If a rod of length , very small area of $$cross-section$$ A, Young’s modulus of elasticity Y is acted upon by two parallel forces $$3F$$ and F respectively (as$$ $$shown) and placed on a smooth horizontal plane. If the elastic limit is not crossed and then to study the change in length of rod Δl and it’s elastic potential energy (U) the rod is segmented into four equal parts where magnitude of change in lengths are Δ$$l1$$,Δ$$l2$$,Δ$$l3$$,Δ$$l4$$ and elastic potential energy stored in each segment are $$U1$$,$$U2$$,$$U3$$,$$U4$$ respectively as shown then which $$is/are$$ correct?

Question

If a rod of length  , very small area of $$cross-section$$ A, Young’s modulus of elasticity Y is acted upon by two parallel forces $$3F$$ and F respectively (as$$ $$shown) and placed on a smooth horizontal plane. If the elastic limit is not crossed and then to study the change in length of rod Δl and it’s elastic potential energy (U) the rod is segmented into four equal parts where magnitude of change in lengths are Δ$$l1$$,Δ$$l2$$,Δ$$l3$$,Δ$$l4$$ and elastic potential energy stored in each segment are $$U1$$,$$U2$$,$$U3$$,$$U4$$ respectively as shown then which $$is/are$$ correct?

Infinity Learn · Accepted Answer

In steady state, the F.B.D. of the rod can be drawn by dividing it into four  equal parts as shown. We have  YΔ$$ll=stressaverage$$⇒Δ$$l=lY$$.$$FavgA=lY$$.$$F1+F22A$$  For any rod . If   Now,  Δ$$l1compression=Favg(length1)AY=F+02l4AY=Fl8AY$$;Δ$$l2elongation=Fl8AY$$ acceleration of the $$system=a=3F+F4m=FmFrom$$ FBD of all blocks we get $$N1$$,$$N2$$ and $$N3$$ are the normal reactions betweeen $$1st$$ and $$2nd$$ blocks,  $$2nd$$  and  $$3rd$$ blocks, and  $$3rd$$  and  $$4th$$ blocks on solving we get $$N1=0$$, $$N2=F$$ and $$N3=2F$$Δ$$l3($$ elongation $$)=2F+F2l4AY=3Fl8AY$$,Δ$$l4($$ elongation $$)=2F+3F2l4AY=5Fl8AY$$Δ$$l=8FI8AY=FlAY$$ ΔU∝Strain $$2$$⇒ΔU∝$$($$Δ$$l)2$$

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