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× (b/a) for box not to topple before moving is

Question

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× (b/a) for box not to topple before moving is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-ij6ipuyruhjdupav.jpg" /> (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

For no toppling F((a/2) + b)≤ mg(a/2) μ (a/2)+μ b≤ (a/2) 0.2a+0.4b≤ 0.5a 0.4b≤ 0.3a b≤ (3 a/4) b ≤ 0.75 a (in limiting case) But it is not possible as b can maximum be equal to 0.5a (100 (b/a))max=50

For no toppling F(2a​+b)≤mg2a​ μ2a​+μb≤2a​ 0.2a+0.4b≤0.5a 0.4b≤0.3a b≤43a​ b≤0.75a (in limiting case) But it is not possible as b can maximum be equal to 0.5a (100ab​)max​=50

For no toppling
$F (\frac{a}{2} + b) \leq m g \frac{a}{2}$
$μ \frac{a}{2} + μ b \leq \frac{a}{2}$
$0.2 a + 0.4 b \leq 0.5 a$
$0.4 b \leq 0.3 a$
$b \leq \frac{3 a}{4}$
$b \leq 0.75 a$ (in limiting case)
But it is not possible as $b$ can maximum be equal to $0.5 a$
$(100 \frac{b}{a})_{ma x} = 50$