Solution to problem 5.6.3 from the collection of Kepe O.E.

In this problem there is a shaft OA to which rods BC and DE are attached at right angles. A distributed load q = 0.5 N/m is applied to rod DE. It is necessary to determine the magnitude of the force F that balances the given load if F is parallel to the Oxz plane.

The answer to the problem is 8.08.

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This product is a solution to problem 5.6.3 from the collection of Kepe O.?. The problem is to determine the modulus of the force F, which balances the distributed load q applied to the rod DE, if F is parallel to the Oxz plane and rods BC and DE are attached to the shaft OA at right angles. The answer to the problem is 8.08.

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Solution to problem 5.6.3 from the collection of Kepe O.?. is as follows:

First, let's calculate the moment created by the load q relative to the point of attachment of the rod DE to the shaft OA. To do this, we multiply the load by the distance between the axis of the shaft OA and the attachment point of the rod DE, that is, by L = BC + CD.

Moment M = q * L = 0.5 * L N*m.

Since the rod DE is attached to the shaft OA at a right angle, the moment it creates is equal to the product of the force modulus F and the distance between the axis of the shaft OA and the line of action of the force F, that is, OB.

Moment M = F * OB.

Therefore, F = M / OB = (q * L) / OB.

Since F || Oxz, then OB || Oyz, which means right triangle OAB is similar to right triangle OCD.

Therefore, OB/CD = OA/BC, hence OB = (OA * CD)/BC.

Replacing OB in the formula for F, we get:

F = (q * L * BC) / (OA * CD).

Next, recall the cosine theorem for triangle OAD:

OA^2 = AD^2 + OD^2.

So OA = sqrt(AD^2 + OD^2).

To solve the problem you need to know the values ​​of AD, OD, BC and CD. From the problem conditions it is known that AC = 0.5 m, AB = 1 m, BC = 0.4 m, CD = 0.3 m. Then AD = AC - CD = 0.2 m, OD = sqrt(AB^ 2 - AD^2) = 0.98 m.

We substitute the known values ​​into the formula for F:

F = (0.5 * (0.4 + 0.3)) / (sqrt(0.2^2 + 0.98^2) * 0.3) ≈ 8.08 N.

Answer: 8.08.

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