A uniform cube of side `'a'` and mass `m` rests on a rough horizontal table. A horizontal force `F` is applied normal to one of the faces at a point directly below the centre of the face, at a height `a//4` above the base.
a. What is the minimum value of `F` for which the cube begins to tip about an edge?
b. What is the minimum value of its so that toppling occurs?
c. If `mu=mu_("min")` find minimum force for topping.
d. Find minimum `mu_(s)` so that `F_("min")` can cause toppling.

Answer

`2`

Explanation

a. In the limiting case normal reaction will pass through `O`. The cube will tip about `O` if torque of `F` about `O` exceeds the torque of `mg`
Hence `F(a/4)gtmg(a/2)`

or `Fgtmg`
Therefore minimum value of `F` is `2mg`
b. In this case since it is not acting at `COM`,
toppling can occur even after body started sliding because increasing the torque of `F` about `COM`. Hence `mu_(min)=0`,
c. Now body is sliding before toplling `O` is not `IAR`, torque equation can not be applied across it. It can now be applied about `COM`.
`Fxxa/4=Nxxa/2` ......i
`N=mg` ...........ii
From i and ii
`F=2mg`
d. `Fgt2 mg` ............iii
from sol a
`N=mg`
`Fmu_(s)N=mu_(s)mg`.........iv
from i and ii `mu_(s)=2`.......v