How Far Should a 100kg Mass Be Placed from Point A to Keep a Rod Horizontal?

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A light rod is suspended by two wires of equal length but different cross-sectional areas, with one wire elongating due to temperature changes. To keep the rod horizontal with a 100kg mass, the tensions in both wires must be equal, and their total lengths must also match after elongation. The Young's modulus equation is applied to relate the tensions and extensions of the wires. The necessary condition for equilibrium involves ensuring both wires have the same effective length after accounting for elongation. Solving these equations will determine the appropriate distance for the mass placement to maintain horizontal alignment.
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A light rod is hung by two wires having same length of 2m and the same young's modulus of 8\times1011Pa,but having different cross sectional areas of 1mm2 and 2mm2.
The wire having the cross sectional area of 1mm2,gets an increment of 2mm in length because of a temperature increment.But the other wire remains same..
A mass of 100kg is meant to be put on the rod to keep the rod horizontal..How much distance has it to be on the rod from point A...?

PLEASE SOLVE THIS CITING REASONS...thanks !
 
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welcome to pf!

hi shalikadm! welcome to pf! :wink:

show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
Several things

i want to know what the necessity that is to be fulfilled in order to keep it horizontally.
whether it is to have same extension(e) for the two wires
or to have the same total length(l+e) for both of the wires...
or to fulfill e1mm2+2mm=e2mm2
want to know whether to get the original length(l) of the 1mm2 wire as 2m or 2m+2mm

Y=\frac{Fl}{Ae}

please help me...
 
after the temperature increment, one wire is longer

apply the young's modulus equation to those lengths :smile:

(of course, you need to extend both wires, to make them the same length)
 
is it like this ?

\downarrow
for the 1mm2wire,
Y=\frac{T1*2.002}{1*10^-6*e1}\Rightarrow1

for the 2mm2wire,
Y=\frac{T2*2}{2*10^-6*e2}\Rightarrow2

hence 1=2 ,
\frac{T1*2.002}{1*10^-6*e1}=\frac{T2*2}{2*10^-6*e2}\RightarrowA

also,
T1+T2=1000\RightarrowB

and also ,
taking moment around the center of 100kg's gravity\RightarrowC

and then solving the A,B,C equations ?
is that right ?
 
yes, but you'll also need an equation relating e1 and e2 :smile:

(btw, please don't make the latex equations larger …

it's not necessary …

each reader can permanently adjust equation size by right-clicking on any equation and choosing "Scale All Math" :wink:)​
 
how can i get a equation relating e1 and e2 ?
sorry for larger Latex...:smile:
 
shalikadm said:
how can i get a equation relating e1 and e2 ?

it'll be the equation saying the two wires are the same length :smile:
 
Is it this ?(i have no idea)
e1mm2+2mm=e2mm2
 
  • #10
shalikadm said:
Is it this ?(i have no idea)
e1mm2+2mm=e2mm2

yes, so it's the same total length for both wires :smile:
 
  • #11
hope this help me...thanks a lot...!
 
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