91Ó°ÊÓ

${\mathrm{M}}_{\mathrm{b}}+{\mathrm{M}}_{\mathrm{p}}=61.22\text{ }\mathrm{kg}$

By using the equilibrium condition along a vertical direction, we can write the equation in terms of tension T. Then compare that equation with the standard equation of line. From that, we can write the slope and y intercept. Once we know the slope and y intercept, then we can find the mass of rod, package and angle.

Equations:

$∑{\mathrm{F}}_{\mathrm{x}}=0$

$∑{\mathrm{F}}_{\mathrm{y}}=0$

$∑\mathrm{τ}=0$

Using condition of equilibrium, we can write
$∑{\mathrm{F}}_{\mathrm{y}}=0$ (i)

Consider point A as pivot point. We have weight of the beam and weight of the package acting downwards at distance$\mathrm{L}/2$and $\mathrm{x}$ respective. The torque caused by these two forces would be balanced by the vertical component of the tension. Using equation for the torque, we have

$\mathrm{TL}\text{sin}\mathrm{θ}=\frac{{\mathrm{M}}_{\mathrm{b}}\mathrm{gL}}{2}+{\mathrm{M}}_{\mathrm{p}}\mathrm{gx}$ (ii)

Now divide both sides of the equation (ii) by $\mathrm{³¢²õ¾\pm ²Ôθ}$

$\begin{array}{c}\mathrm{T}=\frac{{\mathrm{M}}_{\mathrm{b}}\mathrm{gL}}{2\text{sin}\mathrm{θ³¢}}+\frac{{\mathrm{M}}_{\mathrm{p}}\mathrm{gx}}{\mathrm{L}\text{sin}\mathrm{θ}}\\ =\frac{{\mathrm{M}}_{\mathrm{p}}\mathrm{gx}}{\mathrm{L}\text{sin}\mathrm{θ}}+\frac{{\mathrm{M}}_{\mathrm{b}}\mathrm{g}}{2\text{sin}\mathrm{θ}}\end{array}$

We can write the slope as,

role="math" localid="1663335112681" $\mathrm{Slope}=\frac{{\mathrm{M}}_{\mathrm{p}}\mathrm{g}}{\text{sin}\mathrm{θ}}$ (iii)

Therefore, the equation for $\text{T}$ can be written as,

$\mathrm{T}=\mathrm{slope}\left(\frac{\mathrm{X}}{\mathrm{L}}\right)+\frac{{\mathrm{M}}_{\mathrm{b}}\mathrm{g}}{2\text{sin}\mathrm{θ}}$

Now, the slope can be calculated from the graph as,

$\begin{array}{c}\mathrm{Slope}=\frac{700−500}{1}\\ =200\end{array}$

Substitute the value of slope in equation (iii), we get

$200=\frac{{\mathrm{M}}_{\mathrm{p}}\mathrm{g}}{\text{sin}\mathrm{θ}}$

$\text{sin}\mathrm{θ}=\frac{{\mathrm{M}}_{\mathrm{p}}\mathrm{g}}{200}$ (iv)

The y intercept on the graph is written as,

$\mathrm{y}\mathrm{intercept}=\frac{\left({\mathrm{M}}_{\mathrm{b}}\mathrm{g}\right)}{2\text{sin}\mathrm{θ}}$

(v)

But from the graph,$\mathrm{y}\mathrm{intercept}=500$ . Substitute the value in equation (v)

$500=\frac{\left({\mathrm{M}}_{\mathrm{b}}\mathrm{g}\right)}{2\text{sin}\mathrm{θ}}$

Simplifying this equation, we get

$\text{sin}\mathrm{θ}=\frac{\left({\mathrm{M}}_{\mathrm{b}}\mathrm{g}\right)}{2×500}$ (vi)

