In Fig. 12-67a, a uniform$\mathbf{40}\mathbf{.0}\mathbf{}\mathit{k}\mathit{g}$beam is centered over two rollers. Vertical lines across the beam mark off equal lengths. Two of the lines are centered over the rollers; a $\mathbf{10}\mathbf{.0}\mathbf{}\mathit{k}\mathit{g}$package of tamales is centered over roller B.What are the magnitudes of the forces on the beam from (a) roller Aand (b) roller B? The beamis then rolled to the left until the right-hand end is centered over roller B(Fig. 12-67b).What now are the magnitudes of the forces on the beam from (c) roller Aand (d) roller B? Next, the beam is rolled to the right. Assume that it has a length of 0.800 m. (e) what horizontal distance between the package and roller Bputs the beam on the verge of losing contact with rollerA?

$Massofthebeam=40kg$

$Massofpackage=10kg$

$Lengthofbeam=0.8m$

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:

${\sum }_{}^{}{\text{F}}_{\text{y}}=0$

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

$\begin{array}{c}Weightofbeam\left(W_B\right)=m_{beam}\times gravitationalacceleration\\ =40kg\times 9.8m/{\text{s}}^2\\ =392\text{\hspace{0.17em}N}\end{array}$

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

From the equilibrium conditions,

${\sum }_{}^{}{\text{F}}_{\text{y}}=0$

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

$\begin{array}{l}{\sum }_{}^{}Momen{t}_B=0\\ (0.4m\times {\text{R}}_{\text{A}})-(392N\times 0.2m)=0\\ {\text{R}}_{\text{A}}=\frac{78.4N\cdot m}{0.4m}\\ {\text{R}}_{\text{A}}=196N\end{array}$

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

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,

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

Free body diagram for (a),

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

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

${\text{R}}_{\text{B}}=49N$

Free body diagram:

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

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

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

So we can write,

$\begin{array}{l}weightofpackage\times x-(weightofbeam\times (\frac{L}{4}-x))=0\\ 98N\times x=392N\times (\frac{L}{4}-x)\\ 98N\times x=\left(98N\right)L-\left(392N\right)x\\ \left(98N\right)x+\left(392N\right)x=98N\times 0.8m\\ \left(490N\right)x=78.4N\cdot m\\ x=\frac{78.4N\cdot m}{490N}\\ x=0.16m\end{array}$

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