A boy pulls a lawn mower by exerting a force of 700N to the handle. Calculate the (a) vertical and horizontal components of his force on the mower if it is inclined at an angle of 55° to the horizontal. (b) reaction of the ground on the mower if its weight is 1000 N.

$$F_x = F \cos \theta = 700 \text{ N} \times \cos(55^\circ)$$ $$F_x \approx 700 \text{ N} \times 0.5736$$ $$F_x \approx 401.52 \text{ N}$$

$$F_y = F \sin \theta = 700 \text{ N} \times \sin(55^\circ)$$ $$F_y \approx 700 \text{ N} \times 0.8192$$ $$F_y \approx 573.44 \text{ N}$$

The horizontal component is $\boxed{401.52 \text{ N}}$ and the vertical component is $\boxed{573.44 \text{ N}}$.

The weight ($W$) acts downwards, the vertical component of the pulling force ($F_y$) acts upwards, and the normal reaction force ($N$) from the ground acts upwards.
For vertical equilibrium, the sum of upward forces equals the sum of downward forces:

$$N + F_y = W$$

$$N = 1000 \text{ N} - 573.44 \text{ N}$$ $$N = 426.56 \text{ N}$$

The reaction of the ground on the mower is $\boxed{426.56 \text{ N}}$.