Question

When the mass is removed, the length of the cable is found to be l0 = 4.76 m. After the mass is added, the length is remeasured and found to be l1 = 5.43 m. Determine Young's Modulus Y in N/m2 for the steel cable if the weight has a mass m = 35 kg and the cable has a radius r = 0.025 m.

Answer 1

Answer:1265841.097N/m^2

Explanation:LO = 4.76M

L1 = 5.43M, young's modulus (Y) = ?

Mass (M) = 35kg, radius(R) = 0.025

Young's modulus = tensile stress/tensile strain

But tensile stress = force(F)/area(A)

A = the area of the circle of the cable = πr^2

Weight of mass 35kg = force.  Hence W = my = ma

g = 10m/s^2

Force = 35*10 = 350N

area = 3.142*0.025^2 = 3.142*0.000625 = 0.00196375m^2

Tensile stress = 350/0.00196375 = 178230.4265N/m^2

Tensile strain = extension/L0(original length)

Extension = L1 - L0 = 5.43-4.76 = 0.67m

Tensile strain = 0.67/4.76 = 0.1408

Hence young modulus = 178230.4265/0.1408 = 1265841.097N/m^2

Answer 2

Answer:

$1242337 N/m^2$

Explanation:

The cross section area of the cable is

$A = \pi r^2 = \pi * 0.025^2 = 0.002 m^2$

Let g = 9.81m/s2. The stress acting on the cable when mass is added is

$\sigma = F/A = mg/A = 35*9.81/0.002 = 174867 Pa$

The strain when the cable is stretched from 4.76 to 5.43 m is

$\epsilon = \frac{\Delta L}{L} = \frac{5.43 - 4.76}{4.76} = 0.14$

So the young modulus of the cable is

$E = \sigma / \epsilon = 174867 / 0.14 = 1242337 Pa = 1242337 N/m^2$