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Question Number 18666 by tawa tawa last updated on 26/Jul/17

An elastic material has a length of 36cm when a load of 40N is hung on it and a length of 45cm when a load of 60N is hung on it. what is the Original length of the string ?

$$\mathrm{An}\:\mathrm{elastic}\:\mathrm{material}\:\mathrm{has}\:\mathrm{a}\:\mathrm{length}\:\mathrm{of}\:\:\mathrm{36cm}\:\mathrm{when}\:\mathrm{a}\:\mathrm{load}\:\mathrm{of}\:\mathrm{40N}\:\mathrm{is}\:\mathrm{hung}\:\mathrm{on}\:\mathrm{it}\:\mathrm{and} \\ $$$$\mathrm{a}\:\mathrm{length}\:\mathrm{of}\:\mathrm{45cm}\:\mathrm{when}\:\mathrm{a}\:\mathrm{load}\:\mathrm{of}\:\mathrm{60N}\:\mathrm{is}\:\mathrm{hung}\:\mathrm{on}\:\mathrm{it}.\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{Original}\: \\ $$$$\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{string}\:? \\ $$

Answered by mrW1 last updated on 27/Jul/17

36−((45−36)/(60−40))×40=18 cm

$$\mathrm{36}−\frac{\mathrm{45}−\mathrm{36}}{\mathrm{60}−\mathrm{40}}×\mathrm{40}=\mathrm{18}\:\mathrm{cm} \\ $$

Commented by tawa tawa last updated on 27/Jul/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by tawa tawa last updated on 27/Jul/17

Any formular sir ?

$$\mathrm{Any}\:\mathrm{formular}\:\mathrm{sir}\:? \\ $$

Commented by mrW1 last updated on 27/Jul/17

x=length without load k=elongation for unit load 36−x=40k ...(i) 45−x=60k ...(ii) (i)/(ii): ((36−x)/(45−x))=((40)/(60)) 36×60−60x=45×40−40x (60−40)x=36×60−45×40 x=((36×60−45×40)/(60−40))=18

$$\mathrm{x}=\mathrm{length}\:\mathrm{without}\:\mathrm{load} \\ $$$$\mathrm{k}=\mathrm{elongation}\:\mathrm{for}\:\mathrm{unit}\:\mathrm{load} \\ $$$$\mathrm{36}−\mathrm{x}=\mathrm{40k}\:\:\:...\left(\mathrm{i}\right) \\ $$$$\mathrm{45}−\mathrm{x}=\mathrm{60k}\:\:\:...\left(\mathrm{ii}\right) \\ $$$$\left(\mathrm{i}\right)/\left(\mathrm{ii}\right): \\ $$$$\frac{\mathrm{36}−\mathrm{x}}{\mathrm{45}−\mathrm{x}}=\frac{\mathrm{40}}{\mathrm{60}} \\ $$$$\mathrm{36}×\mathrm{60}−\mathrm{60x}=\mathrm{45}×\mathrm{40}−\mathrm{40x} \\ $$$$\left(\mathrm{60}−\mathrm{40}\right)\mathrm{x}=\mathrm{36}×\mathrm{60}−\mathrm{45}×\mathrm{40} \\ $$$$\mathrm{x}=\frac{\mathrm{36}×\mathrm{60}−\mathrm{45}×\mathrm{40}}{\mathrm{60}−\mathrm{40}}=\mathrm{18} \\ $$

Commented by mrW1 last updated on 27/Jul/17

or solution without equation: load from 40 to 60: length from 36 to 45. ⇒elongation for unit load: ((45−36)/(60−40)) ⇒elongation for load= 40 is ((45−36)/(60−40))×40 ⇒length at load =0 is 36−((45−36)/(60−40))×40=18 cm

$$\mathrm{or}\:\mathrm{solution}\:\mathrm{without}\:\mathrm{equation}: \\ $$$$\mathrm{load}\:\mathrm{from}\:\mathrm{40}\:\mathrm{to}\:\mathrm{60}: \\ $$$$\mathrm{length}\:\mathrm{from}\:\mathrm{36}\:\mathrm{to}\:\mathrm{45}. \\ $$$$\Rightarrow\mathrm{elongation}\:\mathrm{for}\:\mathrm{unit}\:\mathrm{load}: \\ $$$$\frac{\mathrm{45}−\mathrm{36}}{\mathrm{60}−\mathrm{40}} \\ $$$$ \\ $$$$\Rightarrow\mathrm{elongation}\:\mathrm{for}\:\mathrm{load}=\:\mathrm{40}\:\mathrm{is} \\ $$$$\frac{\mathrm{45}−\mathrm{36}}{\mathrm{60}−\mathrm{40}}×\mathrm{40} \\ $$$$ \\ $$$$\Rightarrow\mathrm{length}\:\mathrm{at}\:\mathrm{load}\:=\mathrm{0}\:\mathrm{is} \\ $$$$\mathrm{36}−\frac{\mathrm{45}−\mathrm{36}}{\mathrm{60}−\mathrm{40}}×\mathrm{40}=\mathrm{18}\:\mathrm{cm} \\ $$

Commented by tawa tawa last updated on 27/Jul/17

Wow, God bless you sir.

$$\mathrm{Wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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