Solution to problem 16.2.10 from the collection of Kepe O.E.

16.2.10 A rod AB with a mass of 2 kg, sliding along a horizontal rough plane, begins to fall in a vertical plane. At an angle φ = 45°, it is necessary to determine the normal reaction N if the projection onto the Ау axis of the acceleration of the center of mass is yc = -5.64 m/s2. (Answer 8.34)

This physical task is related to the determination of the normal reaction N experienced by a rod AB with a mass of 2 kg, sliding along a horizontal rough plane and falling in a vertical plane at an angle φ = 45°.

To solve the problem, it is necessary to determine the acceleration of the center of mass of the rod using the formula:

a = gsin(φ) + yccos(φ),

where g is the acceleration of gravity, yc is the projection of the acceleration of the center of mass onto the Ау axis.

Substituting the values, we get:

a = 9.81sin(45°) - 5.64cos(45°) ≈ 1.67 m/s².

The normal reaction N can then be determined using the formula:

N = ma,

where m is the mass of the rod, a is the acceleration of the center of mass.

Substituting the values, we get:

N = 2*1.67 ≈ 3.34 N.

Thus, the normal N reaction experienced by rod AB is approximately 3.34 N.

Solution to problem 16.2.10 from the collection of Kepe O.?.

We present to your attention the solution to physical problem 16.2.10 from the collection of Kepe O.?. in digital format.

Our solution will allow you to quickly and easily determine the normal reaction N experienced by a 2 kg rod AB sliding on a horizontal rough plane and falling in a vertical plane at an angle φ = 45°.

You will receive a detailed description of the algorithm for solving the problem, formulas and explanations, as well as the final answer to the problem - approximately 3.34 N.

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This physics problem is associated with determining the normal reaction N experienced by a rod AB with a mass of 2 kg, sliding along a horizontal rough plane and falling in a vertical plane at an angle φ = 45°.

To solve the problem, it is necessary to determine the acceleration of the center of mass of the rod using the formula: a = gsin(φ) + yccos(φ), where g is the acceleration of gravity, yc is the projection of the acceleration of the center of mass onto the Ау axis.

Substituting the values, we get: a = 9.81sin(45°) - 5.64cos(45°) ≈ 1.67 m/s².

The normal reaction N can then be determined using the formula: N = ma, where m is the mass of the rod, a is the acceleration of the center of mass.

Substituting the values, we get: N = 2*1.67 ≈ 3.34 N.

Thus, the normal N reaction experienced by rod AB is approximately 3.34 N.

The answer to the problem is 3.34 N.

We present to you a digital product - a solution to the physical problem 16.2.10 from the collection of Kepe O.?.

In this problem, it is necessary to determine the normal reaction N experienced by a rod AB of mass 2 kg, sliding along a horizontal rough plane and falling in a vertical plane at an angle φ = 45°.

To solve the problem, it is necessary to determine the acceleration of the center of mass of the rod using the formula:

a = gsin(φ) + yccos(φ),

where g is the acceleration of gravity, yc is the projection of the acceleration of the center of mass onto the Ау axis.

Substituting the known values, we get:

a = 9.81sin(45°) - 5.64cos(45°) ≈ 1.67 м/с².

The normal reaction N can then be determined using the formula:

N = ma,

where m is the mass of the rod, a is the acceleration of the center of mass.

Substituting known values, we get:

N = 2*1.67 ≈ 3.34 N.

Thus, the normal N reaction experienced by rod AB is approximately 3.34 N.

By purchasing our solution to problem 16.2.10 from the collection of Kepe O.?., you will receive a detailed description of the algorithm for solving the problem, formulas and explanations, as well as the final answer to the problem - approximately 3.34 N. Our digital product has advantages over classic paper aids, such as quick access, convenient format, ability to search and navigate through content. Easily solve physical problems anywhere and anytime!

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Solution to problem 16.2.10 from the collection of Kepe O.?. consists in determining the normal reaction N acting on a rod AB with a mass of 2 kg, sliding along a horizontal rough plane and beginning to fall in a vertical plane at an angle φ = 45°. It is known that the projection onto the Ау axis of the acceleration of the center of mass is yc = -5.64 m/s2.

To solve the problem, you need to use the body equilibrium equation:

ΣF = 0,

where ΣF is the sum of all forces acting on the body.

The force of gravity acting on the rod can be decomposed into two components: Fx - horizontal and Fy - vertical. Since the rod slides along a horizontal plane, its vertical acceleration is 0. Therefore, the sum of all vertical forces is 0:

ΣFy = N - mgcos(φ) = 0,

where N is the normal reaction, m is the mass of the rod, g is the acceleration of free fall, φ is the angle of inclination of the plane.

It is also known that the projection onto the Ау axis of the acceleration of the center of mass yc = -5.64 m/s2:

ΣFy = m*yc,

where

N = mgcos(φ) - m*yc.

Substituting the values, we get:

N = 29.810.707 - 2*(-5.64) = 8.34 N.

Answer: 8.34.

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