Two bars of 100 g and 300 g are tied with a weightless and inextensible thread, both bars lie on a smooth table

Two bars of 100 g and 300 g are tied with a weightless and inextensible thread, both bars lie on a smooth table. A horizontal force of 0.4 N was applied to the smaller bar. What is the tensile force of the thread?

m1 = 100 grams = 0.1 kilograms – the mass of the first body;

m2 = 300 grams = 0.3 kilograms – the mass of the second body;

g = 10 m / s ^ 2 – acceleration of gravity.

F = 0.4 Newton – the force applied to the body m1.

It is required to find the tension force of the thread between the bodies T.

According to Newton’s second law:

F – T = m1 * a;

T = m2 * a. Then:

F – m2 * a = m1 * a

F = a * (m1 + m2), hence:

a = F / (m1 + m2);

T = m2 * F / (m1 + m2) = 0.3 * 0.4 / (0.3 + 0.1) = 0.3 * 0.4 / 0.4 = 0.3 Newton.

Answer: the pulling force is 0.3 Newton.