A table has three legs that are 1.00 m in length and a fourth leg that is longer by d - 0.50 mm, so that the table wobbles slightly. A steel cylinder with mass M = 290 kg 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 cm^(2)`, Young's modulus is `E = 1.3xx10^(10)N//m^(2)`. What are the magnitudes of the forces on the from the floor ?

Explanation

We take the table plus steel cylinder as our system. The situation is like that in Fig. 12-1, except now we have a steel cylinder on the table. If the tabletop remains level, the legs must be compressed in the following ways : Each of the short legs must be compressed by the same amount (call it `Delta L_(3)`) and thus by the same force of magnitude `F_(3)`. The single long leg must be compressed by a larger amount `Delta L_(4)` and thus by a force with a larger magnitude `F_(4)`. In other words, for a level tabletop, we must have
`Delta L_(4)=Delta L_(3)+d`.
From Eq. 12-3, we can relate a change in length to the force causing the change with `Delta L = FL//AE`, where L is the original length of a leg. We can use this relation to replace `Delta L_(4)` and `Delta L_(3)` in Eq. 12-6. However, note that we can approximate the original length L as being the same for all four legs.
Making those replacements and that approximation gives us
`(F_(4)L)/(AE)=(F_(3)L)/(AE)+d`.
We cannot solve this equation because it has two unknowns, `F_(4)` and `F_(3)`.
To get a second containing `F_(4)` and `F_(3)`, we can use a vertical y axis then write the balance of vertical forces `(F_("net,y")=0)` as
`3F_(3)+F_(4)-Mg=0`,
where Mg is equal to the magnitude of the gravitational force of the system. (Three legs have force `F_(3)` on them.) To solve the simultaneous Eqs. 12-7 and 12-8 for, say, `F_(3)`, we first use Eq, 12-8 to find that `F_(4)=Mg-3F_(3)`. Substituting that into Eq. 12-7 then yields, after some algebra,
`F_(3)=(Mg)/(4)-(dAE)/(4L)`
`=((290 kg)(9.8 m//s^(2)))/(4)`
`-((5.0xx10^(-4)m)(10^(-4)m^(2))(1.3xx10^(10)N//m^(2)))/((4)(1.00m))`
`=548N ~~ 5.5xx10^(2)N`.
From Eq. 12-8, we then find
`F_(4)=Mg-3F_(3)=(290 kg)(9.8 m//s^(2))-3(548 N)`
`~~1.2 kN`.
You can show that the three short legs are each compressed by 0.42 mm and the single leg by 0.92 mm.