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Can you please show me the corrected version?

Can you show me am fbd showing that ? I tried to draw an fbd for that. I applied force F on both ends of a cross section of the rod one on the top and bottom . Is that correct

I meant to say that the forces on all parts of rod should be 2F.

I feel I now know about FBD and forces . What's force? Force is defined as a push or pull which causes or tries to cause a change in its velocity. (I got to know about my error in the previous question. I added mg on the weighing machine which is actually acting on the body) But am I doing...

Thanks for your reply . I understood the 1st para of your explanation buy failed to completely understand the 2nd part. Here's what I think the 2nd part meant (please correct me if im wrong): Traction = F/A Force acting on free end , A = F Force acting on the constrained end B= F(in opp...

Still the force should be 2F? Isn't it?

Can you please explain it with a fbd?

Could you please explain how you got to the send equation?

So , What's L here? New length?

Stress = F(restoring)/ Cross sectional area Stress = 2F/A

Thanks for your explanation .

Whats L_o? Also , my question was that why is strain =[SIZE=14px][FONT=Google Sans]α Δ T . Also can you please elaborate your derivation a bit more (specifically the second step)

My attempt at deriving the equation: Let a temperature a rod of length L constrained by walls on both ends be changed by [SIZE=14px][FONT=Google Sans]ΔT. So the change in lenghth of rod under no constraints = L α ΔT Length of rod after change in temperature without walls =L+L α ΔT= L(1+L α ΔT)...

[SIZE=17px]You seem to have misinterpreted my question. When a person is standing on it Normal force equal to m(person)*g acts downwards along with m(person)g . So , f(downwards)= N+m(person)g= 2m(person)g.