Q.

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 =1.0cm2  ; Young's modulus is  E =1.3 ×1010N / m2, 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

 3F1+F2=3900
 F=AYlx
3AYlx+AYl(x+0.5×103)=3900
4x+0.5+103=3×103
4x=2.5×103
x=2.54×103
F2=1.3×106(2.54+0.5)103
=1.3×103(4.54)
F2=1.46×103=1460N

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