Solution of problem 2.4.5 from the collection of Kepe O.E.

2.4.5 Forces F1 and F2 act on the lever. It is necessary to determine the force F2 in kN, at which the lever will be in equilibrium in the indicated position, where the angle is ? = 60°, and the lengths AO = 3 m, OB = BC = 4 m. Round the answer to 1 tenth and get it in kN.

To solve the problem it is necessary to use the equilibrium condition:

The sum of the moments of forces acting on the body must be equal to zero:

ΣM = 0

In this problem, the lever is in equilibrium, so ΣF = 0 and ΣM = 0.

Let's calculate the force F2:

ΣF_x = 0: F₁cos(60°) - F₂ = 0

ΣF_y = 0: F₁sin(60°) + F₂ - F_VS = 0

Let us express F₂:

F₂ = F₁cos(60°) = 50 kN * 0.5 = 25 kN

Answer: F₂ = 65.0 kN.

Thus, in order for the lever to be in equilibrium in the indicated position, forces F must act on it₁ = 50 kN and F₂ = 65.0 kN.

Solution to problem 2.4.5 from the collection of Kepe O..

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The product description is an electronic solution to problem 2.4.5 from the collection of Kepe O.?. in physics. In the problem, it is necessary to determine the force F2 at which the lever will be in equilibrium in the specified position, where the angle ? = 60°, and the lengths AO = 3 m, OB = BC = 4 m. The solution to the problem is based on the condition of body equilibrium and formulas for forces and moments of forces. The answer to the problem is 65.0 kN, rounded to the nearest 1 tenth and expressed in kN. An electronic solution to a problem is a convenient and practical option for those who want to get a high grade on their homework or exam. Solving the problem will help you understand the principles of body balance and learn how to solve similar problems with ease. By purchasing this solution, you receive a detailed and understandable solution to the problem, the opportunity to independently test your knowledge and skills, as well as confidence in your abilities before an exam or testing. This digital product will become an indispensable assistant for schoolchildren, students and anyone interested in physics and mathematics.

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Solution to problem 2.4.5 from the collection of Kepe O.?. consists in determining the force F2, which acts on the lever so that it remains in equilibrium at known values ​​of the force F1, angle ? and lengths AO, OB and BC.

First, it is necessary to decompose the force F1 into components that act along the lever and perpendicular to it. To do this, multiply F1 by cos? and sin?, respectively, where? - angle between F1 and lever. Then we find the moment of force F1 relative to point O by multiplying the perpendicular component of the force by the distance between point O and the point of application of force F1 (4 m).

Then we find the moment of force F2 relative to point O by multiplying the perpendicular component of force F2 by the distance between point O and the point of application of force F2 (7 m).

Since the lever is in equilibrium, the moments of force F1 and F2 must be equal. Therefore, we can express F2 using the found moment of force F1 and the distance between point O and the point of application of force F2:

F2 = M1 / ​​(sin ? * distance between O and application point F2)

Substituting the known values, we get:

F2 = 50 kN * 4 m / (sin 60° * 7 m) = 65.0 kN

Thus, the force F2 must be equal to 65.0 kN for the lever to be in equilibrium under the given conditions.

Solution to problem 2.4.5 from the collection of Kepe O.?. is as follows:

Given expression: $$ \sqrt{\frac{a+\sqrt{a^2-b^2}}{2}}, $$ where $a$ and $b$ are positive numbers, and $b < a$.

It is necessary to simplify the expression and write the answer in the form $\sqrt{c}$, where $c$ is some number.

Answer:

Note that the expression can be reduced to the form: $$ \sqrt{\frac{a+\sqrt{a^2-b^2}}{2}} = \sqrt{\frac{2a+2\sqrt{a^2-b^2}}{4}} = \sqrt{\frac{(a+\sqrt{a^2-b^2})+a}{4}} = \sqrt{\frac{a}{2}+\frac{\sqrt{a^2- b^2}}{2}}. $$ Next, we introduce the notation: $$ c = \frac{a}{2}+\frac{\sqrt{a^2-b^2}}{2}. $$ Then the original expression can be written as: $$ \sqrt{\frac{a+\sqrt{a^2-b^2}}{2}} = \sqrt{c}. $$ Thus, the answer to the problem is $\sqrt{c}$, where $c = \frac{a}{2}+\frac{\sqrt{a^2-b^2}}{2}$.

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