Solution to problem 5.6.7 from the collection of Kepe O.E.

5.6.7. Solving the problem of a homogeneous square frame OABC using the equation of moments of forces.

Let's consider a homogeneous square frame OABC with side a = 0.5 m and weight G = 140 N, which is held in a horizontal position under the influence of imposed bonds. It is necessary to create an equation for the moments of forces relative to line OB and determine the reaction ZA of hinge A, provided that the angle α = 60°.

To begin with, let's draw a diagram of the frame and designate all the known and unknown quantities:

Using the equation of moments of forces, we create the following equation for the frame:

Ma - A * a/2 - G * a/2 * cos(α) = 0,

where Ma is the moment of force relative to the line OB, ZA is the reaction of the hinge A, α is the angle of inclination of the frame, a is the length of the side of the square, G is the weight of the frame.

From the problem conditions it is known that α = 60°, a = 0.5 m and G = 140 N. Substituting these values ​​into the equation, we obtain:

Ma - ZA * 0.25 - 70 = 0.

To determine the reaction ZA, it is necessary to solve the equation for ZA:

ZA = (Ma - 70) / 0.25.

Thus, we have obtained the equation of moments of forces for the frame and determined the reaction ZA of hinge A.

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Solution to problem 5.6.7 from the collection of Kepe O.?. is as follows:

Given a homogeneous square frame OABC with a side a = 0.5 m and a weight G = 140 N, held in a horizontal position with applied bonds. It is necessary to create an equation for the moments of force relative to line OB and determine the reaction ZA of hinge A at an angle α = 60°.

First you need to draw a diagram of the frame and indicate the known values:

https://i.imgur.com/1YD2EaM.png

Here M is the moment of force relative to line OB, ZA is the reaction of hinge A, G is the weight of the frame, α is the angle between the horizontal and line OA.

Let's create an equation for the moments of forces relative to the line OB:

M = -ZA × OA × sinα + G × a/2 × cosα

where OA = a/√2 is the length of the diagonal of the square OABC.

Substituting the known values ​​and the angle α = 60°, we obtain:

M = -ZA × 0.25 + 70

To determine the reaction ZA, we use the equilibrium condition at vertex A:

ZA = G/2 × cosα + M/0,5 × sinα

Substituting the known values ​​and the angle α = 60°, we obtain:

ZA = 70√3 + 35

Thus, the reaction ZA of hinge A is equal to 70√3 + 35 N, and the equation of the moments of force relative to the line OB has the form M = -ZA × 0.25 + 70.

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