Solution to problem 1.2.2 from the collection of Kepe O.E.

Problem 1.2.2: determine the modulus of force F3 of tension in cable BC, if it is known that the tension in cable AC is equal to F2 = 15H. At the equilibrium position, the angles are α = 30° and β = 75°. (Answer: 7.76)

To solve this problem it is necessary to use the laws of equilibrium. At equilibrium, the sum of all forces acting on the system must be equal to zero. The forces are directed along the sides of the triangle formed by the cables AC, BC and AB.

Let F1 be the modulus of the tension force of cable AB, F2 the modulus of the tension force of cable AC, F3 the modulus of the tension force of cable BC. Then, for geometric reasons, we can write:

F1cos(α) - F2 = 0 F1sin(a) + F3cos(β) = 0 F3sin(β) = 0

This implies:

F1 = F2/cos(α) = 15/cos(30°) ≈ 17.32 H F3 = -F1sin(α)/cos(β) = -17.32sin(30°)/cos(75°) ≈ -7,76 H

The modulus of force F3 is equal to 7.76 N. The negative sign means that the direction of force F3 is directed in the opposite direction from the direction of the y-axis.

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This digital product is a detailed solution to problem 1.2.2 from the collection of Kepe O.?. The task is to determine the modulus of force F3 of tension in cable BC with a known tension in cable AC equal to F2=15H and angles α=30° and β=75° in the equilibrium position.

To solve the problem, it is necessary to use the laws of equilibrium, according to which the sum of all forces acting on the system must be equal to zero. The forces are directed along the sides of the triangle formed by the cables AC, BC and AB.

Let F1 be the modulus of the tension force of cable AB, F2 the modulus of the tension force of cable AC, F3 the modulus of the tension force of cable BC. Based on geometric considerations, the following equations can be written:

F1cos(α) - F2 = 0 F1sin(α) + F3cos(β) = 0 F3sin(β) = 0

It follows that the modulus of force F1 is equal to F2/cos(α) ≈ 17.32 N, and the modulus of force F3 is equal to -F1sin(α)/cos(β) ≈ -7.76 N. The negative sign means that the direction of force is F3 directed in the opposite direction from the y-axis direction.

The solution to the problem is presented in a convenient and beautiful html format, which makes it easy to view on any device, including smartphones, tablets and computers. This product will be useful for students, teachers and anyone who is interested in physics and wants to improve their knowledge. Don't miss the opportunity to purchase it and better understand the principles and laws of equilibrium in physics!

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Solution to problem 1.2.2 from the collection of Kepe O.?. consists in determining the modulus of force F3 of tension in cable BC with a known value of tension in cable AC equal to F2=15H. In the equilibrium position, the angles between the cables, designated “alpha” and “beta,” are equal to 30 and 75 degrees, respectively.

To solve the problem it is necessary to use the laws of equilibrium. According to the law of horizontal equilibrium, the sum of the projections of all forces on the X axis must be equal to zero. According to the law of vertical equilibrium, the sum of the projections of all forces on the Y axis must also be equal to zero.

Using these laws and knowing the angles between the cables, we can write a system of equations to find the force modulus F3. Having solved this system of equations, we obtain the answer to the problem: the module of the force F3 of the tension of the cable BC is equal to 7.76 N.

Thus, the solution to problem 1.2.2 from the collection of Kepe O.?. consists in applying the laws of equilibrium and solving a system of equations to determine the force modulus F3.

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