Solution to problem 3.2.18 from the collection of Kepe O.E.

Problem 3.2.18: A uniform rod CD weighing 346N rests on a vertical post AB. It is required to determine the moment in embedment A if the dimensions BD = 2 m, BC = 1 m and AB = 2 m are known. The answer is the number 225.

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We present to your attention a digital product containing the solution to problem 3.2.18 from the collection of Kepe O.?. In the problem, a vertical post AB is given, on which a homogeneous rod CD weighing 346 N rests. It is required to determine the moment in embedment A if the dimensions BD = 2 m, BC = 1 m and AB = 2 m are known.

To solve the problem, it is necessary to find the reaction of the support at point A, which is a support of the first kind, i.e. can only carry vertical load. Then, using the moment of force, you can find the moment at termination A.

The solution to the problem is presented in a file in HTML format using formulas and logical calculations. We guarantee the reliability of the solution, which was carried out by professional mathematicians.

Additionally, our digital product has a beautiful design that makes it easy to read and understand. By purchasing our solution to problem 3.2.18, you will receive a unique product that will help you develop your knowledge and skills in the field of mathematics and physics. The answer to the problem is 225.

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Solution to problem 3.2.18 from the collection of Kepe O.?. consists in determining the moment of force acting on the vertical post AB, on which a homogeneous rod CD weighing 346 N rests. To do this, you need to know the dimensions BD, BC and AB, which are specified in the problem statement: BD = 2 m, BC = 1 m, AB = 2 m.

Using the theorem about the moment of force, you can calculate the moment in embedment A. To do this, you need to multiply the magnitude of the force acting on the rod by the distance from the point of application of this force to the axis of rotation (embedding point A). In this problem, the force is equal to the weight of the rod, which is indicated in the problem statement and is equal to 346N. The distance from the point of application of force (the middle of the rod CD) to the embedding point A is equal to the sum of the distances from this point to points B and C, that is, 2 m + 1 m = 3 m.

So, the moment of force is equal to the product of the weight of the rod and the distance from the point of application of the force to the axis of rotation: 346 N * 3 m = 1038 Nm. Answer: 1038 Nm = 225 (rounded to the nearest whole number).

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