91Ó°ÊÓ

The mass of gymnasts,$\mathrm{m}=46.0\text{ k²µ}$

The length of the beam,$\mathrm{L}=5.00\text{″¾}$

The mass of the beam,$\mathrm{M}=250\text{ k²µ}$

The supports have given to the beam from both ends at a distance, $\mathrm{d}=0.540\text{″¾}$.

You draw the free body diagram. The system is at equilibrium. For such a system, the vector sum of the forces acting on it is zero. The vector sum of the external torques acting on the ladder at the pivot point can be taken as zero. The formulas are given below.

$\begin{array}{c}\mathbf{Î\pounds }{\stackrel{\mathbf{→}}{\mathbf{F}}}_{\mathbf{net}}\mathbf{=}\mathbf{0}\\ \mathbf{Î\pounds }{\stackrel{\mathbf{→}}{\mathbf{Ï„}}}_{\mathbf{net}}\mathbf{=}\mathbf{0}\end{array}$

The ${\mathrm{F}}_1$ and ${\mathrm{F}}_2$ are the forces acting on the beam due to the supports as shown in the figure. We can choose the axis of rotation at support$2$. The beam can be rotated at support$2$in a counter-clockwise direction; hence the torque will be positive otherwise it will be negative. This is a static equilibrium condition for the beam. The force${\mathrm{F}}_2$is acting at the axis of rotation hence the moment of the arm for it is zero.

For the static equilibrium condition, the vector sum of the external torque acting on the ladder at about any point is zero.

localid="1662098501677" $\begin{array}{l}{\mathrm{Î\pounds Ï„}}_{\mathrm{net}}=0\\ \mathrm{Mg}\left[\left(\frac{\mathrm{L}}{2}−\mathrm{d}\right)\right]−\mathrm{mg}\left(\mathrm{d}\right)−{\mathrm{F}}_1\left[2\left(\frac{\mathrm{L}}{2}−\mathrm{d}\right)\right]=0\\ \mathrm{Mg}\left[\left(\frac{\mathrm{L}}{2}−\mathrm{d}\right)\right]−\mathrm{mg}\left(\mathrm{d}\right)={\mathrm{F}}_1\left[2\left(\frac{\mathrm{L}}{2}−\mathrm{d}\right)\right]\\ {\mathrm{F}}_1=\frac{\mathrm{Mg}\left[\left(\frac{\mathrm{L}}{2}−\mathrm{d}\right)\right]−\mathrm{mg}\left(\mathrm{d}\right)}{\left[2\left(\frac{\mathrm{L}}{2}−\mathrm{d}\right)\right]}\\ {\mathrm{F}}_1=\frac{(250\mathrm{kg}×9.8{\text{m/s}}^{\text{2}})\left[\left(\frac{5.00\mathrm{m}}{2}−0.540\mathrm{m}\right)\right]−(46.0\mathrm{kg}×9.8{\text{″¾/s}}^{\text{2}}×0.540\text{ }\mathrm{m})}{\left[2\left(\frac{5.00\mathrm{m}}{2}−0.540\mathrm{m}\right)\right]}\\ \\ {\mathrm{F}}_1=1.16×{10}^3\text{ N}\end{array}$

In unit vector notation,

${\mathrm{F}}_1=(1.16×{10}^3\text{ }\mathrm{N})\widehat{\mathrm{j}}$

According to the static equilibrium condition, the sum of the vertical forces acting on the beam is zero. Hence:

$\begin{array}{c}{\mathrm{Î\pounds ¹ó}}_{\mathrm{y}\mathrm{net}}=0\\ {\mathrm{F}}_1+{\mathrm{F}}_2−\mathrm{Mg}−\mathrm{mg}=0\\ {\mathrm{F}}_2=\mathrm{Mg}+\mathrm{mg}−{\mathrm{F}}_1\\ {\mathrm{F}}_2=(250\text{ }\mathrm{kg}×9.8{\text{m/s}}^{\text{2}})+(46.0\text{ }\mathrm{kg}×9.8{\text{″¾/s}}^{\text{2}})−(1.16×{10}^3\text{ }\mathrm{N})\\ {\mathrm{F}}_2=1.74×{10}^3\text{ N}\end{array}$

In unit vector notation,

${\mathrm{F}}_1=(1.74×{10}^3\text{ }\mathrm{N})\widehat{\mathrm{j}}$