R4 - Newton's third law and connected particles

Third-law pairs on different bodies

If a horse pulls a cart and the cart pulls back equally, how can the cart accelerate? Start by asking which body each force acts on. The horse's pull acts on the cart; the cart's pull acts on the horse. Only the first belongs in the cart's own resultant.

A stationary book on a table gives two different interactions to keep apart. The table pushes the book upwards, and the book pushes the table downwards. The Earth attracts the book downwards, and the book attracts the Earth upwards. These are two pairs.

The book's upward reaction and downward weight can balance, but they are not a third-law pair: both act on the book and arise from different interactions. Third-law partners have matching subject and object reversed. “The force of the table on the book” pairs with “the force of the book on the table”.

The law also applies during acceleration. If a heavy lorry and a light car exert contact forces on each other, the force magnitudes are equal. Their accelerations need not have equal magnitudes: Newton's second law divides each resultant by a different mass. Nor does third law say each body's total resultant is zero.

A string pulling two bodies needs the same care. The force of the string on $A$ pairs with the force of $A$ on the string. The string's force on $B$ has its own partner. The two forces of the string on the two bodies are not themselves a third-law pair, even when an ideal-string model makes their magnitudes equal.

Return to the horse and cart: an equal opposite pull on the horse does not balance the forward pull on the cart. For the cart, compare its forward pull with the other forces actually acting on the cart, such as resistance.

Ideal strings and pulleys

Two masses hanging over a fixed pulley can have equal acceleration magnitudes while their acceleration vectors point in opposite directions. The string links distances along its path, not one common vertical direction.

Suppose $A$ rises $0.10\,\mathrm m$ while a straight, taut string passes over a fixed pulley. Its vertical section shortens by $0.10\,\mathrm m$; the section on the other side must lengthen by $0.10\,\mathrm m$, so $B$ descends by that distance. Applying this length constraint throughout a smooth stage of motion links the speeds and acceleration magnitudes too.

Each assumption has a particular job. Do not use “inextensible” as a reason for common tension.

For two vertical ends, let $y_A$ and $y_B$ measure their downward distances from the pulley. Fixed total length gives $y_A+y_B=c$, where $c$ is constant. Differentiating gives $v_A+v_B=0$, then $a_A+a_B=0$. Their signed downward accelerations are opposite. These constraint relationships are derived from the model and are not in the formula booklet: learn how to obtain them.

You may instead take upwards positive for $A$ and downwards positive for $B$, then use one scalar $a$ in their separate equations. A negative answer means the actual accelerations oppose those chosen positive directions; it does not make a tension negative automatically.

The ideal string pulls each end towards itself. Its tension must be nonnegative. When it becomes slack, its tension is zero and the previous linked-acceleration condition no longer applies. A negative tension from a proposed taut-string solution means the string cannot enforce that proposed motion by pushing.

Use the assumptions to explain a stated numerical model, rather than repeating the words without their consequences.

Separate equations for connected bodies

Choose a body, draw only its forces, and apply its own resultant equation. Repeat for the other body before eliminating an unknown. This keeps an internal connector force distinct from a driving force or a weight.

A car and trailer moving along one straight horizontal line can share the same acceleration vector. In that case, adding their force equations also produces a legitimate whole-system equation: the internal rope forces cancel. Different-direction pulley motion needs the separate equations first.

Worked example 2 changes the geometry of the connection: the particles move on opposite sides of a pulley. Use the same separate-equation method, but choose a convenient positive direction for each body.