it cannot have an acceleration worth a force. In the picture, two parts of string pull the movable pulley up, each with tension $T$, and the load pulls it down. So the pulley and the load can be treated as one body that is pulled up by $2T$.

A pulley that has mass needs the torque equation, which is in the chapter on rotational motion. We leave it out here. The strings are always light and do not stretch.

A worked example: the movable pulley

Block A of mass $4.0\ \text{kg}$ is on a smooth table. A string goes from A over a pulley at the edge and down round a movable pulley that carries a load B of mass $2.0\ \text{kg}$. It then goes up to a fixed point under the table, as in the picture. Find the accelerations of A and B and the tension in the string. Take $g = 9.8\ \text{m/s}^2$.

1. The constraint. From the string, $a_A = 2a_B$.
2. Block A. Only the string pulls it along the table, so $T = m_A a_A = 4.0\ \text{kg} \times a_A$.
3. Load B with its pulley. The string pulls up with $2T$, and the weight pulls down. Taking down as positive, $m_B g - 2T = m_B a_B$, which is $19.6\ \text{N} - 2T = 2.0\ \text{kg} \times a_B$.
4. Put the three together. From step 2 and step 1, $T = 4.0\ \text{kg} \times 2a_B = 8.0\ \text{kg} \times a_B$. In step 3, $19.6\ \text{N} - 16.0\ \text{kg} \times a_B = 2.0\ \text{kg} \times a_B$. So $18.0\ \text{kg} \times a_B = 19.6\ \text{N}$, and

5. The other answers. $a_A = 2a_B = 2.178\ \text{m/s}^2$ and $T = m_A a_A = 4.0\ \text{kg} \times 2.178\ \text{m/s}^2 = 8.711\ \text{N}$.

Check. The net downward force on B is $19.6\ \text{N} - 2 \times 8.711\ \text{N} = 2.178\ \text{N}$. This equals $m_B a_B = 2.0\ \text{kg} \times 1.089\ \text{m/s}^2$. It fits. Also see that $2T = 17.42\ \text{N}$ is smaller than the weight $19.6\ \text{N}$. B is not at rest, so the string does not hold its full weight.

When the load is much heavier than A, the string hardly holds it back. Its acceleration comes close to $g$, and A, with twice that, comes close to $2g$. A block can accelerate faster than $g$ along a table, because the string pulls it sideways with a tension that is larger than its own weight.

A rope at an angle: only the part along the rope counts

A string that is tight and does not stretch has one more property. The two ends of the string have equal velocity components along the string. If they did not, the string between them would stretch or go slack.

Here is the standard case. A trolley runs on a smooth floor. A rope from it goes up to a small pulley at a height $h$ above the floor, directly above the point $O$ on the floor. A person pulls the free end of the rope with a constant speed $u$. The trolley is at a distance $x$ from $O$, and the rope has length $s$ between the trolley and the pulley.

The lengths obey $s^2 = x^2 + h^2$, with $h$ fixed. Differentiate with time: $s\,\dfrac{ds}{dt} = x\,\dfrac{dx}{dt}$. The rope shortens at the rate $u$, so $\left|\dfrac{ds}{dt}\right| = u$, and the trolley moves at $v = \left|\dfrac{dx}{dt}\right|$. Then

where $\theta$ is the angle between the rope and the floor. This is the same as saying that the component of the trolley's velocity along the rope, $v\cos\theta$, equals the rope speed $u$.

The trolley is faster than the rope. It also speeds up as it comes closer, even though the person pulls at a constant speed. Differentiating once more, with $u$ constant, gives the size of its acceleration:

An example. Take $u = 2.0\ \text{m/s}$, $h = 1.8\ \text{m}$ and $x = 2.4\ \text{m}$. Then $s = \sqrt{2.4^2 + 1.8^2}\ \text{m} = 3.0\ \text{m}$ and $\cos\theta = \dfrac{2.4\ \text{m}}{3.0\ \text{m}} = 0.80$. The trolley moves at $v = \dfrac{2.0\ \text{m/s}}{0.80} = 2.5\ \text{m/s}$. Its acceleration is $a = \dfrac{(2.0\ \text{m/s})^2 \times (1.8\ \text{m})^2}{(2.4\ \text{m})^3} = 0.9375\ \text{m/s}^2$.

