Apply Newton's second law to systems of connected bodies, including tension in light inextensible strings over light frictionless pulleys and trains of carts on horizontal and inclined surfaces

What this dot point is asking

VCAA wants you to analyse connected-body systems by writing Newton's second law for each body separately and applying the constraints that ideal strings impose: same tension throughout, same magnitude of acceleration for all bodies connected by a single inextensible string.

Ideal string and pulley assumptions

Light string. Mass-less, so tension is the same throughout. The string is inextensible, so all bodies connected by it share the same magnitude of acceleration.

Light frictionless pulley. Mass-less and rotates without friction, so tension is identical on both sides of the pulley. The pulley redirects the force without changing its magnitude.

These are idealisations; real strings have mass and real pulleys have friction and inertia, but VCE problems use the ideal case unless explicitly stated.

Method for any connected-body system

1. Identify each body separately.
2. Choose a positive direction for each body (typically the direction of motion).
3. Draw a free-body diagram for each.
4. Write $F = ma$ for each body in the chosen direction.
5. Use the constraints (same $T$, same $|a|$) to combine the equations.
6. Solve for the unknowns.

Atwood machine (two hanging masses)

For $m_2 > m_1$ on opposite sides of a pulley:

The tension is between $m_1 g$ (if $m_2 \to \infty$, tension capped at supporting $m_1$) and $m_2 g$ (if $m_1 \to \infty$, capped at supporting $m_2$).

Train of carts on a horizontal surface

If three carts are linked and pulled by a force $F$ at the front:

Treat the system as a whole to find acceleration: $a = F / (m_1 + m_2 + m_3)$.

For the tension between cart $1$ and $2$, isolate cart $1$ and write $F - T_1 = m_1 a$. The tension between cart $2$ and $3$ pulls cart $3$: $T_2 = m_3 a$.

Block on table connected to hanging mass over a pulley

Mass $m_2$ hangs off the table; mass $m_1$ sits on the table, connected over a pulley. If the table is frictionless:

If kinetic friction is present on the table: replace the numerator with $m_2 g - \mu_k m_1 g$.

Common traps

Treating tension as a vector applied twice

Tension is a single number for each segment of an ideal string. Apply it once to each connected body.

Forgetting that magnitudes of acceleration are equal even when directions differ

In an Atwood machine, one body accelerates up while the other accelerates down. The magnitudes are equal.

Confusing tension with weight

The tension is whatever the string supplies. Tension equals weight only if the body is in equilibrium or accelerating purely due to gravity along a different direction.

In one sentence

For connected bodies linked by light inextensible strings over light frictionless pulleys, the tension is the same throughout the string and the magnitudes of acceleration are equal; write Newton's second law for each body separately, apply these constraints, and solve simultaneously for the acceleration and tension.

Examples in context

Example 1. Snowy 2.0 incline-rail haul system. Snowy 2.0's tunnel construction uses an incline rail to haul a $20$ tonne shuttle wagon up a $14^\circ$ gradient via a counterweight. Treating both masses as connected by a light cable over a frictionless sheave, with $m_1 = 20\,000$ kg shuttle on the incline and $m_2 = 15\,000$ kg counterweight in a vertical shaft, Newton's second law gives $m_2 g - m_1 g \sin 14^\circ = (m_1 + m_2) a$. So $a = (15\,000 \times 9.8 - 20\,000 \times 9.8 \times 0.242)/35\,000 = (147\,000 - 47\,400)/35\,000 = 2.85$ m s$^{-2}$ upward for the shuttle. Cable tension $T = m_2 (g - a) = 15\,000 \times (9.8 - 2.85) = 1.04 \times 10^5$ N.

Example 2. Eureka Tower window-cleaner gondola counterbalance. Eureka Tower's window-cleaning gondola uses a counterweighted cable system. A $400$ kg gondola is connected via a single cable to a $380$ kg counterweight; the difference drives lifting motors that supplement gravity. Treating as Atwood-type system, the natural acceleration without motor input is $a = (m_1 - m_2)g/(m_1 + m_2) = 20 \times 9.8/780 = 0.25$ m s$^{-2}$ downward for the heavier gondola. The motor adds force to control descent at $0.3$ m s$^{-1}$ constant velocity, where tension $T = m_2 g = 380 \times 9.8 = 3724$ N in the cable counterbalances most of the load.

Try this

Q1. State the assumptions made when analysing light inextensible strings over light frictionless pulleys. [2 marks]

• Cue. Massless string and pulley, no friction, tension is the same throughout the string.

Q2. Two masses, $4$ kg and $6$ kg, hang from opposite ends of a light cord over a frictionless pulley. Calculate (a) the acceleration of the system, and (b) the tension in the cord. [4 marks]

• Cue. (a) $a = (6-4)g/(6+4) = 1.96$ m s$^{-2}$. (b) $T = 4(g+a) = 4 \times 11.76 = 47.0$ N.

Q3. Refer to the Snowy 2.0 incline-rail example. (a) Set up Newton's second law equations for both masses. (b) Calculate the acceleration. (c) Determine the cable tension. [2+3+2 marks]

• Cue. (a) $m_2 g - T = m_2 a$ for counterweight, $T - m_1 g \sin\theta = m_1 a$ for shuttle. (b) $2.85$ m s$^{-2}$. (c) $T = 1.04 \times 10^5$ N.