JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Rotational Mechanics

Consider a uniform cubical box of side a on a rough floor that is to be moved by applying minimum possible force

F

at a point

b

above its centre of mass (see figure). If the coefficient of friction is

μ = 0.4

, the maximum possible value of

100 \times \frac{b}{a}

for box not to topple before moving is ______.

Answer: 50

Solution

The least force that just sets the box moving is

F = μ m g

. Taking torques about the leading bottom edge, toppling is avoided only while

F (\frac{a}{2} + b) \leq m g \frac{a}{2}

. Substituting

F = μ m g

with

μ = 0.4

gives

0.4 (\frac{a}{2} + b) \leq \frac{a}{2}

, i.e.

b \leq 0.75 a

. But the point of application cannot sit higher than the top face, which is only

\frac{a}{2}

above the centre, so geometry caps

b

\frac{a}{2}

. The largest usable height is therefore

b = \frac{a}{2}

, and

100 \times \frac{b}{a} = 50

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Integer answer