91Ó°ÊÓ

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

$\text{h}=0.490\text{ }m$

At$\text{x}=0,{\text{F}}_{\text{h}}=20\text{ }kN\left(\frac{{10}^3\text{ N}}{1\text{ k±·}}\right)=2.0×{10}^4\text{ N}$

At$\text{x}=4.0,{\text{F}}_{\text{h}}=25\text{ }kN\left(\frac{{10}^3\text{ N}}{1\text{ k±·}}\right)=2.5×{10}^4\text{ N}$

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:

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

$Weight=mg$

Length of the ramp from Pythagoras theorem,

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

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

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

${∑}_{}^{}{\text{M}}_{\text{x=0}}=0$

$\begin{array}{c}(Weightoframp×\frac{L}{2}×cos6.98°)-({\text{F}}_{\text{h}}×0.490\text{ }m)=0\\ (Weightoframp×\frac{\text{L}}{\text{2}}\text{׳¦´Ç²õ6.98}°)=({\text{F}}_{\text{h}}×0.490\text{ }m)\\ Weightoframp=\frac{\left({\text{2×¹ó}}_{\text{h}}\text{×0.490″¾}\right)}{\text{L׳¦´Ç²õ6.98°}}\\ m_{ramp}g=\frac{\left({\text{2×¹ó}}_{\text{h}}\text{×0.490″¾}\right)}{\text{L׳¦´Ç²õ6.98°}}\end{array}$

$\begin{array}{c}{m}_{ramp}=\frac{\left({\text{2×¹ó}}_{\text{h}}\text{×0.490″¾}\right)}{\text{g×L׳¦´Ç²õ6.98°}}\\ =\frac{(2×2.0\text{ }×{10}^4\text{ }N×0.490\text{ }m)}{9.8\text{ }m/s^2×4.0\text{ }m×cos6.98°}\\ =500\text{ }kg\end{array}$

Mass of the ramp is $500\text{ }kg$.

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

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

${∑}_{}^{}\mathrm{τ}=0$

$((weightoframp+weightofman)×2\text{″¾})−({\text{F}}_{\text{h}}×0.490\text{″¾})=0$

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

Mass of the man is $62.5\text{ }kg$.