A rectangular block of square base (`1 mxx1 m`) and height `1.8 m` is moving with constant speed on a rough horizontal surface (`mu_(k)=0.5`) due to horizontal force `F` as shown. If the mass of block is `40 kg`, find
(a) the value of `F_(1)`
(b) the value of `h` at which the block just starts to tip
(c) If force `F` is applied at `h//2`, find the perpendicular distance of normal reaction from `O` (center of mass).

Explanation

(`a`) Since the block is moving with constant speed
`F=f_(k)=mu_(k)N=mu_(k)mg=0.5xx40xx10=200N`
(`b`) When the block is on the verge of tipping, normal reaction will shift in the direction of the applied force and will pass through the extreme end.
The block will tip over `A` if
`Fhgemg((1)/(2))implies 200hge(40xx10)/(2)`
`hge1mimplies h=1m`
(c) If `F` is applied at `h//2`, block is in equilibrium, the normal reaction will not pass through the extreme point `A`.

`tau_(A)=0`
`N((1)/(2)-x)+F(h)/(2)=(mg)(1)/(2)`
`mg((1)/(2)-x)+200xx(1)/(2)=(mg)/(2)`
`mgx=200xx(1)/(2)`
`40xx10xxx=100implies x=0.25 m`