1. The problem statement, all variables and given/known data Two bodies with masses m1 and m2 are connected by a massless thread that supports a maximum force |Ft|max . On mass m1 the force F1 = -β t x and on mass m2 the force F2 = 3 β t x (t: time, β: constant) acts. The motion is frictionless. 2. Relevant equations When does the thread rip? 3. The attempt at a solution According to Netwon's 3rd law if m2 pulls m1 with a force Ft (equal to the string tension), m2 is then pulled by m1 with a force equal and opposite -Ft. Therefore, Netwon's 2nd law for m1 is: (net force=mass x acceleration): Ft-F1 = m1*a and for m2: F2-Ft = m2*a The acceleration should be the same. We do not know which force "wins", in other words the direction of the acceleration. I guess it depends on the numerical values for m1 and m2 too. For example, let's fix the one-dimensional reference system with the positive x axis in the same direction of F2 as suggested by the problem text, but I could also do the other way. In this system, supposing a is in the positive direction, each vectorial equation becomes only one scalar equation, since we are in one dimension: Ft-F1 = m1*a F2-Ft = m2*a This is a linear system: two equations, two unknowns: Ft and a. I am only interested in finding out Ft. E.g. I solve the second equation for a, substitute into the first and solve for Ft. I get: Ft = (m2 F1 + m1 F2)/(m1 + m2) The tension is the weighted sum of the two forces, weighted by the mass values, but not the respective mass values, rather they are inverted: m2, not m1 multiplies F1, and similarly for F2. So a sort of inverted weighted sum. Since in our reference system: F1=-β t F2=3 β t we have: Ft = β t (3 m1 - m2) / (m1 + m2) The string will break at time t = trip when Ft is equal to Ftmax trip = Ftmax (m1 + m2) / (β (3 m1 - m2)) which should be the solution to the problem. Is this the correct way to solve it?