91影视

Net external force on the body must be zero for the center of mass of a body at rest to remain at rest.

$\underset{\mathbf{}}{\mathbf{鈭�}}\stackrel{\mathbf{鈫�}}{\mathbf{F}}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Here we have given that Two friends are carrying a 200 kg crate up a flight of stairs.

So, m = 200 kg

The stairs make an $45掳$angle with respect to the floor

So, $\mathrm{胃}=45掳$.

The crate is 1.25 m long and 0.500 m high.

The two-person applies forces ${\mathrm{F}}_1$and ${\mathrm{F}}_2$upward direction in opposite to the weight of the crate.

These forces apply equilibrium in the y-direction. So, by equation (1),

$\begin{array}{r}鈭�F_y=0\\ F_1+F_2+W=0\\ F_1+F_2=mg\\ F_1=mg鈭�F_2\end{array}$ (2)

As the equilibrium is achieved, torque is now applied at the bottom person, and the torque sum is zero. So we can obtain an expression for $F_2$,

$\begin{array}{r}鈭�{蟿}_{\text{bottom}}=0\\ {\mathrm{F}}_2{\mathrm{L}}_2鈭�{\mathrm{WLW}}_{\mathrm{W}}=0\\ {\mathrm{F}}_2=\mathrm{W}\left(\frac{{\mathrm{L}}_{\mathrm{W}}}{{\mathrm{L}}_2}\right)\\ {\mathrm{F}}_2=\left(\mathrm{mg}\right)\left(\frac{{\mathrm{L}}_{\mathrm{W}}}{{\mathrm{L}}_2}\right)\end{array}$

Now, substitute all values in the above equation,

$\begin{array}{r}{F}_2=\left(200\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(\frac{0.375脳\mathrm{m}脳\cos 鈦�{45}^{鈭�}}{1.25脳\mathrm{m}脳\cos 鈦�{45}^{鈭�}}\right)\\ =590\mathrm{N}\end{array}$

Now, put the value of $F_2$in equation (2),

$\begin{array}{r}{F}_1=\left(200\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)鈭�590\mathrm{N}\\ =1370\mathrm{N}\end{array}$

Hence, the value of both forces is $F_1$= 1370 Nand ${\mathrm{F}}_2=590\mathrm{N}$

So, is it better to be the below person.