JEE MAIN - Physics (2020 - 7th January Evening Slot - No. 1)
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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 $$\mu $$ = 0.4, the maximum possible value of 100 × $${b \over a}$$
for box not to topple before moving
is .......Answer
50
Explanation
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Torque about Center Of Mass
Fb + f$${a \over 2}$$ = N$${a \over 2}$$ ...(1)
For not sliding
F = f = $$\mu $$N ....(2)
N = Mg ...(3)
$$ \therefore $$ $$\mu $$N$$\left( {b + {a \over 2}} \right)$$ = Mg$${{a \over 2}}$$
$$ \Rightarrow $$ 4b = 3$$a$$
$$ \Rightarrow $$ b = 0.75$$a$$
which is not possible as as b can maximum be equal to 0.5$$a$$
$$ \therefore $$ $${b \over a}$$ = 0.5
So 100 × $${b \over a}$$ = 50
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