Question

A 1000-n weight is hanging from a 2.0 m long aluminum rod. A 500-n weight is hanging from a 1.0 m long aluminum rod. The two aluminum rods have different cross sectional areas. Which rod is under the greater tensile stress?

Answer 1

Answer:

Both rod have the same tensile stress

Explanation:

Given information,

The weight first rod, $W_{1}$ = 1000 N

The length of first rod, $l_{1}$ = 2.0 m

The weight second rod, $W_{2}$ = 500 N

The length of second rod, $l_{2$ = 1.0 m

The equation of tensile stress, σ = $\frac{F}{A}$

where

σ = tensile stress (N/$m^{2}$ or Pa)

F = Force (N)

A = Area (N/$m^{2}$ or Pa)

so

σ1 =  $\frac{W_{1} }{A_{1} }$, A = 2πl

    = $\frac{1000}{2\pi(2) }$

    = $\frac{250}{\pi }$ N/$m^{2}$

now calculate σ2

σ2 =  $\frac{W_{2} }{A_{2} }$

     = $\frac{500}{2\pi(1) }$

     = $\frac{250}{\pi }$ N/$m^{2}$

σ1/σ2 = $\frac{250}{\pi }$ / $\frac{250}{\pi }$

σ1/σ2 = 1

σ1 = σ2

Hence, the tensile stress of first and second rod are the same.