Solution of problem 2.4.4 from the collection of Kepe O.E.

2.4.4 Determination of the moment M of a pair of forces at which the reaction of the support B is equal to 250 N, if the intensity of the distributed load q = 150 N/m, and the dimensions AC = CB = 2 m. (Answer 200)

To solve this problem, it is necessary to use the moment equilibrium equation. From the conditions of the problem, the intensity of the distributed load q and the dimensions AC = SV are known. The reaction of the support B is also known and is equal to 250 N. Let us denote the unknown moment of the pair of forces by the letter M.

Applying the moment equilibrium equation, we obtain the following expression:

М = q*L^2/2 - Rv*L

where L is the length of the section over which the load q is distributed, Rv is the reaction of support B.

Substituting the known values, we get:

M = 150 * 2^2 / 2 - 250 * 2 = 200

Thus, the moment M of a pair of forces at which the reaction of support B is equal to 250 N is equal to 200 N * m.

Solution to problem 2.4.4 from the collection of Kepe O.?.

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Solution to problem 2.4.4 from the collection of Kepe O.?. consists in determining the moment M of a pair of forces at which the reaction of the support B is equal to 250 N. To solve the problem, you need to know that the intensity of the distributed load is q = 150 N/m, and the dimensions AC = CB = 2 m.

According to the conditions of the problem, it is known that the support reaction B is equal to 250 N. Since the support reaction force is equal to the sum of the vertical forces acting on the structure, we can write the vertical equilibrium equation:

В + qAC - Fsinα = 0,

where q is the intensity of the distributed load, AC are the dimensions of the structure, F is the force created by a pair of forces, α is the angle between the direction of force F and the horizon.

Since the structure is in equilibrium, the moment equilibrium condition must also be satisfied:

M = FFOODsinα,

where M is the moment of a pair of forces acting on the structure.

Substituting known values, we get:

250 + 1502 - Fsinα = 0,

M = F2sinα.

Solving this system of equations, we find:

F = 200 N,

M = 400 N*m.

Answer: moment M of a pair of forces, at which the reaction of support B is equal to 250 N, is equal to 400 N*m.

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