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Physics

Mechanics

Easy

Question

A horizontal force F is applied at the top of an equilateral triangular block of mass m and side L as shown. The minimum value of coefficient of friction between block and the horizontal surface to topple the block before it slides.

$\frac{2}{\sqrt{3}}$
$\frac{1}{2}$
$\frac{1}{\sqrt{3}}$
$\frac{1}{3}$

The correct answer is: $fraction numerator 1 over denominator square root of 3 end fraction$

$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em rowspacing 3 pt end attributes row cell text About point end text P subscript 2 text if we find torques end text F fraction numerator square root of 3 L over denominator 2 end fraction equals m g L over 2 not stretchy rightwards double arrow F equals fraction numerator m g over denominator square root of 3 end fraction end cell row cell text for equilibrium without sliding, end text F equals f less than mu m g end cell row cell not stretchy rightwards double arrow fraction numerator m g over denominator square root of 3 end fraction less than mu m g not stretchy rightwards double arrow mu greater than fraction numerator 1 over denominator square root of 3 end fraction end cell end table$

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