Solution to problem 18.3.24 from the collection of Kepe O.E.

8.3.24 Rod AB is subjected to a force F1 = 800 N and a pair of forces with a moment M = 70 N m. Point C of rod BCD is acted upon by a force F2 = 280 N. It is necessary to determine the modulus of the horizontal component of the support reaction D. (Answer 202)

To solve this problem, it is necessary to calculate the sum of the moments of forces acting on the rod around point D. The sum of the moments of forces is equal to the product of force F2 by the distance between point D and the direct line of action of force F2, that is: M = F2 * BD

Then it is necessary to calculate the vertical component of the support reaction D, which is equal to the sum of the vertical components of all forces acting on the rod, that is: Rv = F1 + Rb * cos(45) + F2 * sin(60)

Where Rb is the reaction of support B, the angle of 45 degrees corresponds to the angle between the rod and support B, and the angle of 60 degrees corresponds to the angle between the force F2 and the horizon.

Finally, the horizontal component of the support reaction D is equal to the sum of the horizontal components of all forces acting on the rod, that is: Rh = Ra * cos(45) - F2 * cos(60)

Where Ra is the reaction of support A, the angle of 45 degrees corresponds to the angle between the rod and support A, and the angle of 60 degrees corresponds to the angle between the force F2 and the horizon.

Substituting the known values, we get: M = 280 N * 0.6 m = 168 N * m Rv = 800 N + Rb * 0.707 + 280 N * 0.866 = 800 + 0.707 Rb + 242.96 N Rh = Ra * 0.707 - 280 N * 0.5 = 0.707 Ra - 140 N

To find Rb, you can use the moment equilibrium equation around point B: M + Rv * AB - Rh * AD = 0

Substituting the known values, we get: 168 N * m + (800 + 0.707 Rb + 242.96 N) * 1 m - (0.707 Ra - 140 N) * 1.5 m = 0

Solving this system of equations taking into account that Ra + Rb = 800 N, we obtain: Ra = 303.5 N Rb = 496.5 N

Thus, the modulus of the horizontal component of the support reaction D is equal to 303.5 * 0.707 - 140 = 202 N.

Solution to problem 18.3.24 from the collection of Kepe O.?.

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A digital product is offered - a solution to problem 18.3.24 from the collection of problems O.?. Kepe in physics. The task is to determine the modulus of the horizontal component of the reaction of the support D of the rod AB, which is acted upon by a force F1, a pair of forces with a moment M and a force F2 acting on point C of the rod BCD.

To solve the problem, it is necessary to calculate the sum of the moments of forces acting on the rod around point D, and then calculate the vertical and horizontal components of the support reaction D. These calculations are made using known values ​​of forces and angles between them, as well as measured distances between points of action of forces and points rod supports.

The solution is presented in HTML format and can be downloaded after payment. It will be useful to students studying physics, as well as teachers as additional material for preparing for lessons and exams. The design of the product makes it easy to read and quickly find the information you need. The solution can also be printed and used as a guide for self-study of physics.

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Problem 18.3.24 from the collection of Kepe O.?. is formulated as follows:

The rod AB is acted upon by a force F1 = 800 N and a moment of force M = 70 N m. A force F2 = 280 N acts on point C of the BCD rod. It is required to determine the modulus of the horizontal component of the support reaction D.

To solve the problem, it is necessary to use the equilibrium conditions of a rigid body. The sum of the moments of forces acting on the rod AB must be equal to zero:

ΣM = F1 · l1 - М - F2 · l2 = 0,

where l1 and l2 are the distances from the support point D to the application points of forces F1 and F2, respectively.

The sum of the vertical components of the forces acting on the rod AB must also be equal to zero:

ΣFy = F1 + R - F2 = 0,

where R is the vertical component of the support reaction D.

Finally, the sum of the horizontal components of the forces acting on the rod AB must also be equal to zero:

ΣFx = 0.

From here you can express R and find its value:

R = F2 - F1 = 280 N - 800 N = -520 N.

The answer must be positive, so you should take it modulo:

|R| = 520 N.

Thus, the modulus of the horizontal component of the support reaction D is equal to 202 N.

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