A weight of 200 g was suspended on the right arm of the lever 40 cm long. What force must be applied

A weight of 200 g was suspended on the right arm of the lever 40 cm long. What force must be applied to the left arm of the lever 25 cm long in order to balance the lever?

rp = 40 cm = 0.4 m.

m = 200 g = 0.2 kg.

g = 10 m / s2.

rl = 25 cm = 0.25 m.

Fl -?

The condition for the balance of the lever is the equality of the moments of forces on the left and right sides: Ml = Mp.

The moment of the force M is the product of the force F by the smallest distance from the line of its action to the axis of balance r, which is called the shoulder: M = F * r.

Fl * rl = Fp * rp.

On the right side, the weight acts on the lever with its weight. The weight of the load Fп is expressed by the formula: Fп = m * g, where m is the mass of the load, g is the acceleration of gravity.

Fl * rl = m * g * rp.

Fl = m * g * rp / rl.

Fl = 0.2 kg * 10 m / s2 * 0.4 m / 0.25 m = 3.2 N.

Answer: for the balance of the lever, a force Fl = 3.2 N. must act on the left side.