Trains of carts, a block pulled by a hanging weight, two boxes pushed together: the objects move together, so they share one acceleration. The trick is choosing the right system for each question, then using Newton's Second Law Along an Axis.
Key idea
To find the shared acceleration, treat all the linked objects as one system: the forces between them are internal and cancel. To find a rope tension or a contact force, take one object alone: then that force is external and appears in its equation.
Why it works
Each choice of system leaves out different forces. The whole train hides the ropes between the cars, so its equation has only the outside pull and the total mass. A single car shows its rope as an ordinary force. Both equations are true at once, and using them in that order gives each unknown with the least algebra.
Worked examples
Worked example: Two carts and a rope
On a frictionless track you pull a $3\ \text{kg}$ cart with $20\ \text{N}$; a rope links it to a $2\ \text{kg}$ cart behind. Find $a$ and the tension.
- Whole system: $a = \dfrac{20}{3 + 2} = 4\ \text{m/s}^2$.
- Rear cart alone: the rope is its only horizontal force, so $T = (2)(4) = 8\ \text{N}$.
Answer: $a = 4\ \text{m/s}^2$, $T = 8\ \text{N}$
Worked example: A hanging weight
A $4\ \text{kg}$ block on a frictionless table is tied over a light pulley to a hanging $1\ \text{kg}$ block. Find $a$ and $T$.
- System along the string: the only unbalanced force is the hanging weight, $9.8\ \text{N}$, moving $5\ \text{kg}$ in all.
- $a = \dfrac{9.8}{5} = 1.96\ \text{m/s}^2$.
- Table block alone: $T = (4)(1.96) = 7.84\ \text{N}$, less than the hanging block's $9.8\ \text{N}$ weight.
Answer: $a = 1.96\ \text{m/s}^2$, $T = 7.84\ \text{N}$
Common mistake
Thinking a rope's tension equals the applied force, or equals the hanging weight. The rope only has to accelerate what is behind it, so its tension is a fraction of the pull; and if a hanging block is accelerating downward, the rope must pull up on it with less than its weight. Check: compare your tension with the forces on each side of the rope; it should fall between them.