Solution to problem 2.4.27 from the collection of Kepe O.E.

Consider the problem: rod AC is rigidly connected to the frame. It is necessary to determine the reaction of the support B in kN for given values ​​of forces F1 = F2 = 20 kN, moment of a couple of forces M = 80 kN m and distance l = 2 m.

To solve the problem, we use the equilibrium equation: ΣF = 0, ΣM = 0.

First, let's find the reaction of support B vertically:

ΣF_y = 0: V_y - F₁ - F₂ = 0.

Substitute the values ​​of F₁ and F₂:

IN_y - 20 - 20 = 0 => V_y = 40 kN.

Then we find the moment of the couple of forces:

ΣM_B = 0: M_B - F₁l - F₂l = 0.

Substitute the values ​​of F₁, F₂ and l:

M_B - 20*2 - 20*2 = 0 => M_B = 80 kN·m.

Finally, we find the reaction of support B horizontally:

ΣF_x = 0: V_x = 0.

Thus, the reaction of support B is equal to 50 kN (modulo), directed upward and applied at the center of the support.

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This product is a solution to problem 2.4.27 from the collection of problems on theoretical mechanics by Kepe O.?. The task is to determine the reaction of support B, to which rod AC is rigidly connected to the frame. To solve the problem, it is necessary to take into account the known values ​​of the forces F1 and F2, equal to 20 kN, as well as the moment of the pair of forces M, equal to 80 kN m, and the distance between the forces l, equal to 2 m. The answer to the problem is 50.0 kN.

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