Laws of Motion

JEE Main level, numerical answer, medium

172 of 210

Question 172 of 210



Consider a uniform cubical box of side a on a rough floor that is to be moved by applying minimum possible force
FF at a point bb above its centre of mass (see figure). If the coefficient of friction is μ=0.4\mu =0.4 , the maximum possible value of 100×ba100\times \dfrac{b}{a} for a box not to topple before moving is ________

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Answer: 50


For no toppling

F(a2+b)≤mga2F\left(\dfrac{a}{2}+b\right)\leq mg\dfrac{a}{2}
μa2+μb≤a2\mu \dfrac{a}{2}+\mu b\leq \dfrac{a}{2}
0.2a+0.4b≤0.5a0.2a+0.4b\leq 0.5a
0.4b≤0.3a0.4b\leq 0.3a
b≤3a2b\leq \dfrac{3a}{2}
b≤0.75ab\leq 0.75a (in limiting case)
But it is not possible as
bb can maximum be equal to 0.5a0.5a
∴(100ba)max=50.00∴{\left(100\dfrac{b}{a}\right)}_{max}=50.00

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