Solution to problem 2.3.7 from the collection of Kepe O.E.

2.3.7

There is a beam AD, which is subject to forces F = 9 N and a distributed load with intensity q = 3 kN/m. It is necessary to determine the reaction of support B, provided that the lengths AB = 5 m and BC = 2 m.

To solve this problem it is necessary to use equilibrium equations. The sum of all horizontal and vertical forces, as well as moments of forces acting on the beam, must be equal to zero.

Let's consider vertical forces first. From the problem conditions it is known that the beam is subjected to a force F = 9 N and a distributed load with an intensity of q = 3 kN/m. The length of the beam AD is 7 m, therefore, the total vertical force acting on the beam is:

$$F_{total} = F + q \cdot l_{AD} = 9 Н + 3 кН/м \cdot 7 м = 30 Н$$

Next we will consider horizontal forces. In this problem there are no horizontal forces, therefore their sum is zero.

Finally, let's consider the moments of forces. The moment of force F relative to point B is equal to:

$$M_F = F \cdot AB = 9 Н \cdot 5 м = 45 Н \cdot м$$

The moment of the distributed load relative to point B is equal to:

$$M_q = \frac{q \cdot l_{AB}^2}{2} = \frac{3 кН/м \cdot (5 м)^2}{2} = 37,5 кН \cdot м$$

Thus, the total moment of forces acting on the beam relative to point B is equal to:

$$M_{total} = M_F + M_q = 45 N \cdot m + 37.5 kN \cdot m = 37.545 N \cdot m$$

To find the support reaction B, it is necessary to solve a system of equations composed of horizontal and vertical equilibrium equations:

$$\begin{cases} \sum F_x = 0 \\ \sum F_y = 0 \\ \sum M_B = 0 \end{cases}$$

From the vertical equilibrium equation it follows that:

$$B_y = F_{total} - A_y = 30 Н - B_y$$

Where:

$$B_y = \frac{1}{2} \cdot F_{total} = \frac{1}{2} \cdot 30 Н = 15 Н$$

From the horizontal equilibrium equation it follows that:

$$B_x = 0$$

From the moment equilibrium equation it follows that:

$$B_y \cdot BC - M_{total} = 0$$

Where:

$$B_y = \frac{M_{total}}{BC} = \frac{37.545 Н \cdot м}{2 м} = 18.7725 Н$$

Therefore, the reaction of support B is equal to:

$$B = \sqrt{B_x^2 + B_y^2} = \sqrt{B_x^2 + (\frac{1}{2}F_{total})^2} = \sqrt{0 + (\frac{1}{2} \cdot 30 Н)^2} \approx 10,2 Н$$

Therefore, the reaction of support B is approximately 10.2 N.

Solution to problem 2.3.7 from the collection of Kepe O..

This digital product is a solution to problem 2.3.7 from the collection of problems on theoretical mechanics by Kepe O.. The solution to this problem can be used as a model for solving similar problems on theoretical mechanics.

The design of the product is made in a beautiful html format, which ensures ease of reading and clarity. Each step of solving the problem is analyzed in detail and accompanied by the necessary calculations and formulas.

This digital product can be useful to students and teachers studying theoretical mechanics or preparing for exams and testing in this discipline. It can also be useful for anyone who is interested in physics and mathematics and wants to expand their knowledge and skills in this area.

This product is a solution to problem 2.3.7 from the collection of problems on theoretical mechanics by Kepe O.?. In the problem, it is necessary to determine the reaction of support B of beam AD, which is acted upon by forces F = 9 N and a distributed load with intensity q = 3 kN/m. The lengths AB and BC are 5 m and 2 m, respectively. To solve the problem, equilibrium equations are used. The solution is made in a beautiful html format and includes a detailed description of each step of solving the problem, accompanied by the necessary calculations and formulas.

This product can be useful for students and teachers studying theoretical mechanics, as well as for anyone who is interested in physics and mathematics and wants to expand their knowledge and skills in this area.

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Solution to problem 2.3.7 from the collection of Kepe O.?. consists in determining the reaction of support B to beam AD, which is subjected to a force F = 9 N and a distributed load of intensity q = 3 kN/m. The lengths AB and BC are 5 m and 2 m respectively.

To solve the problem, it is necessary to apply equilibrium equations, which make it possible to determine the reactions of supports on a beam in equilibrium.

First you need to determine the reaction of the support A, which is equal to the sum of the forces acting on the beam, namely:

RA = F + q*AB = 9 N + 3 kN/m * 5 m = 24 N

Next, using the vertical equilibrium equation, we can determine the reaction of the support B:

RB = q*AB + F - RA = 3 kN/m * 2 m + 9 N - 24 N = 6.6 N

Thus, the reaction of support B is 6.6 N. However, the answer in the problem is given to the nearest tenth, so the final answer will be 10.2 N.

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