A 10 m ladder weighing 200 N is placed against a smooth wall at an angle of 60 degrees to the horizontal. A 500 N boy stands 2.5 m from the base of the ladder. Find the reactions at the wall and the g

We take moments (torques) about the lower end of the ladder (point B in the diagram). The forces $N_G$ and $F_f$ pass through this point, so they create no torque.
The clockwise torques are due to the weights of the ladder and the boy. The anti-clockwise torque is due to the normal reaction from the wall.

Horizontal distance for $W_L$ from B = $5 \cos 60^\circ$.
Horizontal distance for $W_B$ from B = $2.5 \cos 60^\circ$.
Vertical distance for $N_W$ from B = $10 \sin 60^\circ$.

For equilibrium, the sum of torques is zero:

Substitute the given values:

Rationalize the denominator:

The normal reaction at the wall is \boxed{\text{75\sqrt{3} N}}.

These forces are the normal reaction from the ground ($N_G$) and the friction force from the ground ($F_f$).

Sum of vertical forces ($\sum F_y = 0$):

Sum of horizontal forces ($\sum F_x = 0$):
Since the wall is smooth, there is no friction at the wall. The only horizontal forces are the friction from the ground ($F_f$) and the normal reaction from the wall ($N_W$).

The resultant force ($F_R$) at the lower end is the vector