# The weightless block is fixed at the end of the table. A thread is thrown through the block, to one end

**The weightless block is fixed at the end of the table. A thread is thrown through the block, to one end of which a weight equal to 1 kg is suspended. Attached to the other end of the thread is a weight of the same mass lying on the table. The coefficient of friction of the load on the table is 0.18. Find the thread tension.**

Given:

m1 = m2 = m = 1 kilogram is the mass of loads suspended from the block and lying on the table;

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

k = 0.18 – coefficient of friction of the load on the table surface.

It is required to determine the thread tension T.

Let the bodies move downward with some acceleration a. Then:

m * g – T = m * a;

T – Ftr = m * a;

m * g – T = T – Ftr;

m * g + Ftr = T + T;

m * g + k * m * g = 2 * T;

T = m * g * (1 + k) / 2 = 1 * 10 * (1 + 0.18) / 2 = 1 * 10 * 1.18 / 2 = 11.8 / 2 = 5.9 Newton.

Answer: the tensile force of the thread is 5.9 Newtons.