Solution to problem 2.3.12 from the collection of Kepe O.E.

Let's solve problem 2.3.12:

The ADB frame is subject to vertical forces F1 = 9 kN and F2 = 4 kN. It is necessary to determine the reaction of the support B in kN if the distances AC = 2.5 m and AB = 6 m are known.

Answer:

The sum of the moments of forces acting on the frame around point A:

M_A = F₁ * AC - F₂ * AV - R_B * AB = 0

where R_B - support reaction B.

From this equation we find R_B:

R_B = (F₁ * AC - F₂ * AB) / AB = (9 * 2.5 - 4 * 6) / 6 = 7.75 kN.

Answer: 7.75 kN.

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In our digital goods store you can purchase the solution to problem 2.3.12 from the collection of Kepe O.?. This digital product is a beautifully designed document in HTML format containing a step-by-step solution to the problem. In this problem, it is necessary to determine the reaction of the support B in kN if the frame ADB is acted upon by vertical forces F1 = 9 kN and F2 = 4 kN, and the distances AC = 2.5 m and AB = 6 m. To solve the problem, the equation of equilibrium moments of forces is used acting on the frame around point A. From this equation we can express the reaction of the support B RB and get the answer: RB = (F1 * AC - F2 * AB) / AB = (9 * 2.5 - 4 * 6) / 6 = 7.75 kN. This digital product will be useful both for solving this problem and for educational purposes or preparing for exams in mathematics and physics. Don't miss the opportunity to expand your knowledge and purchase this useful digital product!

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Solution to problem 2.3.12 from the collection of Kepe O.?. is associated with determining the reaction of support B on frame ADB, which is acted upon by vertical forces F1 = 9 kN and F2 = 4 kN. The distances AC and AB are 2.5 m and 6 m, respectively, and it is required to find the reaction of the support B in kN.

To solve the problem, it is necessary to use the equilibrium conditions of the body. Since the frame is in equilibrium, the sum of all forces acting on it is zero. Therefore, the sum of the vertical forces acting on the frame is equal to the sum of the vertical reactions of the supports:

F1 + F2 = VB

Next, it is necessary to determine the moments of forces acting on the frame. The moment of force is defined as the product of force and the distance to the axis of rotation. In this problem, you can choose point A as the axis of rotation.

The moment of force F1 is equal to:

M1 = F1 * AC

The moment of force F2 is equal to:

M2 = F2 * AB

Since the frame is in equilibrium, the sum of the moments of forces acting on it is also zero:

M1 + M2 - VB * AB = 0

Substituting the values ​​of the moments and vertical reaction of support B, we obtain the equation:

F1 * AC + F2 * AB - VB * AB = 0

Solving this equation for VB, we get:

VB = (F1 * AC + F2 * AB) / AB

VB = (9 kN * 2.5 m + 4 kN * 6 m) / 6 m

VB = 7.75 kN

Thus, the reaction of support B on frame ADB is 7.75 kN.

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