Question

A vertical metal cylinder of radius 2 cm and length 2 m is fixed at the lower end and a load of 100 kg is put on it. Find (a) the stress (b) the strain and (c) the compression of the cylinder. Young modulus of the metal = 2 × 1011 N m−2.

Answer 1

Conclusion:

Stress$= 7.8 \times 10^{5}\frac{N}{m^{2}}$

Strain$=3.9\times 10^{-6}$

Compression $= 7.8 \times 10^{-6} meter$

Explanation:

Given that, Radius of the cylinder $r =\textrm{2 cm}$ $= 2\times 10^{-2} meter$

                  Length of the cylinder $l =\textrm{2 m}$

                  Load $m = \textrm{100 kg}$

Now we have to find out the Area and Force for results

Area $A = \pi r^{2}$ $= 4\pi \times10^{-4}$ $m^{2}$

                 Force $F = mg$ $= 100 \times 9.8$

             ⇒$F =\textrm980 Netwon$

As we know that, $Stress =\frac{Force}{Area}$

                           1.  $Stress =\dfrac{980}{4\pi \times10^{-4}}$

                             ⇒ $Stress =\frac{2450000}{\pi} = 7.8 \times 10^{5}$

                           2. $\textrm{Young Modulus} = \frac{Stress}{Strain}$

                              ⇒ $2\times 10^{11} =\frac{7.8\times 10^{5}}{Strain}$

                            ⇒ $Strain =\frac{7.8 \times 10^{5}}{2 \times 10^{11}}$

                           ⇒ $Strain = 3.9\times 10^{-6}$

                        3. $Strain =\frac{\textrm{Change in length}}{\textrm{Original length}}$

                       ⇒ $\textrm {Change in length}$ $= 3.9\times10^{-6} \times 2$

            ⇒ Change in length $= 7.8\times 10^{-6} meter$

Answer 2

Stress is Proportional to Strain

Explanation:

Length of the cylinder L = 2 m  

Area of cross-section $A = \pi \times r^2 = \pi (\frac {2}{100})^2m^2$

= $\frac {4\times \pi}{10000} m^2$  

Force on the cylinder $F = 100\times 10 N = 1000 N$  

Hence the stress = $\frac {F}{A} = \frac {1000}{(4\pi)} / (10000) N/m^2$

= $(\frac {1}{4\pi})\times 10^7$ N/m²  

= $0.0796 \times 10^7$ N/m²  

= $7.96 \times 10^5$ N/m²    

(b) Strain = $\frac {Stress}{Y} = F/A/Y$

= $\frac {7.96 \times 10^5}{2.0 \times 10^1^1}$

= $3.98 \times 10^-^6  = 4 \times 10^-^6$    

(c) Let the compression of the cylinder = l m  

The $strain =\frac {l}{L} = \frac {l}{2}$

⇒ $4 \times 10^-^6 = \frac {l}{2}$

$l = 8 \times 10^-^6 m$