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A uniform plank of Young’s modulus Y is moved over a smooth horizontal surface by a constant horizontal force F. The area of cross-section of the plank is A. the compressive strain on the plank in the direction of the force is:

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$\frac{F}{A Y}$

$\frac{2 F}{A Y}$

$\frac{1}{2} (\frac{F}{A Y})$

$\frac{3 F}{A Y}$

answer is A.

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Detailed Solution

Compressive strain

$= \frac{S t r e s s}{Y o u n g' s m o d u l u s} = \frac{F / A}{Y} = \frac{F}{A Y}$

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