Question

According to one set of measurements, the tensile strength of hair is 196 MPa , which produces a maximum strain of 0.380 in the hair. The thickness of hair varies considerably, but let's use a diameter of 50.0 μm
Part A
What is the magnitude of the force giving this tensile stress?
F = ? N

Part B
If the length of a strand of the hair is 12.0 cm at its breaking point, what was its unstressed length? original length = ?cm

Answer 1

Answer:

(a). The magnitude of the force is 0.38416 N.

(b). The original length is 0.0869 m.

Explanation:

Given that,

Tensile strength = 196 MPa

Maximum strain = 0.380

Diameter = 50.0 μm

Length = 12.0 cm

We need to calculate the area

Using formula of area

$A=\dfrac{\pi}{4}\times d^2$

Put the value into the formula

$A=\dfrac{\pi}{4}\times(50.0\times10^{-6})^2$

$A=1.96\times10^{-9}\ m^2$

We need to calculate the magnitude of the force

Using formula of force

$F=\sigma A$

Put the value into the formula

$F=196\times10^{6}\times1.96\times10^{-9}$

$F=0.38416\ N$

(b). If the length of a strand of the hair is 12.0 cm at its breaking point

We need to calculate the unstressed length

Using formula of strain

$strain=\dfrac{\Delta l}{l_{0}}$

$\Delta l=strain\times l_{0}$

Put the value into the formula

$\Delta l=0.380\times l_{0}$

Length after expansion is 12 cm

We need to calculate the original length

Using formula of length

$l=l_{0}+\Delta l$

Put the value into the formula

$I=l_{0}+0.380\times l_{0}$

$l=1.38l_{0}$

$l_{0}=\dfrac{l}{1.38}$

$l_{0}=\dfrac{12\times10^{-2}}{1.38}$

$l_{0}=0.0869\ m$

Hence, (a). The magnitude of the force is 0.38416 N.

(b). The original length is 0.0869 m.