The tricycle weighing 20 kg has a small wheel symmmetrically placed 1m behind the two large wheels, which are also 1m apart. If the center of gravity of machine is at a horizontal distance of 25 cm behind the front wheels and the rider whose weight is 40 kg is 10 cm behind the front wheels. Then, the thrust on each front wheel is

Options

A. 255 N
B. 90 N
C. 200 N
D. 400 N

Explanation

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