A copper wire of length 2.2 M and steel wire of length 1.6 M both of diameter 3.0 millimetre are connected end to end when stretched by a load the net along the sun is found to be 0.70 CM obtain the load applied? pls ans this
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Length of copper wire=2.2 cm
Length of steal wire=1.6 cm
Diameter cl=3 mm=3x10*-3m
cross section are=/piR*2=/pi(d/2)*2
3.14x2.25x10*-6m*2
7.1x10*-6m*2
Net elongated =0.70mm
=7x10*-4m
Let load applied is F
y=2.0x10*11N/m*2
T=1.3x10*11N/m*2
Ys=stress/strain=F/A/(ls/Ls)
Ys=(F/A)/(lc/Lc)
Ys/Yc=(F/A)/(ls/Ls)x(lc/Lc)/(F/A)
=(Ls/Lc)*(lc/ls)
2.0x10*11/1.3x10*11
(2.2)/(1.16)x(lc/ls)
lc/ls=2x1.6/1.3x2.2=1.12
lc=1.12ls
1.12ls+ls=7x10*-4m
(1.12ls)=7x10*4m
ls=3.30x10*-4m
Ys=Fxls/Axls
F=Ys x A x ls/LS
=2x10*11x7.16x10*-6x3.30x10*-4/1.6
=2x7.1x33/1.6
=292.87N
Length of steal wire=1.6 cm
Diameter cl=3 mm=3x10*-3m
cross section are=/piR*2=/pi(d/2)*2
3.14x2.25x10*-6m*2
7.1x10*-6m*2
Net elongated =0.70mm
=7x10*-4m
Let load applied is F
y=2.0x10*11N/m*2
T=1.3x10*11N/m*2
Ys=stress/strain=F/A/(ls/Ls)
Ys=(F/A)/(lc/Lc)
Ys/Yc=(F/A)/(ls/Ls)x(lc/Lc)/(F/A)
=(Ls/Lc)*(lc/ls)
2.0x10*11/1.3x10*11
(2.2)/(1.16)x(lc/ls)
lc/ls=2x1.6/1.3x2.2=1.12
lc=1.12ls
1.12ls+ls=7x10*-4m
(1.12ls)=7x10*4m
ls=3.30x10*-4m
Ys=Fxls/Axls
F=Ys x A x ls/LS
=2x10*11x7.16x10*-6x3.30x10*-4/1.6
=2x7.1x33/1.6
=292.87N
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