anonymous
  • anonymous
A 135-N force is applied at a 50-degree angle to the surface of the end of a square bar. The surface is 2.50 cm on a side. What is the compressional stress?
Physics
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Hmm, I think I remember how to do this from a previous materials analysis course. Stress can be defined as: \[\huge \sigma = \frac{\text{F}}{\text{A}}\]The issue here is that we want compressional stress, which is the force that compresses the square bar. I think what we have to do is take the component that is parallel to the "area vector" AKA perpendicular to the cross sectional area surface. Visually, we want: |dw:1449652070420:dw| (assuming that I drew the right angle.. The component that is perpendicular to the cross sectional area would be \(\text{135} \cos(50)\). So \[\huge \sigma =\frac{135 \cos(50)}{\text{side}^2}\]
anonymous
  • anonymous
Let's see what @IrishBoy123 @rvc @Michele_Laino says X)
anonymous
  • anonymous
Pretty sure this makes sense though. The component of the force that is parallel to the "area vector" causes compression/tensile stress and the component of the force that is perpendicular to the "area vector" will cause shear stress.

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anonymous
  • anonymous
And an area vector is not a true vector, just one that helps us see if an area is perpendicular/parallel/etc to another measurable quantity, such as force |dw:1449652554285:dw|
anonymous
  • anonymous
Well I guess I shouldn't say it's not a true vector. It's just not a normal vector that we would think of.
Michele_Laino
  • Michele_Laino
you are right! @CShrix

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