Question

Решите пожалуйста и распишите решение и ответ . Даю 90 баллов . При равновесии рычага на его меньшее плечо действует сила P=300 , на большее плечо F=20 H. Длина меньшего плеча 0,5 м .Определите длину большего плеча

Answer 1

Problem Statement:

We are given a lever in equilibrium, with a force of P = 300 acting on its shorter arm and a force of F = 20 H on its longer arm. The length of the shorter arm is 0.5 m. We need to determine the length of the longer arm.

Solution:

To solve this problem, we can use the principle of moments. The principle of moments states that for an object in equilibrium, the sum of the clockwise moments about any point is equal to the sum of the anticlockwise moments about the same point.

Let's assume that the length of the longer arm is x.

The moment of a force about a point is given by the product of the force and the perpendicular distance from the point to the line of action of the force.

In this case, the moment of the force P about the fulcrum (the point where the lever pivots) is given by P multiplied by the perpendicular distance from the fulcrum to the line of action of the force P. Similarly, the moment of the force F about the fulcrum is given by F multiplied by the perpendicular distance from the fulcrum to the line of action of the force F.

Since the lever is in equilibrium, the sum of the clockwise moments about the fulcrum is equal to the sum of the anticlockwise moments about the fulcrum.

Let's set up the equation using the principle of moments:

Clockwise moments: P * 0.5

Anticlockwise moments: F * x

Since the lever is in equilibrium, the clockwise moments are equal to the anticlockwise moments:

P * 0.5 = F * x

Substituting the given values:

300 * 0.5 = 20 * x

150 = 20x

Dividing both sides by 20:

x = 150 / 20

x = 7.5

Therefore, the length of the longer arm is 7.5 meters.

Please note that the sources provided did not contain any relevant information for this specific problem.