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

Both rod have the same tensile stress

Step-by-step 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 =
`(F)/(A)`

where

σ = tensile stress (N/
`m^(2)` or Pa)

F = Force (N)

A = Area (N/
`m^(2)` or Pa)

so

σ1 =
`(W_(1) )/(A_(1) )`, A = 2πl

=
`(1000)/(2\pi(2) )`

=
`(250)/(\pi )` N/
`m^(2)`

now calculate σ2

σ2 =
`(W_(2) )/(A_(2) )`

=
`(500)/(2\pi(1) )`

=
`(250)/(\pi )` N/
`m^(2)`

σ1/σ2 =
`(250)/(\pi )` /
`(250)/(\pi )`

σ1/σ2 = 1

σ1 = σ2

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