Question

Can you solve it descriptively . thanks

Answer 1

Answer:

$|M_y| = 170.82 \ N.mm$

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 $F_x \ , F_y \ , F_z$

So;

$F_x$ = positive x axis

$F_y =$ Negative y axis

$F_z$ = positive z axis

Then;

$|M_x| = F_y *27-F_z*11 = 77 ----- equation(1) \\ \\ |M_z| = F_y*4 - F_x*11 = 81 ---- equation (2) \\ \\ |M_y| = F_x *27 - F_z *4 = ? ---- equation (3)$

From equation (1); Let's make $F_y$ the subject of the formula ; then :

$F_y = \frac{77+11F_z}{27}$

Substituting  the value for $F_y$ into equation (2) ; we have:

$(\frac{77+11F_z}{27})4-F_x*11=81 \\ \\ 11(\frac{7+F_z}{27} ) 4- F_x -11 =81 \\ \\ 28+4 F_z - 27F_x = \frac{81*27}{11} \\ \\ 4F_z - 27F_x = 198.82 -28 \\ \\ 4F_z - 27F_x = 170.82 \\ \\ Since \ |M_y| = 4F_z-27F_x \\ \\ Then: \\ \\ \\ |M_y| = 170.82 \ N.mm$