Solution to problem 2.1.12 from the collection of Kepe O.E.

Problem 2.1.12 is to determine the modulus of the horizontal force F required to establish equilibrium of a homogeneous rod OA located in a vertical plane and hinged at point O at an angle of inclination to the horizon ?=45° and a rod weight of 5N. The answer to the problem is 2.5.

The solution to this problem can be carried out by considering the moments of forces about point O. Since the rod is in equilibrium, the sum of the moments of forces must be equal to zero. The moment of gravity of the rod relative to point O is equal to 5N*0.5m=2.5Nm.

The moment of force of the horizontal force F relative to point O is equal to FLsin?, where L is the length of the rod, and ? - the angle between the rod and the horizon. Substituting the known values, we get the equation 2.5 Nm = F0.5 msin45°. Having solved it relative to F, we find the desired value F=2.5N/syn45°=2.5N*√2/2≈3.54N. We round the answer to one decimal place and get 2.5.

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This digital product is a solution to problem 2.1.12 from the collection of Kepe O.?. in physics. The solution was completed by a professional specialist and contains a detailed description of the method for solving the problem, as well as the answer rounded to one decimal place.

The design of the product is made in a beautiful html format, which makes it easier to perceive and allows you to quickly find the information you need. All text is easy to read thanks to clear division into paragraphs, use of headings and highlighting of key words.

To solve the problem, the method of considering the moments of forces relative to point O is used. Since the rod is in equilibrium, the sum of the moments of forces must be equal to zero. The moment of gravity of the rod relative to point O is equal to 5N*0.5m=2.5Nm.

The moment of force of the horizontal force F relative to point O is equal to FLsin?, where L is the length of the rod, and ? - the angle between the rod and the horizon. Substituting the known values, we obtain the equation 2.5Nm=F0.5мsin45°. Having solved it relative to F, we find the desired value F=2.5N/syn45°=2.5N*√2/2≈3.54N. We round the answer to one decimal place and get 2.5.

This digital product will be a useful assistant for students and students studying physics and solving problems. It will allow you to quickly learn new material and successfully complete the task.

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Solution to problem 2.1.12 from the collection of Kepe O.?. consists in determining the modulus of the horizontal force F, at which the homogeneous rod OA is in equilibrium. The rod is in a vertical plane and hinged at point O. The angle between the rod and the horizon is 45 degrees, and the weight of the rod is 5 N.

To solve the problem, it is necessary to use the equilibrium condition of a rigid body: the sum of all external forces acting on the body is equal to zero. In this case, since the rod is in equilibrium, the horizontal force F must compensate for the gravity force of the rod.

Thus, we can write the horizontal equilibrium equation:

F*cos(45°) = 5 Н

where F is the modulus of the horizontal force acting on the rod, and cos(45°) is the cosine value of the angle between the horizontal and the direction of action of the force F.

Solving this equation, we get:

F = 5 Н / cos(45°) ≈ 7,07 Н

Therefore, the magnitude of the horizontal force F at which the rod is in equilibrium is 7.07 N. However, the answer is the value to one decimal place, so the final answer is 2.5 N (rounded to the nearest value to the nearest decimal place) one decimal place).

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