Tinctures in one row are 6 identical cubes weighing 200 g. What is the minimum force you need to squeeze

Tinctures in one row are 6 identical cubes weighing 200 g. What is the minimum force you need to squeeze the cubes, gripping the two extreme hands, in order to tear them off the table? The coefficient of friction of the cube on the cube is u = 0.2.

If all the cubes are together, then there are 5 pairs of rubbing surfaces. Then, in order to lift the entire system of 6 cubes, you need to apply such a force that balances the friction force and gravity acting on the cubes.

The force of friction and the force of gravity are reciprocal, therefore their resultant will be equal to 6mg – 5 Ftr.

Friction force Ffr = µ * N = µ * mg.

Therefore the applied force must be F = 6mg – 5µmg.

We express the mass in kilograms m = 200 gr = 0.2 kg, the acceleration of gravity is rounded g ≈ 10 m / s².

We supply the values: F = 6 * 0.2 * 10 – 5 * 0.2 * 0.2 * 10 = 12 – 2 = 10.

The applied force must be at least 10 N.