Question

A uniform aluminium beam 9.00 m long, weighing 300 N, rests symmetrically on two supports 5.00 m apart (Fig. E11.12). 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 ${\mathit{F}}_{\mathbf{A}}\mathbf{}\mathbf{and}\mathbf{}{\mathit{F}}_{\mathbf{B}}$ 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?

Short answer

1. 
2. The boy can go to a distance of 1.25 m beyond point B before the beam tips.
3. Point B should be positioned at 1.50 m from the right side.

Step-by-step solution

Equilibrium

The condition for translational equilibrium can be written as : $\sum _{}{F}_{ext}=0$

And that for rotational equilibrium can be written as:$\sum _{}{\tau }_{ext}=0$

Graph the Forces (a)

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

$\begin{array}{r}\sum {\tau }_A=0\\ F_B(5.00)=\left(300\right)(2.50)+\left(600\right)\left(x\right)\\ F_B=\left(150+120x\right)N.....................\left(1\right)\end{array}$

Again, for the torque at point B, considering the criteria for rotational equilibrium, we have

$\begin{array}{r}\sum {\tau }_B=0\\ F_A(5.00)=\left(300\right)(2.50)+\left(600\right)(5.00-x)\\ F_A=\left(750-120x\right)N.....................\left(1\right)\end{array}$

Now, the graph showing forces as a function of coordinate x of the boy can be as shown below:

Find the Position(b)

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

$\begin{array}{r}{F}_A=0\\ (750-120x)=0\\ x=\frac{750}{120}\\ =6.25\mathrm{m}\end{array}$

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.

Find the Position (c)

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,

$\begin{array}{r}\sum {\tau }_B=0\\ F_A\left(5.00\right)=\left(300\right)(4.50-x)-\left(600\right)\left(x\right)\end{array}$

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

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

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