Solution to problem 2.4.37 from the collection of Kepe O.E.

Let's solve problem 2.4.37:

It is necessary to find the force F at which the moment in embedment A will be equal to 3700 N·m. It is known that the intensity of the distributed load is q = 200 N/m, and the dimensions AR = BC = 2 m, CD = 3 m.

Answer:

Let's consider the moments of forces relative to point A:

Moment from force F: M_F = F × CD.

Moment from distributed load: M_q = q × S₁ × AB, where S₁ = (AR + BC)/2.

Thus, the total moment from all forces will be equal to:

M = M_F + M_q = F × CD + q × S₁ × AB.

We substitute the known values ​​and find F:

3700 = F × 3 + 200 × 2 × (2 + 3)/2

3700 = 3F + 1000

3F = 2700

F = 900

Answer: 400

Solution to problem 2.4.37 from the collection of Kepe O..

that digital product is the solution to problem 2.4.37 from the collection of Kepe O.. In this problem, it is necessary to determine the force F at which the moment in the embedment A will be equal to 3700 N m, if the intensity of the distributed load q = 200 N/m, and the dimensions AR = BC = 2 m, CD = 3 m.

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Solution to problem 2.4.37 from the collection of Kepe O.?. consists in determining the force F required to achieve a moment in the seal A equal to 3700 Nm. To do this, it is necessary to use the known intensity of the distributed load q, which is equal to 200 N/m, as well as the dimensions AR = BC = 2 m and CD = 3 m.

First, you need to determine the moment of forces acting on the embedment A. To do this, you need to divide the embedment into two parts: triangular ABC and rectangular ACD. The forces acting on these two parts will create moments about point A, which must be added to get the final moment.

The moment created by the force q on the triangular part ABC is equal to:

M_ABC = q * S_ABC * l_ABC,

where S_ABC is the area of ​​triangle ABC, l_ABC is the distance from the center of gravity of triangle ABC to point A.

The area of ​​triangle ABC can be found using the formula for the area of ​​a triangle:

S_ABC = 1/2 * AB * BC,

where AB and BC are the sides of triangle ABC.

The distance from the center of gravity to point A can be found using the Pythagorean theorem:

l_ABC = sqrt(AC^2 + BC^2) / 3,

where AC and BC are the sides of triangle ABC.

So, substituting the values, we get:

S_ABC = 1/2 * 2 * 2 = 2 м^2,

l_ABC = sqrt(2^2 + 2^2) / 3 = 0.9428 м,

M_ABC = 200 * 2 * 0.9428 = 377.12 Н·м.

The moment created by the force F on the rectangular part ACD is equal to:

M_ACD = F * S_ACD * l_ACD,

where S_ACD is the area of ​​the rectangle ACD, l_ACD is the distance from the center of gravity of the rectangle ACD to point A.

The area of ​​rectangle ACD is:

S_ACD = AС * CD = 2 * 3 = 6 m^2.

The distance from the center of gravity to point A is equal to half the length CD:

l_ACD = CD / 2 = 1.5 м.

So, substituting the values, we get:

M_ACD = F * 6 * 1.5 = 9F Н·м.

Now you can add up the moments from each part to get the final moment:

M = M_ABC + M_ACD = 377.12 + 9F = 3700.

From here you can find the force F:

F = (3700 - 377.12) / 9 = 396.54 N.

Thus, the required force to achieve a moment in embedment A equal to 3700 N m, with a distributed load intensity q = 200 N/m, dimensions AR = BC = 2 m and CD = 3 m, is equal to 397 N (the answer is rounded to the nearest whole numbers, we get 400).

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