Figure 12-65ashows a uniform ramp between two buildings that allows for motion between the buildings due to strong winds.At its left end, it is hinged to the building wall; at its right end, it has a roller that can roll along the building wall. There is no vertical force on the roller from the building, only a horizontal force with magnitude $F_h$. The horizontal distance between the buildings is $D=4.00m$. The rise of the ramp is$D=4.00m$. A man walks across the ramp from the left. Figure 12-65bgives $F_h$as a function of the horizontal distance xof the man from the building at the left. The scale of the $F_h$axis is set by a= 20kN and $b=25kN$. What are the masses of (a) the ramp and (b) the man?

Distance between buildings$\left(\text{D}\right)=4.0m$

$\text{h}=0.490m$

At$\text{x}=0,{\text{F}}_{\text{h}}=20kN\left(\frac{{10}^3\text{\hspace{0.17em}N}}{1\text{\hspace{0.17em}kN}}\right)=2.0\times {10}^4\text{\hspace{0.17em}N}$

At$\text{x}=4.0,{\text{F}}_{\text{h}}=25kN\left(\frac{{10}^3\text{\hspace{0.17em}N}}{1\text{\hspace{0.17em}kN}}\right)=2.5\times {10}^4\text{\hspace{0.17em}N}$

We find the length of the ramp first, using trigonometry (Pythagoras theorem). After finding the length, we take the value for the horizontal force at the displacement of the man on the ramp. Using those conditions, we find the mass of the man and the ramp.

Formula:

${\sum }_{}^{}{\text{M}}_{\text{anypointonthebeam}}=0$

$Weight=mg$

Length of the ramp from Pythagoras theorem,

$\begin{array}{c}L=\sqrt{D^2+h^2}\\ =\sqrt{{\left(4.0m\right)}^2+{\left(0.49m\right)}^2}\\ =\sqrt{16.24m^2}\\ =4.03m\end{array}$

$\begin{array}{l}tan\theta =\frac{h}{D}\\ \theta =\left(\frac{0.49}{4.0}\right)\\ \theta =6.98°\end{array}$

Taking moment at the hinge point and assuming$\text{x}=0$,

${\sum }_{}^{}{\text{M}}_{\text{x=0}}=0$

$\begin{array}{c}(Weightoframp\times \frac{L}{2}\times cos6.98°)-({\text{F}}_{\text{h}}\times 0.490m)=0\\ (Weightoframp\times \frac{\text{L}}{\text{2}}\text{×cos6.98}°)=({\text{F}}_{\text{h}}\times 0.490m)\\ Weightoframp=\frac{\left({\text{2×F}}_{\text{h}}\text{×0.490\hspace{0.17em}m}\right)}{\text{L×cos6.98°}}\\ m_{ramp}g=\frac{\left({\text{2×F}}_{\text{h}}\text{×0.490\hspace{0.17em}m}\right)}{\text{L×cos6.98°}}\end{array}$

$\begin{array}{c}{m}_{ramp}=\frac{\left({\text{2×F}}_{\text{h}}\text{×0.490\hspace{0.17em}m}\right)}{\text{g×L×cos6.98°}}\\ =\frac{(2\times 2.0\times {10}^{4}N\times 0.490m)}{9.8m/s^2\times 4.0m\times cos6.98°}\\ =500kg\end{array}$

Mass of the ramp is $500kg$.

Taking moment at the hinge point and assuming x = 2 m,

From the graph, we can say that
$\text{At}\text{\hspace{0.17em}x}=2,{\text{F}}_{\text{h}}=2.25\times {10}^{4}N$

${\sum }_{}^{}\tau =0$

$((weightoframp+weightofman)\times 2\text{\hspace{0.17em}m})-({\text{F}}_{\text{h}}\times 0.490\text{\hspace{0.17em}m})=0$

$\begin{array}{c}((500kg\times 9.8m/s^2)+(m_{man}\times 9.8m/s^2))\times 2m=(2.25\times {10}^{4}N\times 0.490m)\\ 4900N+m_{man}\times 9.8m/s^2=5512N\\ m_{man}\times 9.8m/s^2=612N\\ m_{man}=\frac{612N}{9.8m/s^2}\\ m_{man}=62.5kg\end{array}$

Mass of the man is $62.5kg$.