Solution to problem 5.1.12 from the collection of Kepe O.E.

Let's calculate the moment of the distributed load relative to the Oy axis if q || Oz. We have:

q = 3N/m, OA = 2m, AB = 3m.

Answer: 31.5.

Solution to problem 5.1.12 from the collection of Kepe O..

This solution is a digital product intended for those who solve problems in theoretical mechanics. The solution to problem 5.1.12 from the collection of Kepe O.. was carried out by a professional specialist with extensive experience in this field and checked for accuracy.

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A digital product is offered in PDF format - a solution to problem 5.1.12 from the collection of Kepe O.?. in theoretical mechanics. The solution was carried out by a professional specialist with extensive experience in this field and checked for accuracy.

By purchasing this product, you will receive a detailed solution to the problem with a step-by-step description of each step, as well as illustrations and graphics necessary for a more clear understanding of the solution. The document is designed in a beautiful and understandable html format, which allows you to conveniently and quickly find the necessary information and easily navigate the document.

In this problem, it is required to determine the moment of the distributed load relative to the Oy axis if the load is q ||Oz, and the values ​​OA = 2m and AB = 3m, q = 3N/m are given. The answer to the problem is 31.5.

This digital product can be used to prepare for exams, tests, and self-study. By purchasing it, you save your time and get a ready-made solution to the problem from a professional.

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Solution to problem 5.1.12 from the collection of Kepe O.?. consists in determining the moment of the distributed load relative to the Oy axis under given conditions. It is known that the load q is parallel to the Oz axis, the length of the segment OA is 2 meters, and the length of the segment AB is 3 meters. The task is to determine the moment of this load relative to the Oy axis.

To solve the problem, you need to use the formula to determine the moment of force about a given axis. In this case, the moment of force M will be equal to the product of force q and the distance from the Oy axis to the center of gravity of the distributed load. The distance from the Oy axis to the center of gravity of the distributed load can be determined by dividing the length of the segment OA by 2 and adding to this result the length of the segment AB.

So, applying the formula, we get:

M = q * ((OA/2) + AB) = 3 N/m * ((2 m / 2) + 3 m) = 3 N/m * 4 m = 12 Nm

Thus, the moment of the distributed load relative to the Oy axis is equal to 12 Nm. Answer: 12 Nm.

Problem 5.1.12 from the collection of problems by Kepe O.?. is formulated as follows:

“On a smooth horizontal surface located at an angle of inclination α to the horizon, a small cylinder of radius r and mass m lies. Find the period of vertical oscillations of the cylinder along the vertical axis passing through its center of mass.”

To solve the problem, it is necessary to use the laws of dynamics and formulas of solid body mechanics. First, the forces acting on the cylinder are determined: the weight m*g, directed vertically downward, and the normal reaction force of the surface, directed perpendicular to the surface. Then we can write the equation of motion of the cylinder relative to the vertical axis in the form:

mr^2(d^2θ/dt^2) = -mgrsin a + Nr*sin a

where θ is the angle of rotation of the cylinder, N is the normal reaction force of the surface.

By solving this equation, we can obtain the period of oscillation of the cylinder:

T = 2πsqrt(m/(mg*sin α - N))

To find the normal reaction force N, it is necessary to use the equilibrium condition along an axis perpendicular to the inclined surface:

Ncos α = mg*cos a

From here we can express the strength of the normal reaction N and substitute it into the formula for the oscillation period.

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