The same idea works for a rigid rod. For any two points of a rigid body, the velocity components along the line that joins them are equal.

Here is an example. A ladder of length $5.0\ \text{m}$ stands with its foot $3.0\ \text{m}$ from the wall and its top $4.0\ \text{m}$ up. The foot is at distance $x$ from the wall and the top at height $y$, with $x^2 + y^2 = 25\ \text{m}^2$. So $x\dot{x} + y\dot{y} = 0$. If the top slides down at $1.5\ \text{m/s}$, the foot moves out at $\dfrac{y}{x} \times 1.5\ \text{m/s} = \dfrac{4.0}{3.0} \times 1.5\ \text{m/s} = 2.0\ \text{m/s}$.

What does the support carry?

A pulley is held by a support, and we often want the force on the support. The rule is that the external force on a system equals the sum of $m\vec{a}$ of its parts. The weight is not the whole story when parts accelerate.

An example. An Atwood machine has $m_1 = 3.0\ \text{kg}$ and $m_2 = 5.0\ \text{kg}$ on the two sides of a light pulley that hangs from a hook. The common acceleration is $a = \dfrac{(m_2 - m_1)g}{m_1 + m_2} = \dfrac{2.0\ \text{kg} \times 9.8\ \text{m/s}^2}{8.0\ \text{kg}} = 2.45\ \text{m/s}^2$. The tension is $T = \dfrac{2m_1m_2g}{m_1 + m_2} = \dfrac{2 \times 3.0\ \text{kg} \times 5.0\ \text{kg} \times 9.8\ \text{m/s}^2}{8.0\ \text{kg}} = 36.75\ \text{N}$. The hook carries $2T = 73.5\ \text{N}$.

The total weight is $8.0\ \text{kg} \times 9.8\ \text{m/s}^2 = 78.4\ \text{N}$. The hook carries $4.9\ \text{N}$ less. This is $(m_2 - m_1)a = 2.0\ \text{kg} \times 2.45\ \text{m/s}^2 = 4.9\ \text{N}$, which is what the heavier block gains over the lighter one in downward $ma$.

Practise with new numbers

Try it in Python

When a system has several bodies, writing the equations as a table of numbers is neater. The program below writes the movable pulley system as three equations in three unknowns, and numpy solves them in one step. Each row of the table holds the numbers in front of $a_A$, $a_B$ and $T$.

Common mistakes

• Giving every body the same acceleration. That is true only for a string over a fixed pulley, and then the directions are opposite. A movable pulley changes the sizes.
• Forgetting that two parts of string hold the movable pulley. The pulley is pulled by $2T$, not by $T$.
• Using $2T = m_B g$ for a load that accelerates. That holds only for equilibrium.
• Using the angle of the rope as if the trolley moved along it. The trolley moves along the floor. Only its component along the rope equals the rope speed, so $v = u/\cos\theta$, not $u\cos\theta$.
• Taking the support force to be the total weight. If the parts accelerate, the support carries $\sum m g - \sum m a$ with the right signs.
• Counting unknowns too late. Before you solve, count the equations and the unknowns. If you are short by one, you have missed a constraint.

Remember these

• Write the string length as a constant, then differentiate twice. This gives the constraint between the accelerations.
• A fixed pulley gives $a_1 = -a_2$. A movable pulley held by two parts of one string gives a factor of $2$.
• A light pulley: the same tension on both sides if it is fixed, and a net force of zero if it is movable.
• The velocity components along a tight string, or along a rigid rod, are equal at the two ends.
• The force on a support is the weight minus the sum of $m a$, with signs.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A string goes from a ceiling hook down round a movable pulley and up over a fixed pulley to a block. The movable pulley carries a load. The block moves down by $60\ \text{cm}$. How far does the load move, and in which direction?

2. In the worked example, the table block A is replaced by a block of twice the mass. Does the acceleration of B become larger or smaller, and by how much?

3. A trolley is pulled by a rope over a pulley, and the person pulls the rope at a constant speed. Is the acceleration of the trolley zero? Give the reason.

4. A man stands on a weighing scale in a lift. In which direction is the net force on the whole system of lift, man and scale if the lift speeds up upward, and why does the cable carry more than the weight?

References

• OpenStax, University Physics Volume 1, chapters 5 and 6 (Newton's laws of motion and their applications).
• NCERT, Physics Part I for Class 11, the chapter Laws of Motion.