A table has three legs that are 1.00 m in length and a fourth leg that is longer by d = 0.50mm, so that the table wobbles slightly. A steel cylinder with mass M = 390kg is placed on the table (which has a mass much less than M) so that all four legs are compressed but unbuckled and the table is level but no longer wobbles. The legs are wooden cylinders with cross-sectional area $A = 1.0 c m^{2}$ ; Young's modulus is $E = 1.3 × 10^{10} N / m^{2}$ , the magnitude of the force applied by the longer leg on the table in newtons is

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answer is 1443.

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Detailed Solution

$3 F_{1} + F_{2} = 3900$
$F = \frac{A Y}{l} x$
$3 \frac{A Y}{l} x + \frac{A Y}{l} (x + 0.5 \times 10^{- 3}) = 3900$
$4 x + 0.5 + 10^{- 3} = 3 \times 10^{- 3}$
$4 x = 2.5 \times 10^{- 3}$
$x = \frac{2.5}{4} \times 10^{- 3}$
$F_{2} = 1.3 \times 10^{6} (\frac{2.5}{4} + 0.5) 10^{- 3}$
$= 1.3 \times 10^{3} (\frac{4.5}{4})$
$F_{2} = 1.46 \times 10^{3} = 1460 N$