When lifting a load weighing 50 kg using a movable block, the cable is pulled with a force of 530 N.

When lifting a load weighing 50 kg using a movable block, the cable is pulled with a force of 530 N. Find the value of the friction force

Given:

m = 50 kilograms – the mass of the load, which is lifted using a movable block;

F = 530 Newton – the force with which the cable is pulled.

It is required to determine Ftr (Newton) – the magnitude of the friction force.

The movable block gives a 2-fold gain in strength. That is, the actual magnitude of the force will be equal to:

F1 = 2 * F = 2 * 530 = 1060 Newton.

Then, taking into account the assumption that the load is pulled evenly, according to Newton’s first law:

F1 = F gravity + Ftr;

Ftr = F1 – F gravity;

Ftr = F1 – m * g (where g = 10 Newton / kilogram is an approximate value);

Ftr = 530 – 50 * 10 = 530 – 500 = 30 Newtons.

Answer: the magnitude of the friction force is 30 Newtons.