Solution to problem 5.5.9 from the collection of Kepe O.E.

5.5.9. In three vertical rods 1, 2 and 3, a square plate ABCD with a weight G = 115 N is fixed in a horizontal position. At point A, it is acted upon by a vertical force Q = 185 N. It is necessary to determine the force in rod 2 from the equilibrium equation of moments of forces relative to the BD axis. (Answer: -185).

A square plate ABCD weighing 115 N is hinged in three vertical rods 1, 2 and 3. At point A, a vertical force Q = 185 N is applied to it. To determine the force in rod 2, it is necessary to use the equilibrium equation of moments of forces relative to the BD axis. The solution is -185.

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Our store presents a digital product - a solution to problem 5.5.9 from the collection of Kepe O.?. This product is a detailed solution to the problem, which is to determine the force in rod 2, provided that a square plate ABCD weighing 115 N is hinged in three vertical rods 1, 2 and 3, and at point A it is acted upon by a vertical force Q = 185 N. To solve the problem, it is necessary to use the equation of equilibrium of moments of forces relative to the BD axis. The result of the solution is the value of the force in rod 2, which is -185 N. The product is designed in a beautiful HTML format, which will allow you to quickly and easily understand the solution to this problem without having to waste time searching for a solution on the Internet. By purchasing this digital product in our store, you can save your efforts and quickly solve the problem.

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Solution to problem 5.5.9 from the collection of Kepe O.?. consists in determining the force in the vertical rod 2, provided that a square plate ABCD weighing 115 N is hinged in three vertical rods 1, 2 and 3, and at point A a vertical force Q = 185 N is applied.

To solve the problem, it is necessary to create an equation for the equilibrium of moments of forces relative to the BD axis. Since the plate is in a horizontal position, the sum of the vertical forces acting on it is zero. Therefore, the force in rod 2 is also zero.

From the equation of equilibrium of moments of forces relative to the BD axis it follows that:

Q * AB - G * BC = 0

where AB and BC are the distances from point B to points A and C, respectively.

Substituting the known values, we get:

185 * AB - 115 * BC = 0

AB/BC = 115/185 = 23/37

Also, from the equilibrium equation of moments of forces it follows that:

F2 * BD - Q * AD - G * CD = 0

where F2 is the force in rod 2, AD and CD are the distances from point D to points A and C, respectively.

Substituting the known values ​​and expressing F2, we get:

F2 = Q * AD / BD + G * CD / BD

Substituting the values, we get:

F2 = 185 * (2/3) / (2 * (23/37)) + 115 * (1/2) / (2 * (23/37)) = -185 N

Answer: the force in rod 2 is -185 N.

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