a)
In the given figure, the tricycle with two front wheels A and B with one rear wheel C with center of gravity of machine `G_(1)` and that of rider `G_(2)`.
Now, force at `G_(1) = 20 xx 10 = 200N`
Force at `G_(2)`=400 N
Let `F_(C)` = Force on ground under rear wheel C
Force at D = `F_(D)=2F_(A)` acts on mid-point of AB.
Taking torque about D, for equilibrium,
`F_(c) xx 100 = 200 xx 25 + 400 xx 10= 9000`
`therefore` Thrust on each front wheel is N.
`therefore` 2N + F_(c)` = Total weight = 600 N
`therefore` N = (600-F_(C))/2`= 255 N