Solution to problem 5.5.7 from the collection of Kepe O.E.

Horizontal homogeneous square slab ABCD with weight G = 500N

The plate is suspended at points A, D, E from three vertical rods 1, 2, 3. It is necessary to determine the force in rod 1 if AD = 2AE.

To solve the problem, we use the equilibrium condition. Since the plate is at rest, the sum of all forces acting on it is zero. Therefore, the sum of the vertical components of the forces acting on the slab must be equal to its weight G.

Let the forces in rods 1, 2 and 3 be equal to F1, F2 and F3, respectively. Then:

F1 + F2 = G / 2, (1)

F3 = G / 2. (2)

Since AD ​​= 2AE, point E is located at a distance h = AD / 3 from point D. In this case, the angle between the plate and rod 1 is 45 degrees. Consequently, rod 1 is acted upon by the vertical component of the force F1 and the horizontal component of the force F1 * tg(45°).

From the horizontal equilibrium condition it follows that:

F1 * tg(45°) = F2 / 2. (3)

From the condition of vertical equilibrium it follows that:

F1 + F2 + F3 = G. (4)

From equations (1), (2) and (4) we obtain:

F1 + 2 * F1 + G / 2 = G,

whence F1 = G / 3 = 500 N.

Thus, the force in rod 1 is 500 N.

Solution to problem 5.5.7 from the collection of Kepe O.?.

This digital product is a solution to problem 5.5.7 from a collection of problems in physics, authored by Kepe O.?. The problem involves determining the forces in the rods when suspending a horizontal homogeneous square slab from three vertical rods.

The solution to the problem is presented in the form of a beautifully designed html document that is easy to read and understand. It uses various HTML elements, such as headings, paragraphs, lists and formulas, which makes the text more structured and visual.

By purchasing this digital product, you receive a high-quality solution to the problem that will help you better understand the topic and successfully cope with similar tasks in the future.

This digital product is a solution to problem 5.5.7 from the collection of problems in physics by the author Kepe O.?. The problem is to determine the forces in the rods when suspending a horizontal homogeneous square slab from three vertical rods. The solution to the problem is presented in the form of a beautifully designed HTML document that is easy to read and understand. It uses various HTML elements such as headings, paragraphs, lists and formulas, which makes the text more structured and visual.

To solve the problem, an equilibrium condition was used, according to which the sum of all forces acting on the plate must be equal to zero. From this condition, equations were obtained to determine the forces in the rods. As a result, it was found that the force in rod 1 is 500 N.

By purchasing this digital product, you receive a high-quality solution to the problem that will help you better understand the topic and successfully cope with similar tasks in the future.

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Solution to problem 5.5.7 from the collection of Kepe O.?. is to determine the force in the vertical rod 1, which holds a horizontal homogeneous square plate ABCD weighing 500 N, suspended from points A, D, E. To solve the problem, you need to know that the plate is in equilibrium, that is, the sum of all vertical forces acting on plate is equal to zero.

Since the slab is suspended from points A, D, E, three vertical forces act on it: F1 at point A, F2 at point D and F3 at point E. The sum of these forces must be equal to the weight of the slab G = 500 N, that is F1 + F2 + F3 = G.

It is also known from the problem conditions that AD = 2AE. This means that point E is at a distance of AE/3 from point A, and point D is at a distance of 2AE/3 from point A.

To determine the force in rod 1, it is necessary to decompose the forces F1, F2 and F3 into components parallel and perpendicular to rod 1. The force parallel to rod 1 will be equal to the force F1, since it is only directed along rod 1. The forces perpendicular to rod 1 will be are equal to the projections of forces F2 and F3 on the vertical axis, since they are directed perpendicular to rod 1.

By equating the sum of forces perpendicular to rod 1 to zero, we can determine what fraction of the weight of the plate is carried by rod 1. Since the sum of forces perpendicular to rod 1 is equal to the projection of force F1 on the vertical axis, we can express F1 in terms of G: F1 = (F2 + F3) * (AE/3) / (2AE/3) = (F2 + F3) / 2.

Substituting this expression for F1 into the equation F1 + F2 + F3 = G, we obtain: (F2 + F3) / 2 + F2 + F3 = G, whence F1 = F2 + F3 = G / 2 = 500 N.

Thus, the force in rod 1 is 500 N.

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