Question

Assume that Young’s modulus for a new material is known to be 2.25 × 1011 /2

. The material
breaks when it is subjected to a stress greater than 2.1 × 107 /2
. What is the maximum force
that can be applied to the material if the sample being tested has a diameter of 50.5 mm?

Answer 1

Maximum force,
`F_(max) = 42 kN`

Step-by-step explanation:

Maximum Stress,
`\sigma_(max) = 2.1 * 10^(7) N/m^(2)`

The diameter of the material, d = 50.5 mm = 0.0505 m

The area of the material,
`A = (\pi d^(2) )/4`

`A = (\pi * 0.0505^(2) )/(4) \\A = 0.002 m^(2)`

The formula for the maximum stress that the material can take is given by:

`\sigma_(max) = (F_(max) )/(A) \\2.1 * 10^7 = (F_(max) )/(0.002)\\F_(max) = 2.1 * 10^7 * 0.002\\F_(max) = 4.2 * 10^4 N\\F_(max) = 42 kN`