Solution to problem 1.2.20 from the collection of Kepe O.E.

Solution to problem 1.2.20 from the collection of Kepe O..

We present to your attention the solution to problem 1.2.20 from the collection of Kepe O.. in electronic form.

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We present to your attention the solution to problem 1.2.20 from the collection of Kepe O.?. The electronic version of this solution is an excellent choice for students and schoolchildren who study physics and want to understand the material more deeply and thoroughly.

To solve the problem, we need to determine the tension of the rope BC if the weight of the load G2 is 90N and the angles α=45°, β=60°.

Let's use the well-known formula to find the tension force of a rope:

T = (G2 + G1) / (sin α + sin β)

Let's substitute the known values ​​and get:

T = (90Н + G1) / (sin 45° + sin 60°)

To solve the equation, we need to find the weight of load G1. We use the load equilibrium condition:

G1 = G2 * sin α / sin β

We substitute the known values ​​and find:

G1 = 90N * sin 45° / sin 60° ≈ 51.96N

Now we can substitute the found values ​​into the original formula to find the rope tension:

T = (90N + 51.96N) / (sin 45° + sin 60°) ≈ 73.5N

Answer: 73.5N.

By purchasing this digital product, you receive a detailed solution to problem 1.2.20 with step-by-step instructions that make it easy to understand and remember the material. The solution can be used for independent work and preparation for exams. It will help improve your physics problem-solving skills and strengthen your understanding of theoretical concepts. In addition, this digital product is a convenient and affordable tool for studying physics and passing exams, which is always at hand on your computer or tablet. Don't miss the opportunity to purchase this product and improve your knowledge of physics!

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Solution to problem 1.2.20 from the collection of Kepe O.?. requires calculating the tension in the rope BC, which is necessary to keep two weights in balance. It is known that one of the loads has a weight G2 = 90 N and the angle of inclination of the rope BC to the horizon is equal to? = 45°, and the other load has an unknown weight G1 and the angle of inclination of the rope BC to the horizon is equal to ? = 60°.

To solve the problem it is necessary to use the laws of body equilibrium and the law of sines. According to the laws of equilibrium, the sum of all forces acting on a system of bodies must be equal to zero. It is also known that the tension of the rope BC is directed along the rope, and therefore, the tension vector of the rope BC and the gravity vector G2 form a right angle.

Using the law of sines, we can express the weight of the load G1 through the weight of the load G2 and the angles of inclination of the rope BC to the horizon:

G1/sin(60°) = G2/sin(45°)

From here we get:

G1 = G2 * sin(60°) / sin(45°) = 90 * sin(60°) / sin(45°) ≈ 104.1 Н

And finally, we calculate the tension in the rope BC using the equilibrium law:

BC = √(G1² + G2² + 2 * G1 * G2 * cos(60°)) ≈ 73.5 N

Thus, to keep the two weights in balance, it is necessary to apply a tension in the rope BC equal to approximately 73.5 N.

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