91Ó°ÊÓ

$Massofthebeam=40kg$

$Massofpackage=10kg$

$Lengthofbeam=0.8m$

From the Free Body Diagrams for the beam and using equilibrium conditions, we can find the reactions on rollers A and B.

Using the equilibrium conditions for net torque, we can find the horizontal distance between the package and roller B.

Equations:

${∑}_{}^{}{\text{F}}_{\text{y}}=0$

${∑}_{}^{}{\text{Moment}}_{\text{atanypointonthebeam}}=0$

$\begin{array}{c}Weightofbeam\left(W_B\right)=m_{beam}×gravitationalacceleration\\ =40\text{ }kg×9.8\text{ }m/{\text{s}}^2\\ =392\text{ N}\end{array}$

$\begin{array}{c}Weightofpackage\left(W_P\right)=m_{package}×gravitationalacceleration\\ =10\text{ }kg×9.8\text{ }m/s^2\\ =98\text{ }N\end{array}$

From the equilibrium conditions,

${∑}_{}^{}{\text{F}}_{\text{y}}=0$

$\begin{array}{l}{\text{R}}_{\text{A}}+{\text{R}}_{\text{B}}−392\text{ }N−98\text{ }N=0\\ {\text{R}}_{\text{A}}+{\text{R}}_{\text{B}}=490\text{ }N\end{array}$

$\begin{array}{l}{∑}_{}^{}Momen{t}_B=0\\ (0.4\text{ }m×{\text{R}}_{\text{A}})−(392\text{ }N×0.2\text{ }m)=0\\ {\text{R}}_{\text{A}}=\frac{78.4\text{ }Nâ‹\dots m}{0.4\text{ }m}\\ {\text{R}}_{\text{A}}=196\text{ }N\end{array}$

${\begin{array}{c}{\text{R}}_{\text{A}}+{\text{R}}_{\text{B}}=490\text{ }N\\ 196\text{ }N+R_B=490\text{ }N\\ R_B=294\text{ }N\end{array}}^{}$

Assuming that axis of rotation passes through point B, we can write the equations as

$\frac{L}{4}{W}_P+\frac{L}{2}{W}_B−\frac{L}{2}{R}_A=0$

Simplifying this we get,

$\begin{array}{c}{\text{R}}_{\text{A}}={\text{W}}_{\text{B}}+\frac{{\text{W}}_{\text{P}}}{\text{2}}\\ =392\text{ }N+\frac{98\text{ }N}{2}\\ =441\text{ â¶Ä‰}N\end{array}$

Free body diagram for (a),

$\begin{array}{l}{∑}_{}^{}{\text{F}}_{\text{y}}=0\\ R_A+R_B−392\text{ }N−98\text{ }N=0\\ R_A+R_B=490\text{ }N\end{array}$

Putting the value of${\text{R}}_{\text{A}}$we get,

${\text{R}}_{\text{B}}=49\text{ }N$

Free body diagram:

Assuming the axis of rotation passes through point B, we can write

${∑}_{}^{}{\text{Moment}}_{\text{B}}=0$

When the beam is about to lose contact with point A,${\text{R}}_{\text{A}}=0$,

So we can write,

$\begin{array}{l}weightofpackage×x-(weightofbeam×(\frac{L}{4}-x))=0\\ 98\text{ }N×x=392\text{ }N×(\frac{L}{4}−x)\\ 98\text{ }N×x=\left(98\text{ }N\right)\text{ }L−\left(392\text{ }N\right)x\\ \left(98\text{ }N\right)x+\left(392\text{ }N\right)x=98\text{ }N×0.8\text{ }m\\ \left(490\text{ }N\right)x=78.4\text{ }Nâ‹\dots m\\ x=\frac{78.4\text{ }Nâ‹\dots m}{490\text{ }N}\\ x\text{ }=0.16\text{ }m\end{array}$

$0.16\text{ }m$ Is the horizontal distance between the package and roller B that puts the beam on the verge of losing contact with roller A