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anonymous
 3 years ago
A uniform steel wire X and Y, suspended at the same horizontal level, are connected by a light and crosssectional area A, A weight of 30 N is placed at O, on the road OX : OY = 1 : 2. Both wires are stretched and the rod XY then remains horizontal.
If he wire X has a young modulus E1 of 1.0 * 10^11 pa, calculate the yound modulus of E2 of the wire Y assuming the elastic limit is not exceeded for both wires.
anonymous
 3 years ago
A uniform steel wire X and Y, suspended at the same horizontal level, are connected by a light and crosssectional area A, A weight of 30 N is placed at O, on the road OX : OY = 1 : 2. Both wires are stretched and the rod XY then remains horizontal. If he wire X has a young modulus E1 of 1.0 * 10^11 pa, calculate the yound modulus of E2 of the wire Y assuming the elastic limit is not exceeded for both wires.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1358779489492:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Hero @Preetha @ghazi @AravindG

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0I think I saw a similiar question in NCERT text chapter 11 under topic of Young's modulus and all..take a look

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can u give me the link of the book?

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0you are not from India?

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0i suggest you google it..its easy to get NCERT CLASS 11 PHYSICS PART 2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0do they have the solutions

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0infact the book is very nice for reference

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0http://www.vidhyarjan.com/news/ncertsolutionsforclass11physics

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can u change the language?

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0ofcourse eng is below scroll down

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the question isnt there

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok.. so since the xy is horizontal.. it means after stretching, both the bars have exactly the same length.. so i can say \[l1+\Delta l1 = l2 + \Delta l2 \] next step we find the stress that is coming on each of them.. that should be easy.. since the ration of ox: oy is given to us..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can u show the working please:)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0try doing it :P.. first you tell me how much will be the stress that comes on X and Y ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0stress = force/ area how do i distrbte the force

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0first find the force.. force equals the weight that each one would experience.. the weight is at a ratio of 2:1 .. hence the distribution will be in inverse ratio right? 1:2.. cause if weight is far away, its effect is smaller!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so now calculate Fx and Fy

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Fx = 2/3 Fy Fy = 1/3 Fx ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0total force is known to you.. so find the ratios now!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Fx = 2/3 Ftotal :P and Fy = 1/3 Ftotal

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Fx = 20N and Fy = 10N

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok.. :D.. next.. find the stress on X and since young's modulus of X is given.. find the strain.. Similarly do the same thing for Y.. but young's modulus is not given.. divide the two equations to get rid of A

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thanks alot Strain = F/A*E1 strain in X = 20/A* 1.0 * 10^11 strain is equal so E2 = stress/ strain E2 = 10/A *A*1.0*10^11/20 E2 = 5 * 10^11 pa

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0woah wait.. how did you get the strain for Y ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0nope :/.. why would the strains be equal?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now you have to use the first equatoin.. that i put.. but use lx and ly notation instead of l1 and l2 notation

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0they are NOT SAME MATERIAL.. IF THEY WERE.. THEY WOULD HAVE SAME YOUNG"S MODULUS :P

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i dont undrstnd >< cn u shw the working

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0use the first equation .. try doing it.. see you need to get rid of \[\Delta ly/ ly\] in that equation i asked you to build by dividing the two equations

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so here are the two equations \[Fx / Fy = Ex \Delta x ly / Ey \Delta ylx \] and \[lx + \Delta x = ly + \Delta y\] somehow use them

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok il use thnks fr ur help :)
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