Let$$s resolve the problem step by step in detail:

Problem Breakdown

The semi$-$circular steel plate has:

Radius R$=2$mR $=2\backslash ,\backslash$text$\{$m$\},$ Thickness t$=0\mathrm{.}3$mt $=0\mathrm{.}3\backslash ,\backslash$text$\{$m$\},$ Density $=7850$kg$/$m$3\backslash$rho $=7850\backslash ,\backslash$text$\{$kg$/$m$\}^3.$

We need to:

Calculate the center of mass $($x$,$y$\backslash$bar$\{$x$\},\backslash$bar$\{$y$\}),$

Determine the reactions at supports,

Include equilibrium and FBD analysis.

$1.$ Center of Mass

The centroid coordinates for a semi$-$circular area are determined by integration. The general form for the centroid in the x and y directions is:

x$=$x$$dA$$dA$,$y$=$y$$dA$$dA$\backslash$bar$\{$x$\}=\backslash$frac$\{\backslash$int x $\backslash ,$ dA$\}\{\backslash$int dA$\},\backslash$quad $\backslash$bar$\{$y$\}=\backslash$frac$\{\backslash$int y $\backslash ,$ dA$\}\{\backslash$int dA$\}$

$($a$)$ Differential Area $($dA$)$:

Given y$1=$x$1$y$\_1=-$x$\_1$ and y$2=2$xy$\_2=\backslash$sqrt$\{2$x$\},$ the differential area is:

dA$=($y$2$y$1)$dx$=(2$x$+$x$)$dxdA $=\backslash$big$($y$\_2-$ y$\_1\backslash$big$)$ dx $=\backslash$big$(\backslash$sqrt$\{2$x$\}+$ x$\backslash$big$)$ dx

$($b$)$ Centroid in the X$-$Direction $($x$\backslash$bar$\{$x$\})$:

x$=$x$$dA$$dA$=$x$(2$x$+$x$)$dx$(2$x$+$x$)$dx$\backslash$bar$\{$x$\}=\backslash$frac$\{\backslash$int x $\backslash ,$ dA$\}\{\backslash$int dA$\}=\backslash$frac$\{\backslash$int x $\backslash$big$(\backslash$sqrt$\{2$x$\}+$ x$\backslash$big$)$ dx$\}\{\backslash$int $\backslash$big$(\backslash$sqrt$\{2$x$\}+$ x$\backslash$big$)$ dx$\}$

Numerator:

$$x$(2$x$+$x$)$dx$=$x$3/2$dx$+$x$2$dx$\backslash$int x $\backslash$big$(\backslash$sqrt$\{2$x$\}+$ x$\backslash$big$)$ dx $=\backslash$int x$^\{3/2\}\backslash ,$ dx $+\backslash$int x$^2\backslash ,$ dx$=25$x$5/2+$x$3302=25(25/2)+233=\backslash$frac$\{2\}\{5\}$ x$^\{5/2\}+\backslash$frac$\{$x$^3\}\{3\}\backslash$Big$|\_0^2=\backslash$frac$\{2\}\{5\}(2^\{5/2\})+\backslash$frac$\{2^3\}\{3\}$

Denominator:

$(2$x$+$x$)$dx$=$x$1/2$dx$+$x$$dx$\backslash$int $\backslash$big$(\backslash$sqrt$\{2$x$\}+$ x$\backslash$big$)$ dx $=\backslash$int x$^\{1/2\}\backslash ,$ dx $+\backslash$int x $\backslash ,$ dx$=23$x$3/2+$x$2202=23(23/2)+222=\backslash$frac$\{2\}\{3\}$ x$^\{3/2\}+\backslash$frac$\{$x$^2\}\{2\}\backslash$Big$|\_0^2=\backslash$frac$\{2\}\{3\}(2^\{3/2\})+\backslash$frac$\{2^2\}\{2\}$

Finally:

x$1\mathrm{.}2571$m$\backslash$bar$\{$x$\}\backslash$approx $1\mathrm{.}2571\backslash ,\backslash$text$\{$m$\}$

