A 6.5 m long ladder rests against a vertical wall, reaching a...

Let the length of the ladder be $L = 6.5 \text{ m}$ and the height of the wall it reaches be $h = 6.0 \text{ m}$.

The distance of the base of the ladder from the wall is $x = \sqrt{L^2 - h^2} = \sqrt{6.5^2 - 6.0^2} = 2.5 \text{ m}$.

The forces acting on the ladder are the weight of the man $W = mg = 60 \times 9.8 = 588 \text{ N} \approx 590 \text{ N}$, the normal reaction from the ground $N_V$, the horizontal friction from the ground $N_H$, and the normal reaction from the smooth wall $N_W$.

For vertical equilibrium, the vertical force exerted by the ground is:

$N_V = W \approx 590 \text{ N}$

For horizontal equilibrium, the horizontal force exerted by the ground is equal to the normal reaction from the wall:

$N_H = N_W$

Taking torque about the base of the ladder to find $N_W$:

$N_W \times h = W \times \dfrac{x}{2}$

$N_W \times 6.0 = 588 \times \dfrac{2.5}{2}$

$N_W = \dfrac{588 \times 1.25}{6.0} = 122.5 \text{ N} \approx 120 \text{ N}$

Thus, the force exerted by the ground on the ladder is $590 \text{ N}$ vertical and $120 \text{ N}$ horizontal.