# At the ends of the thread, thrown over a block suspended from the ceiling, hang two weights of the same mass.

**At the ends of the thread, thrown over a block suspended from the ceiling, hang two weights of the same mass. After another weight of 4 kg was placed on the right weight, the weights began to move with an acceleration of 0.25g. Find the initial mass of the load, as well as the tension force of the thread when moving. The mass of the block and thread, as well as the friction in the axis of the block, should be neglected.**

Given:

m1 = m2 = m – the mass of the weights at the ends of the thread thrown over the block is the same;

dm = 4 kilograms – the mass of the additional load, which was placed on the right load;

g = 10 Newton / kilogram – acceleration of gravity;

a = 0.25 * g = 2.5 m / s2 is the acceleration with which the loads began to move.

It is required to determine m (kilogram) – the initial mass of the weights, as well as T (Newton) – the tension force of the thread when the system moves.

By the condition of the problem, we neglect the masses of the block and the thread itself, as well as the friction force. Then, we compose the equations of Newton’s second law for the right and left weights:

(1) (m + dm) * g – T = (m + dm) * a;

(2) T – m * g = m * a.

Let us add equations (1) and (2):

(m + dm) * g – T + T – m * g = (m + dm) * a + m * a;

m * g + dm * g – m * g = m * a + dm * a + m * a;

dm * g = 2 * m * a + dm * a;

2 * m * a = dm * (g – a);

m = dm * (g – a) / (2 * a) = dm * (g – 0.25 * g) / (2 * 0.25 * g) = 0.75 * dm * g / 0.5 * g =

= 0.75 * dm / 0.5 = 1.5 * dm = 1.5 * 4 = 6 kilograms.

From equation (2) we find the thread tension force:

T = m * (g + a) = m * (g + 0.25 * g) = 1.25 * m * g = 1.25 * 6 * 10 = 75 Newtons.

Answer: the initial mass of the weights is 6 kilograms, the tension force of the thread during movement is 75 Newtons.