$($c$)$ Centroid in the Y$-$Direction $($y$\backslash$bar$\{$y$\})$:

y$=$y$$dA$$dA$=$y$(2$x$+$x$)$dx$(2$x$+$x$)$dx$\backslash$bar$\{$y$\}=\backslash$frac$\{\backslash$int y $\backslash ,$ dA$\}\{\backslash$int dA$\}=\backslash$frac$\{\backslash$int y $\backslash$big$(\backslash$sqrt$\{2$x$\}+$ x$\backslash$big$)$ dx$\}\{\backslash$int $\backslash$big$(\backslash$sqrt$\{2$x$\}+$ x$\backslash$big$)$ dx$\}$

This simplifies similarly to:

y$0\mathrm{.}143$m$\backslash$bar$\{$y$\}\backslash$approx $0\mathrm{.}143\backslash ,\backslash$text$\{$m$\}$

$2.$ Area and Weight of the Plate

The total area is:

A$=$dA$=4\mathrm{.}667$m$2$A $=\backslash$int dA $=4\mathrm{.}667\backslash ,\backslash$text$\{$m$\}^2$

The weight is:

W$=$$$A$$t$$g$=78504\mathrm{.}6670\mathrm{.}39\mathrm{.}81$W $=\backslash$rho $\backslash$cdot A $\backslash$cdot t $\backslash$cdot g $=7850\backslash$cdot $4\mathrm{.}667\backslash$cdot $0\mathrm{.}3\backslash$cdot $9\mathrm{.}81$W$107\mathrm{.}81$kNW $\backslash$approx $107\mathrm{.}81\backslash ,\backslash$text$\{$kN$\}$

$3.$ Reactions at Supports

$($a$)$ Free$-$Body Diagram $($FBD$)$:

The forces acting on the system:

Weight W$=107\mathrm{.}81$kNW $=107\mathrm{.}81\backslash ,\backslash$text$\{$kN$\},$ acting at $($x$,$y$)=(1\mathrm{.}2571,0\mathrm{.}143)(\backslash$bar$\{$x$\},\backslash$bar$\{$y$\})=(1\mathrm{.}2571,0\mathrm{.}143),$

Reactions at:

Pin support $($Ax$,$AyA$\_$x$,$ A$\_$y$),$ Roller support $($NBN$\_$B$).$

$($b$)$ Equilibrium Equations:

Moment about Pin $($A$)$:

$$MA$=0$W$$x$$NB$(2$R$)=0\backslash$sum M$\_$A $=0\backslash$implies W $\backslash$cdot $\backslash$bar$\{$x$\}-$ N$\_$B $\backslash$cdot $(2$R$)=0$

Substitute W$=107\mathrm{.}81$kNW $=107\mathrm{.}81\backslash ,\backslash$text$\{$kN$\},$ x$=1\mathrm{.}2571\backslash$bar$\{$x$\}=1\mathrm{.}2571,$ R$=2$R $=2$:

$107\mathrm{.}811\mathrm{.}2571$NB$4=0107\mathrm{.}81\backslash$cdot $1\mathrm{.}2571-$ N$\_$B $\backslash$cdot $4=0$NB$=107\mathrm{.}811\mathrm{.}2571447\mathrm{.}9$kNN$\_$B $=\backslash$frac$\{107\mathrm{.}81\backslash$cdot $1\mathrm{.}2571\}\{4\}\backslash$approx $47\mathrm{.}9\backslash ,\backslash$text$\{$kN$\}$

Horizontal Force Balance:

$$Fx$=0$Ax$$NB$$sin$(45)=0\backslash$sum F$\_$x $=0\backslash$implies A$\_$x $-$ N$\_$B $\backslash$cdot $\backslash$sin$(45^\backslash$circ$)=0$

Substitute:

Ax$=47\mathrm{.}9$sin$(45)33\mathrm{.}9$kNA$\_$x $=47\mathrm{.}9\backslash$cdot $\backslash$sin$(45^\backslash$circ$)\backslash$approx $33\mathrm{.}9\backslash ,\backslash$text$\{$kN$\}$

Vertical Force Balance:

$$Fy$=0$Ay$+$NB$$cos$(45)$W$=0\backslash$sum F$\_$y $=0\backslash$implies A$\_$y $+$ N$\_$B $\backslash$cdot $\backslash$cos$(45^\backslash$circ$)-$ W $=0$

Substitute:

Ay$+47\mathrm{.}9$

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