Solution to problem 18.3.11 from the collection of Kepe O.E.

Determination of the modulus of the balancing force F applied to the crank OA

It is necessary to determine the modulus of the balancing force F acting on the crank OA at point A of the articulated four-link OABC, if a pair of forces with a moment M = 40 N • m acts on the connecting rod AB, and the length of the connecting rod AB is 0.4 m.

To solve the problem, we use the equilibrium condition of a mechanical system: the sum of all forces acting on the system is equal to zero.

In this case, two forces act on the crank OA: the balancing force F and a pair of forces acting on the connecting rod AB. A pair of forces can be represented in the form of two forces directed along the axis of the connecting rod AB and equal in magnitude, but opposite in direction. Thus, the sum of all forces acting on the system will be the vector sum of the balancing forces F and one of the two forces forming a pair.

From the equilibrium condition of a mechanical system it follows that the moment of the balancing force F must be equal in magnitude to the moment of the pair of forces acting on the connecting rod AB:

M = F * OA = 40 Н • м

where OA is the distance from point A to the axis of rotation (the center of the crank).

Thus, the modulus of the balancing force F will be equal to:

F = M / OA = 40 Н • м / OA

In order to calculate the distance OA from point A to the axis of rotation, we use the cosine theorem for triangle OAB:

OA^2 = AB^2 + OB^2 - 2 * AB * OB * cos(BOA)

where AB = 0.4 m is the length of the connecting rod, OB = BC = AC is the length of the connecting rod, BOA is the angle between the connecting rod and the connecting rod.

From the figure you can see that triangle OAB is a right triangle, so angle BOA is equal to angle BOC. You can also notice that triangle BOC is isosceles, so OB = BC = AC.

OA^2 = 0,4^2 + OB^2 - 2 * 0,4 * OB * cos(BOC)

OA^2 = 0.4^2 + OB^2 - 2 * 0.4 * OB * cos(2 * pi / 3)

Determination of the modulus of the balancing force F acting on the crank OA

It is necessary to determine the modulus of the balancing force F applied to the crank OA at point A of the articulated four-link OABC, if a pair of forces with a moment M = 40 N•m acts on the connecting rod AB, and the length of the connecting rod AB is 0.4 m.

To solve the problem, you can use the equilibrium condition of a mechanical system: the sum of all forces acting on the system must be equal to zero.

Two forces act on the crank OA: a balancing force F and a pair of forces acting on the connecting rod AB. A pair of forces can be represented in the form of two forces directed along the axis of the connecting rod AB and equal in magnitude, but opposite in direction. Thus, the sum of all forces acting on the system will be the vector sum of the balancing forces F and one of the two forces forming a pair.

From the equilibrium condition of a mechanical system it follows that the moment of the balancing force F must be equal in magnitude to the moment of the pair of forces acting on the connecting rod AB:

M = F × OA = 40 Н•м

where OA is the distance from point A to the axis of rotation (the center of the crank).

Therefore, the modulus of the balancing force F will be equal to:

F = M / OA = 40 Н•м / OA

To determine the distance OA from point A to the axis of rotation, you can use the cosine theorem for triangle OAB:

OA² = AB² + OB² - 2 × AB × OB × cos(BOA)

where AB = 0.4 m is the length of the connecting rod, OB = BC = AC is the length of the connecting rod, BOA is the angle between the connecting rod and the connecting rod.

Triangle OAB is a right triangle, so angle BOA is equal to angle BOC. You can also notice that triangle BOC is isosceles, so OB = BC = AC.

OA² = 0,4² + OB² - 2 × 0,4 × OB × cos(BOC)

OA² = 0.4² + OB² - 2 × 0.4 × OB × cos(2 × pi / 3)

Thus, the modulus of the balancing force F will be equal to 100 N

Solution to problem 18.3.11 from the collection of Kepe O.?.

This digital product is a solution to problem 18.3.11 from the famous collection “Problems in Theoretical Mechanics” by O.?. Kepe.

The solution to the problem was carried out by a qualified specialist and contains a detailed description of the solution process using formulas and graphic illustrations.

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This product is a solution to problem 18.3.11 from the collection “Problems in Theoretical Mechanics” by O.?. Kepe. The solution was completed by a qualified specialist and contains a detailed description of the solution process using formulas and graphic illustrations.

