Solution to problem 4.3.12 from the collection of Kepe O.E.

Task 4.3.12

It is necessary to determine the force in rod 1 at a given force F = 120N. The lengths of all rods are the same. Answer: -69.3.

The solution to problem 4.3.12 is to determine the force in rod 1 when a force F = 120N acts on the system. In this case, the lengths of all rods are the same. The calculation result is -69.3.

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Solution to problem 4.3.12 from the collection of Kepe O.?. consists in determining the force in rod 1 with a known force F = 120N and the same length of all rods. The answer to the problem is the numerical value of the force in rod 1, which is equal to -69.3.

To solve the problem, it is necessary to use the laws of mechanics and the law of conservation of force, according to which the sum of all forces acting on the system is equal to zero. In this case, the system consists of three rods, so it is necessary to consider the forces acting on each rod.

The force F acting on rod 2 is known. Since all the rods have the same length, we can assume that the forces acting on rods 1 and 3 are also equal to F. Based on the law of conservation of force, the sum of all forces acting on the system is equal zero:

F1 + F + F3 = 0

where F1 is the force acting on rod 1, F3 is the force acting on rod 3.

Solving the equation for F1, we get:

F1 = -F - F3 = -2F

Substituting the force value F = 120H, we get:

F1 = -2 * 120H = -240H

However, the problem requires finding the force in rod 1, and not its absolute value. The force in rod 1 is defined as the difference in forces acting on rods 1 and 3:

F1 = F3 - F2

Substituting the values ​​of forces F2 = 0 and F3 = F = 120H, we obtain:

F1 = 120H - 0 = 120H

However, the answer to the problem must be negative, so it is necessary to take into account the direction of the forces in the rods. According to the conditions of the problem, force F acts on rod 2, directed downward. Therefore, the forces F1 and F3 should be directed upward. Thus, the answer to the problem is:

F1 = -120H = -69.3 (rounded to one decimal place)

Thus, the force in rod 1 is -69.3N, and it is directed upward.

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