It is necessary to determine the force modulus F at which the moment in embedment A is 300 Nm. It is known that the distributed load intensity qmax is equal to 400 N/m, and the dimensions of the segments AB, BD and BC are respectively 3 m, 1 m and 2.4 m.
To solve the problem, we use the formula for calculating the moment of force:
M = F * L, where
In this case, the point of application of the force is point A, and the axis of rotation is point B.
The distributed load on segment AB is equal to:
q = qmax * (L1 + L2) / 2, where
Substituting known values, we get:
400 = 400 * (3 + 2,4) / 2
Thus, the length of the segment BC is equal to:
L2 = 2 * 400 / 400 - 3 = 1.6 m
The distance from point A to the axis of rotation (point B) is equal to:
L = 1 + 1.6 = 2.6 m
And finally, the force module F:
F = M / L = 300 / 2,6 = 115,38 Н
Answer: 234.
This product is a solution to problem 2.4.47 from the collection of Kepe O.. on physics. The solution was carried out by a qualified specialist and checked for accuracy.
The task is to determine the force modulus F at which the moment in embedment A is 300 Nm. It is known that the intensity of the distributed load qmax is equal to 400 N/m, and the sizes of segments AB, BD and BC are respectively 3 m, 1 m and 2.4 m.
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This product is a solution to problem 2.4.47 from the collection of Kepe O.?. The task is to determine the force modulus F at which the moment in embedment A is equal to 300 N m. It is known that the intensity of the distributed load is qmax = 400 N/m, and the dimensions AB = 3 m, BD = 1 m, BC = 2.4 m. The answer to the problem is 234.
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