Question

A iron or 4.00m long and 0.500cm2 in cross section stretches 1.00mm when a mass of 225kg is hung from its lower end. Compute the Young's Modulus for the iron

Answer 1

what is formula for youngs modulus

Answer 2

\[E=\frac{ PL }{ Ay }\]
where E is the young's modulus or the modulus of elasticity
P is the applied force
L is the length of the material
A is the cross-sectional area
y is the deformation or elongation
P = 225kg(9.81m/s2)
= 2,207.25 N
L = 4 m
= 4000mm
A = 0.50cm2(100mm2/1cm2)
= 50 mm2
y = 1mm
substituting it will be:
\[E = \frac{ 2,207.25N(4000mm) }{ 50mm^2(1mm) }\]
\[E = 176,580\frac{ N }{ mm^2 }\]
\[E = 176,580 MPa\]
or expressing it in GPa you'll have
\[E = 176.58 GPa\]

Answer 3

y did u block me!!?