# Two weights tied with a thread are lifted vertically upward, acting on the upper one with a force of 14N directed upward.

**Two weights tied with a thread are lifted vertically upward, acting on the upper one with a force of 14N directed upward. The weight of the upper weight is 600 g, the lower weight is 200 g. What is the tensile force of the thread connecting the weights?**

Given:

F = 14 Newton – the force with which the loads are lifted vertically upwards;

m1 = 600 grams = 0.6 kilograms – the mass of the top load;

m2 = 200 grams = 0.2 kilograms – the mass of the lower weight;

g = 10 m / s2 – acceleration of gravity.

It is required to determine the thread tension T (Newton).

Let a be the acceleration with which the loads are lifted. Then, according to Newton’s second law, for the upper body:

F – m1 * g – T = m1 * a;

For the second body:

T – m2 * g = m2 * a, hence:

a = (T – m2 * g) / m2.

Substitute the value of a into the first equation:

F – m1 * g – T = m1 * (T – m2 * g) / m2;

F * m2 – m1 * m2 * g – T * m2 = m1 * T – m1 * m2 * g;

F * m2 – m1 * m2 * g + m1 * m2 * g = m1 * T + T * m2;

F * m2 = T * (m1 + m2);

T = F * m2 / (m1 + m2) = 14 * 0.2 / (0.6 + 0.2) = 2.8 / 0.8 = 3.5 Newton.

Answer: the tensile force of the thread is 3.5 Newtons.