Two bodies with masses m1 = 50 g and m2 = 100 g, connected by a weightless thread
Two bodies with masses m1 = 50 g and m2 = 100 g, connected by a weightless thread, lie on a smooth table. With what force can the first body be pulled parallel to the surface so that a thread that can withstand a load of 5 N does not break?
Given:
m1 = 50 grams = 0.05 kilograms – the mass of the first body;
m2 = 100 grams = 0.1 kilogram – the mass of the second body;
g = 10 m / s ^ 2 – acceleration of gravity;
T = 5 Newton – the maximum load that the thread can withstand.
It is required to determine the force F (Newton) with which the first body can be pulled so that the thread does not break.
Since, according to the condition of the problem, the bodies lie on a smooth table, that is, the friction force can be ignored, and the thread itself is weightless, then:
Newton’s second law for the first body:
F – T = m1 * a;
Newton’s second law for the second body:
T = m2 * a, hence:
a = T / m2. Substitute in the first equation:
F – T = m1 * T / m2;
F = m1 * T / m2 + T = T * (m1 / m2 + 1) = 5 * (0.05 / 0.1 + 1) = 5 * (0.5 + 1) = 5 * 1.5 = 7.5 Newton.
Answer: You can pull with a force equal to 7.5 Newtons.