To solve the problem, it is necessary to use the equilibrium condition of a mechanical system: the sum of all forces acting on the system must be equal to zero. Two forces act on the crank OA: a balancing force F and a pair of forces acting on the connecting rod AB. A pair of forces can be represented in the form of two forces directed along the axis of the connecting rod AB and equal in magnitude, but opposite in direction. Thus, the sum of all forces acting on the system will be the vector sum of the balancing forces F and one of the two forces forming a pair.

From the equilibrium condition of a mechanical system it follows that the moment of the balancing force F must be equal in magnitude to the moment of the pair of forces acting on the connecting rod AB: M = F * OA = 40 N • m, where OA is the distance from point A to the axis of rotation (the center of the crank ). Therefore, the modulus of the balancing force F will be equal to: F = M / OA = 40 N • m / OA.

To determine the distance OA from point A to the axis of rotation, you can use the cosine theorem for triangle OAB: OA² = AB² + OB² - 2 × AB × OB × cos(BOA), where AB = 0.4 m is the length of the connecting rod, OB = BC = AC is the length of the connecting rod, BOA is the angle between the connecting rod and the connecting rod. Triangle OAB is a right triangle, so angle BOA is equal to angle BOC. You can also notice that triangle BOC is isosceles, so OB = BC = AC. Calculating using the formula, we find that the modulus of the balancing force F is equal to 100 N.

This product can be useful for students, teachers and anyone who is interested in theoretical mechanics and seeks to improve their knowledge and skills in this area. After purchasing, you will instantly receive access to the solution to the problem in PDF format.

This product is a solution to problem 18.3.11 from the collection "Problems in Theoretical Mechanics" O.?. Kepe. To solve the problem, it is necessary to use the equilibrium condition of a mechanical system: the sum of all forces acting on the system must be equal to zero. Two forces act on the crank OA: a balancing force F and a pair of forces acting on the connecting rod AB. A pair of forces can be represented in the form of two forces directed along the axis of the connecting rod AB and equal in magnitude, but opposite in direction. Thus, the sum of all forces acting on the system will be the vector sum of the balancing forces F and one of the two forces forming a pair.

From the equilibrium condition of a mechanical system it follows that the moment of the balancing force F must be equal in magnitude to the moment of the pair of forces acting on the connecting rod AB: M = F * OA = 40 N • m, where OA is the distance from point A to the axis of rotation (the center of the crank ). Therefore, the modulus of the balancing force F will be equal to: F = M / OA = 40 N • m / OA.

To determine the distance OA from point A to the axis of rotation, you can use the cosine theorem for triangle OAB: OA² = AB² + OB² - 2 × AB × OB × cos(BOA), where AB = 0.4 m is the length of the connecting rod, OB = BC = AC is the length of the connecting rod, BOA is the angle between the connecting rod and the connecting rod. Triangle OAB is a right triangle, so angle BOA is equal to angle BOC. You can also notice that triangle BOC is isosceles, so OB = BC = AC.

Based on the formulas obtained, the modulus of the balancing force F will be equal to 100 N. After purchasing the product, you will be able to access a file with a detailed description of the process of solving the problem, which includes the use of formulas and graphic illustrations. This product is recommended for students and teachers interested in theoretical mechanics and seeking to improve their knowledge and skills in this field.

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Problem 18.3.11 from the collection of Kepe O.?. consists in determining the modulus of the balancing force F applied to the crank OA at point A of the articulated four-bar OABC. It is given that a pair of forces with a moment M = 40 N • m acts on the connecting rod AB, and the length of the connecting rod is 0.4 m. It is required to find the value of the modulus F.

To solve the problem, it is necessary to use the moment equilibrium condition, which states: the sum of the moments of forces acting on the body is equal to zero. In this case, since the crank is in equilibrium, the moment of the balancing force must be equal to the moment of the pair of forces.

The moment of a pair of forces can be found by the formula M = F * l, where F is the force modulus, l is the distance from the point of application of the force to the axis of rotation. From the problem conditions it is known that M = 40 N • m, and l = 0.4 m.

Thus, substituting the known values ​​into the formula for the moment of a pair of forces, we obtain the equation: 40 N • m = F * 0.4 m, whence F = 40 N • m / 0.4 m = 100 N.

Answer: the modulus of the balancing force F applied to the crank OA at point A of the four-bar hinge OABC is equal to 100 N.

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