# Physics (Stress & Strain)

What is the factor of safety for a steel hanger having an ultimate strength of 70000N per cm^2 and supporting a load of 87500N. The hanger has a cross sectional area of 5cm^2.

Factor of safety =
ultimate stress/ allowable stress
= 70000/87500
= 0.8
Have I done this correctly?

1. Stress = force / area

Load is 87500N which is just a force.

You need to divide the force by the area resisting the tensile stress.

Factor of safety is then
ultimate stress / stress under load

posted by MathMate
2. Stress = force/ area
= 87500N/ 5cm^2
= 17500

= 70000/ 17500
ANS = 4

Is that correct?

posted by Danny
3. Correct!

posted by MathMate

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