JEE MAIN - Physics (2021 - 17th March Evening Shift - No. 26)
A body of mass 1 kg rests on a horizontal floor with which it has a coefficient of static friction $${1 \over {\sqrt 3 }}$$. It is desired to make the body move by applying the minimum possible force F N. The value of F will be ____________. (Round off to the Nearest Integer) [Take g = 10 ms$$-$$2 ]
Answer
5
Explanation
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Minimum possible force $$ \Rightarrow $$
$$F = {{\mu mg} \over {\sqrt {1 + {\mu ^2}} }}$$
$${F_{\min }} = {{{1 \over {\sqrt 3 }} \times 1 \times 10} \over {\sqrt {1 + {1 \over 3}} }}$$
Fmin = 5N
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