Equating equation (iv) and (vi), we get

$\begin{array}{c}\frac{{\mathrm{M}}_{\mathrm{p}}\mathrm{g}}{200}=\frac{\left({\mathrm{M}}_{\mathrm{b}}\mathrm{g}\right)}{2×500}\\ {\mathrm{M}}_{\mathrm{b}}=\text{5}{\mathrm{M}}_{\mathrm{p}}\end{array}$

But we know

$\begin{array}{c}{\mathrm{M}}_{\mathrm{b}}+{\mathrm{M}}_{\mathrm{p}}=61.22\text{ }\mathrm{kg}\\ 5{\mathrm{M}}_{\mathrm{p}}+{\mathrm{M}}_{\mathrm{p}}=61.22\text{kg}\\ {\mathrm{M}}_{\mathrm{p}}=10.2\text{kg}\end{array}$

Therefore, the mass of the package is $10.2\text{ }\mathrm{kg}$.

We can calculate the mass of bar as,

$\begin{array}{c}{\mathrm{M}}_{\mathrm{b}}+{\mathrm{M}}_{\mathrm{p}}=61.22\text{ }\mathrm{kg}\\ {\mathrm{M}}_{\mathrm{b}}=61.22\text{kg}−{\mathrm{M}}_{\mathrm{p}}\\ =61.22\text{kg}−10.2\text{kg}\\ \text{=51kg}\end{array}$

Therefore, the mass of bar is $51\text{ }\mathrm{kg}$.

Now, use equation (vi) to calculate the angle.

$\begin{array}{c}\text{in}\mathrm{θ}=\frac{{\mathrm{M}}_{\mathrm{p}}\mathrm{g}}{200}\\ \mathrm{θ}={\text{sin}}^{−1}\left(\frac{{\mathrm{M}}_{\mathrm{p}}\mathrm{g}}{200}\right)\\ ={\text{sin}}^{−1}\left(\frac{10.2\text{kg}×9.81\text{m}/{\text{s}}^2}{200}\right)\\ =30.02°\end{array}$

Therefore, the angle is $30.02°$.

Equating equation (iv) and (vi), we get

$\begin{array}{c}\frac{{\mathrm{M}}_{\mathrm{p}}\mathrm{g}}{200}=\frac{\left({\mathrm{M}}_{\mathrm{b}}\mathrm{g}\right)}{2×500}\\ {\mathrm{M}}_{\mathrm{b}}=\text{5}{\mathrm{M}}_{\mathrm{p}}\\ {\mathrm{M}}_{\mathrm{p}}=\frac{{\mathrm{M}}_{\mathrm{b}}}{5}\end{array}$

But we know

$\begin{array}{c}{\mathrm{M}}_{\mathrm{b}}+{\mathrm{M}}_{\mathrm{p}}=61.22\text{ }\mathrm{kg}\\ {\mathrm{M}}_{\mathrm{b}}+\frac{{\mathrm{M}}_{\mathrm{b}}}{5}=61.22\text{kg}\\ {\mathrm{M}}_{\mathrm{b}}=51\text{kg}\end{array}$

Therefore, the mass of the bar is $51\text{ }\mathrm{kg}$.

Equating equation (iv) and (vi), we get

$\begin{array}{c}\frac{{\mathrm{M}}_{\mathrm{p}}\mathrm{g}}{200}=\frac{\left({\mathrm{M}}_{\mathrm{b}}\mathrm{g}\right)}{2×500}\\ {\mathrm{M}}_{\mathrm{b}}=\text{5}{\mathrm{M}}_{\mathrm{p}}\end{array}$

But we know

$\begin{array}{c}{\mathrm{M}}_{\mathrm{b}}+{\mathrm{M}}_{\mathrm{p}}=61.22\text{ }\mathrm{kg}\\ 5{\mathrm{M}}_{\mathrm{p}}+{\mathrm{M}}_{\mathrm{p}}=61.22\text{kg}\\ {\mathrm{M}}_{\mathrm{p}}=10.2\text{kg}\end{array}$

Therefore, the mass of the package is $10.2\text{ }\mathrm{kg}$.