A.3.3.TR:5.2 - A fixed connection and two force equations determine acceleration

Two carts move on an ideal frictionless straight track. A massless rigid connector keeps their positions q1 and q2 at q2 - q1 = L. Their positive masses are m1 and m2. Signed external forces along the track are F1 and F2. The connector exerts equal and opposite signed forces, written -lambda on cart 1 and +lambda on cart 2.

The positional constraint alone permits many common motions. Differentiating the constant-separation condition gives v2 = v1 and a2 = a1 = a for compatible motion. Newton’s equations for this ideal model give

m1*a = F1 - lambda
m2*a = F2 + lambda

Adding them obtains a = (F1 + F2)/(m1 + m2). Substituting back gives lambda = (m2*F1 - m1*F2)/(m1 + m2). The connection constraint and both interaction equations were needed to determine the common acceleration and exchanged force.

For m1 = 1 kg, m2 = 3 kg, F1 = 4 N and F2 = 0, the result is a = 1 m/s² and lambda = 3 N. Start with q1 = 0, q2 = L and both velocities zero. While the forces remain constant, q1(t) = t²/2 and q2(t) = L + t²/2 in metres when t is in seconds; both velocities are t metres/second.

These formulas provide a state-transition calculation for any chosen interval within those assumptions. If the forces vary, supply their time or state dependence before computing the motion. If the connector is elastic or has consequential mass, its constitutive and interaction account changes the rule. Choosing smaller numerical time steps cannot supply that missing physical relation.

The calculation combines the physical interaction account, mathematical constraints and an executable rule. It also shows why reducing to one position can simplify motion prediction while retaining the connector force lets a designer determine the load on the connection.