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Q.

A uniform rod of length L and density p is being pulled along a smooth floor with a horizontal acceleration $α$ (see fig). The magnitude of the stress at the
transverse cross-section trough the mid-point of the rod is

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a

$ραL$

b

$ραL / 2$

c

$ρα / 2 L$

d

$\frac{1}{2} ραL$

answer is A.

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Detailed Solution

Let A be the area of cross-section of the rod. The mass m of the rod will be L A p
Force at the end of the rod $= mα = LAρα$
Force at the middle $= \frac{1}{2} [LAρα]$
stress = Force per unit area $= \frac{1}{2} [\frac{LAρα}{A}] = \frac{1}{2} ραL$

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