A load of 10 kg is suspended by a metal wire 3 m long and having a cross-sectional area 4 mm2. Find
(a) the stress
(b) the strain and
(c) the elongation. Young's modulus of the metal is 2.0 x 1011N/m2.
A load of 10 kg is suspended by a metal wire 3 m long and having a cross-sectional area 4 mm^2. Find
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Length of wire L= 3 m,
Load F = 10*10 N =100 N, {taking g = 10 m/s²}
Area of cross-section A = 4 mm²
→A = 4x10⁻⁶ m²
(a) The stress = Load (force) on unit area of cross-section =F/A
= 100/4x10⁻⁶ N/m²
=25 x 10⁶ N/m²
=2.5 x10⁷ N/m²
(b) Let the elongation of the wire under this stress be l,
The strain =l/L, Young's modulus of the metal Y =2.0 x10¹¹ N/m²
We have Stress/strain =Y (constant)
→(F/A)/Strain =Y
→Strain =(F/A)/Y =2.5x10⁷/2.0 x10¹¹
→Strain = 1.25 x10⁻⁴ m
(c) Strain = l/L = 1.25 x10⁻⁴
→l = 3.0 * 1.25 x10⁻⁴ m
=3.75 x10⁻⁴ m