Answer:
[tex]|M_y| = 170.82 \ N.mm[/tex]
Explanation:
From the diagram affixed below completes the question
Now from the diagram; We need to resolve the force at point A into (3) components ; i.e x.y. & z directions which are equivalent to [tex]F_x \ , F_y \ , F_z[/tex]
So;
[tex]F_x[/tex] = positive x axis
[tex]F_y =[/tex] Negative y axis
[tex]F_z[/tex] = positive z axis
Then;
[tex]|M_x| = F_y *27-F_z*11 = 77 ----- equation(1) \\ \\ |M_z| = F_y*4 - F_x*11 = 81 ---- equation (2) \\ \\ |M_y| = F_x *27 - F_z *4 = ? ---- equation (3)[/tex]
From equation (1); Let's make [tex]F_y[/tex] the subject of the formula ; then :
[tex]F_y = \frac{77+11F_z}{27}[/tex]
Substituting the value for [tex]F_y[/tex] into equation (2) ; we have:
[tex](\frac{77+11F_z}{27})4-F_x*11=81 \\ \\ 11(\frac{7+F_z}{27} ) 4- F_x -11 =81 \\ \\ 28+4 F_z - 27F_x = \frac{81*27}{11} \\ \\ 4F_z - 27F_x = 198.82 -28 \\ \\ 4F_z - 27F_x = 170.82 \\ \\ Since \ |M_y| = 4F_z-27F_x \\ \\ Then: \\ \\ \\ |M_y| = 170.82 \ N.mm[/tex]
