A uniform rod of length L and density `rho` is being pulled along a smooth floor with a horizontal acceleration `alpha` (see Fig.) The magnitude of the stress at the transverse cross-section through the mid-point of the rod is……..
Text Solution
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Let `A` be area of cross section of the rod. Mass `mxx"volume"xx"density =(L/2A)rho` `T=-ma=(L/2Arhoa)` Therefore stress `T/A=1/2rhoaL`
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