Friction and Tension in Connected Systems
In a connected system, objects linked by ropes or strings share the same acceleration magnitude, so Newton’s second law is applied to each object (or to the whole system) to find the acceleration and the tension; for an ideal rope over a frictionless pulley the tension is the same throughout. In brief: for two blocks, one on a rough table (mass m1) and one hanging (mass m2), the acceleration is a = (m2g − μm1g) ÷ (m1 + m2), and the tension in the string is T = m2(g − a). For the Atwood machine, a = (m2 − m1)g ÷ (m1 + m2).
Trains of wagons, cranes, elevators and towing arrangements are connected systems. This guide gives a method for solving them, treats the table-and-hanging-mass system with friction and the Atwood machine, compares the systems in a table, works through numerical examples and corrects common mistakes.
What is the method?
- Draw a free-body diagram for each object, with all forces: weight, normal force, tension, friction.
- Choose a positive direction for each object so that it matches the direction of motion (so that the acceleration is positive for all objects).
- Write ΣF = ma for each object. The tension appears in both equations with the same magnitude (for an ideal rope), and the acceleration is the same.
- Solve the equations simultaneously. A shortcut is to treat the whole system as a single body with total mass: the internal tensions cancel, and only external forces are included.
- Substitute back to find the tension.
What are the standard systems?
- Block on a table connected to a hanging mass. Table block m1: T − f = m1a, where f = μm1g. Hanging mass m2: m2g − T = m2a. Adding: m2g − μm1g = (m1 + m2)a.
- Atwood machine: two masses on either side of a pulley. For m2 > m1: a = (m2 − m1)g ÷ (m1 + m2), T = 2m1m2g ÷ (m1 + m2).
- Towing: a car pulling a trailer. The car and trailer have the same acceleration, and the tension in the tow bar must supply the force to accelerate the trailer and overcome its friction.
- Blocks pushed in contact: the contact force between them is the force needed to accelerate the second block.
Ropes are assumed massless and unstretchable, and pulleys frictionless and massless. In real systems, a rope has mass and the pulley has rotational inertia, so the tension differs on the two sides of the pulley.
Standard systems compared
| System | Acceleration | Tension |
|---|---|---|
| Table block + hanging mass, with friction | (m2 − μm1)g ÷ (m1 + m2) | m2(g − a) |
| Same, no friction | m2g ÷ (m1 + m2) | m1a |
| Atwood machine | (m2 − m1)g ÷ (m1 + m2) | 2m1m2g ÷ (m1 + m2) |
Worked example: numerical problems
- Table block and hanging mass. m1 = 4.0 kg on a table with μ = 0.10, connected to a hanging mass m2 = 2.0 kg. Friction f = 0.10 × 4.0 × 9.8 = 3.92 N. Net force on the system = 2.0 × 9.8 − 3.92 = 19.6 − 3.92 = 15.68 N. Total mass = 6.0 kg, so a = 15.68 ÷ 6.0 = 2.61 m/s2. Tension: T = m2(g − a) = 2.0 × (9.8 − 2.61) = 14.4 N. Check on the table block: T − f = 14.4 − 3.92 = 10.5 N = 4.0 × 2.61 = 10.5 N ✓.
- Atwood machine. m1 = 3.0 kg and m2 = 5.0 kg. a = (5.0 − 3.0) × 9.8 ÷ 8.0 = 2.45 m/s2. T = 2 × 3.0 × 5.0 × 9.8 ÷ 8.0 = 36.75 N. Check: for m2: 5.0 × 9.8 − 36.75 = 12.25 N = 5.0 × 2.45 ✓. After 2.0 s from rest the speed is 2.45 × 2.0 = 4.9 m/s.
- Towing. A car of mass 1,000 kg pulls a trailer of mass 400 kg. The driving force is 3,500 N, and a friction/resistance force of 100 N acts on the trailer and 500 N on the car. Net external force = 3,500 − 500 − 100 = 2,900 N. a = 2,900 ÷ 1,400 = 2.07 m/s2. Tension in the tow bar: for the trailer, T − 100 = 400 × 2.07, so T = 928 N.
Who introduced the Atwood machine?
George Atwood, an English mathematician, invented the machine in 1784 to verify the laws of motion with measurements of slow accelerations. With nearly equal masses, the acceleration is small and can be timed with an ordinary clock. Today it is a standard teaching apparatus, and pulley systems on cranes and elevators use counterweights on the same principle.
Which mistakes are common?
Mistake 1: Assuming that the tension equals the weight of the hanging mass
It is equal only if the hanging mass is not accelerating. If it accelerates downward, T < m2g.
Mistake 2: Using different accelerations for connected bodies
With a taut, inextensible rope, the magnitudes of the accelerations are the same.
Mistake 3: Inconsistent sign conventions
Define the positive direction for each body along its motion so that all accelerations are positive.
Mistake 4: Including the internal tension when treating the system as one body
Internal forces cancel in pairs, and should not appear in the equation for the whole system.