91Ó°ÊÓ

The mass of climber is,$m=70\text{ }kg$.

The climber touches the rock wall at a distance,$H=2\text{ }m$.

The center of mass is at,$a=0.20\text{ }m$ .

We can write the equation of the horizontal component of forces. We can then write the torque equation at point O. using these two equations, we can get the horizontal component of force. To find the vertical force component, use the vertical force equilibrium condition.

Free Body Diagram:

Using equilibrium condition, the net force in x direction can be written as,

$∑F_x=0$

Substitute the values in the above expression, and we get,

$∑F_y=0$

(1)

Using equilibrium condition, the net force in y- direction can be written as,

$∑F_y=0$

Substitute the values in the above expression, and we get,

$4F_v−mg=0$

(2)

Torque at point acan be calculated as,

$\begin{array}{c}mga−\left(4F_h\right)H=0\\ F_h=\frac{mga}{4H}\end{array}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}{F}_h=\frac{70\text{ k²µ}×9.8{\text{″¾/²õ}}^{\text{2}}×0.2\text{″¾}}{4×2.0\text{″¾}}\\ =17.16â‹\dots (\frac{1\text{ k²µ}â‹\dots {\text{m/s}}^{\text{2}}×1\text{″¾}}{1\text{″¾}}×\frac{1\text{ }N}{1\text{ k²µ}â‹\dots {\text{m/s}}^{\text{2}}})\\ =17\text{ }N\end{array}$

Thus, the horizontal force component of force is$17\text{ }N$ .

From equation 2, we can write,

$F_v=mg/4$

Substitute the values in the above expression, and we get,

$\begin{array}{c}{F}_v=\frac{70\text{ k²µ}×9.8{\text{″¾/²õ}}^{\text{2}}}{4}\\ =171.675â‹\dots (1\text{ k²µ}â‹\dots {\text{m/s}}^{\text{2}}×\frac{1\text{ }N}{1\text{ k²µ}â‹\dots {\text{m/s}}^{\text{2}}})\\ ~170\text{ }N\end{array}$

Thus, the vertical force component of force is$170\text{ }N$$170\text{ }N$.