Solution to problem 2.3.16 from the collection of Kepe O.E.

2.3.16 It is necessary to determine the vertical force F at which the moment at the embedding point A is equal to 240 N•m. It is known that the intensity of the distributed load q is 40 N/m, and the dimensions of the segments CD, AB and BC are 3 m and 1 m, respectively.

To solve the problem, we use the formula for determining the moment relative to a given point:

$M_A = \int_{x_1}^{x_2} q(x)\cdot x\,dx + F\cdot h$

where $M_A$ is the moment about point A, $q(x)$ is the intensity of the distributed load, $x$ is the distance to point A, $F$ is the vertical force, h is the distance from point A to the straight line passing through point B and S.

Let's replace the values ​​in the formula:

$240 = \int_{0}^{3} 40\cdot x\,dx + F\cdot 1$

Let's solve the integral:

$\int_{0}^{3} 40\cdot x\,dx = 40 \cdot \frac{3^2}{2} = 180$

Let's substitute the value of the integral into the equation:

$240 = 180 + F$

From here we get:

$F = 60$ H

Thus, in order for the moment at point A to be equal to 240 N•m, it is necessary to apply a vertical force F equal to 60 N.

Solution to problem 2.3.16 from the collection of Kepe O..

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The product is the solution to problem 2.3.16 from the collection of Kepe O.?. The task is to determine the vertical force F required to achieve a moment in the embedment A equal to 240 N•m. It is known that the intensity of the distributed load q is equal to 40 N/m, and the dimensions CD = 3 m, AB = BC = 1 m. To solve the problem, it is necessary to use the moments of force and the equilibrium of the moments relative to the embedment A. The answer to the problem is 180 N.

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