A uniform aluminumbeam 9.00 m long, weighing 300 N, rests symmetrically on two supports 5.00 m apart (Fig. E 11.12). A boy weighing 600 N starts at point A and walks toward the right.

(a) In the same diagram construct two graphs showing the upward forces ${\mathit{F}}_{\mathbf{A}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{F}}_{\mathbf{B}}$exerted on the beam at points A and B, as functions of the coordinate x of the boy. Let 1 cm = 100 N vertically, and 1 cm = 1.00 m horizontally.

(b) From your diagram, how far beyond point B can the boy walk before the beam tips?

(c) How far from the right end of the beam should support B be placed so that the boy can walk just to the end of the beam without causing it to tip?

The condition for translational equilibrium can be written as :$\mathbf{\sum }{\mathit{F}}_{\mathbf{e}\mathbf{x}\mathbf{t}}\mathbf{=}\mathbf{0}$ .

And that for rotational equilibrium can be written as:$\mathbf{\sum }{\mathit{\tau }}_{\mathbf{e}\mathbf{x}\mathbf{t}}\mathbf{=}\mathbf{0}$ .

The beam, weighing 300 N, is supported at two points A and B, which are 5.00 meters apart. The boy is weighing 600 N walks from point A towards B.

Let the whole given setup be illustrated as a free body diagram for forces on the board, as shown in the figure below:

Now, for the torque ${\tau }_A$at point A, considering the criteria for rotational equilibrium, we have

$$\begin{gathered} \sum {\tau }_A=0 \\ {F}_B\left(5.00\right)=\left(300\right)\left(2.50\right)+\left(600\right)\left(x\right) \\ {F}_B=\left(150+120x\right)\text{N} \end{gathered}$$ .................(i)

Again, for the torque \({\tau _B}\) at point B, considering the criteria for rotational equilibrium, we have

$$\begin{gathered} \sum {\tau }_B=0 \\ {F}_A\left(5.00\right)=\left(300\right)\left(2.50\right)+\left(600\right)\left(5.00-x\right) \\ {F}_A=\left(750-120x\right)\text{N} \end{gathered}$$ ...............(ii)

Now, the graph showing forces $F_{A}and{F}_B$as a function of coordinate x of the boy can be as shown below:

Clearly, from the graph, we see that the force $F_A$is zero at:

$$\begin{gathered} F_A=0 \\ \left(750-120x\right)=0 \\ x=\frac{750}{120} \\ =6.25\text{m} \end{gathered}$$

This is 1.25 m beyond point B.

Thus, the boy can go to a distance of 1.25 m beyond point B before the beam tips.

Now, considering the boy's weight at the end of the beam, the new free body diagram will be as shown:

Now, for the torque ${\tau }_B$at point B, considering the criteria for rotational equilibrium, we have,

$$\begin{gathered} \sum {\tau }_B=0 \\ {F}_A\left(5.00\right)=\left(300\right)\left(4.50-x\right)-\left(600\right)\left(x\right) \end{gathered}$$

In the limiting case, we have $F_A=0$, so:

$$\begin{gathered} \left(300\right)\left(4.50-x\right)-\left(600\right)\left(x\right)=0 \\ 300\times 4.50=300x+600x \\ 900x=1350 \\ x=1.50\text{m} \end{gathered}$$

Thus, point B should be positioned at 1.50 m from the right side.