Question

Bhai sb velle hi hai yaa koi padhne waala bhi hai yahan..

iska ans bata do koi..
#Physics..
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Answer 1

A force F is being applied along a rod of transverse sectional area A,

Let the Force be = F

Let the Area be = A

The normal stress is given as pcosФ, where p is the pressure

Therefore,

p = F/A

Hence,

Normal stress = F/A × cosФ

Thus, the normal stress to a section PQ inclined o to  transverse section is  F/AcosФ

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