Question

A floor has a mximum loading capacity of 5 kN $m^{-2}$. Which of the following is allowed on this floor? (Take g = 10 $Nk{g}^{-1}$)

• A. A designer's table weighing 1000 N stands on 3 legs, each leg has a contact area of 0.1 $c{m}^2$ with the floor.
• B. A 50 kg lady wears a pair of high heel shoes, each shoe has a contact area of 9 $c{m}^2$ with the floor.
• C. An elephant weighing 50 kN stands on 4 legs with each foot covering 400 $c{m}^2$ of the floor.
• D. A 100 kg washing machine stands on a surface area of 6500 $c{m}^2$.

Answer 1

The correct option is D A 100 kg washing machine stands on a surface area of 6500 $c{m}^2$.

Maximum loading capacity of the floor = 5 kN $m^{-2}$
We know that,
Pressure, P = $\frac{Force}{Area}$

Maximum pressure exerted by elephant:
$P_E$ = $\frac{Force}{Area}$

= $\frac{50kN}{4\times 400c{m}^2}$

= $\frac{50kN}{4\times 0.04m^2}$

= 312.5 kN $m^{-2}$
Maximum pressure exerted by the elephant is more than the allowable limit (5 kN $m^{-2}$).

Maximum pressure exerted by lady:
$P_L$ = $\frac{Force}{Area}$

= $\frac{50\times 10N}{2\times 9c{m}^2}$

= $\frac{500N}{2\times 0.0009m^2}$

= 277.77 kN $m^{-2}$
Maximum pressure exerted by the lady is more than the allowable limit (5 kN $m^{-2}$).

Maximum pressure exerted by washing machine:
$P_W$ = $\frac{Force}{Area}$

= $\frac{100\times 10N}{6500c{m}^2}$

= ​​​​​​​$\frac{1kN}{0.65m^2}$

= 1.54 kN $m^{-2}$
Maximum pressure exerted by the washing machine is less than the allowable limit (5 kN $m^{-2}$).

Maximum pressure exerted by table:
$P_T$ = $\frac{Force}{Area}$

= $\frac{1000N}{3\times 0.1c{m}^2}$

= ​​​​​​​$\frac{1kN}{0.00003m^2}$

= 33,333.3 kN $m^{-2}$
Maximum pressure exerted by the table is more than the allowable limit (5 kN $m^{-2}$).

Therefore, only the washing machine is allowed on this floor.