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    I used the centre of mass in order not to use...

    I used the centre of mass in order not to use integrations (not an expert on this...).
    According to the drawing the force F'=G*M*m/(r/cos(φ))2=G*m*π*d*R2*cos2(φ)/(2*r2), where M=π*R2*d/2 and...
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    The centre of mass of the upper semicircle is at...

    The centre of mass of the upper semicircle is at (x=0, y=4*R/(3*π)), and the centre of mass of the down semicircle is at (x=0, y=-4*R/(3*π)).
    I did not use the vertical cord.
    PS. It is a pity that...
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    !) π: is 3.14... 2)...

    !) π: is 3.14...
    2) https://math.stackexchange.com/questions/1459727/deriving-the-center-of-mass-of-a-semi-circular-disk-with-cylindrical-coordinates
    3) yes (not just double due to the projection...
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    I still feel that there is a mistake somewhere....

    I still feel that there is a mistake somewhere. (F' should be: F'=G*m*π*d/(2*cos(atan(4*R/(3*π*r)), and F=G*m*π*d. But this does not make sense)
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    Finally I found that: 1) we draw the diameter to...

    Finally I found that:
    1) we draw the diameter to the point (P) under consideration, that is considered as the x-axis.
    2) we find the centre of mass of the upper semicircle (0,4*R/(3*π)), where R:...
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    That's what I am asking for. The force would be...

    That's what I am asking for. The force would be directed also towards the centre.
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    You may be kind enough to help me.

    You may be kind enough to help me.
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    Gravity within disc.

    What is the gravity force on a unit mass inside an homogeneous disc of density d and radius r?
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