Connected Carts on a Rope
Two carts connected by a rope, with one pulled by an external force; tension is identical throughout
Objective
Verify that rope tension depends on the mass being pulled and the system's acceleration, following T = m₂ · a, rather than being proportional to the applied force.
Setup
- Set applied force to 20 N, cart 1 mass to 2 kg, and cart 2 mass to 3 kg. Press Start and observe the acceleration and tension values.
- Note the acceleration (a ≈ 4 m/s²) and tension (T ≈ 12 N). Verify that tension is less than the applied force—the rope does not transmit the full external push.
- Reset and increase cart 2's mass to 10 kg while holding force and cart 1's mass constant. Start again and compare the new acceleration and tension to the previous values.
- Observe that increasing m₂ reduces acceleration but increases tension. The tension climbs because the rope must now accelerate a heavier mass, even though the acceleration itself is lower.
- Reset and increase the applied force to 50 N. Start and note that both acceleration and tension rise, but the relationship is indirect: tension follows a, not F directly.
Analytical Prediction
With F = 20 N, m₁ = 2 kg, m₂ = 3 kg, the system acceleration and rope tension are governed by:
The readouts display a ≈ 4.00 m/s² and T ≈ 12.0 N. Note that T < F: the rope's role is to accelerate cart 2 at the system's rate, not to transmit the applied force unchanged. When m₂ is increased to 10 kg (with F and m₁ held constant), a = 20/12 ≈ 1.67 m/s² and T = 10 × 1.67 ≈ 16.7 N—tension rises despite acceleration falling, because the cart 2's mass dominates the T calculation.
Results Analysis
During the simulation, the Acceleration readout shows the system's shared acceleration (computed as a = F/(m₁+m₂)). The Tension readout displays T = m₂ · a. Compare these against hand calculations for the slider settings: both should match to within 0.01 m/s² and 0.1 N respectively, confirming that the rope's job is to impart the force needed to accelerate its own mass at the system's rate. Visually, both carts move together (identical velocity vectors), and the rope's drawn length and colour encode the tension magnitude.
Source of Error
The simulation assumes ideal rigid carts (point masses) and a massless, inextensible rope. Air resistance and friction are neglected; all acceleration comes from the applied force alone. The numerical integration uses fixed substeps, so position and velocity accumulate small rounding errors over the 8-second run. The readouts round to 2 decimal places, hiding sub-centimetre position drift and sub-pascal tension ripple. These idealizations are mirrored in the prediction (which assumes the same neglect), so the residual gap is purely numerical, not physical.
Further Exploration
- What happens if you double the applied force while keeping the cart masses constant? Does the tension double, or does it increase by a different factor?
- If you increase cart 1's mass to 10 kg while leaving force and cart 2's mass at their defaults, how does the tension change? Does cart 1's mass affect the rope at all?
- Set cart 2's mass to 1 kg and cart 1's mass to 10 kg, with force at 20 N. Calculate the expected tension by hand, then run the simulation to verify. Does your prediction match the readout?
- Can you find a configuration where the tension is exactly zero? What must be true about the applied force for this to happen?