Mechanical Properties of Solids
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Question : 18
Total: 21
A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in figure. The cross- sectional areas of wire A and B are 1.0 mm2 2
Solution:
For steel wire A , l 1 = l ; A 1 = 1 m m 2 ;
Y 1 = 2 × 10 11 N m − 2
For aluminium wireB , l 2 = 1 ; A 2 = 2 m m 2 ;
Y 2 = 7 × 10 10 N m − 2
(a) Let mass m be suspended from the rod at distance x from theend where wire A is connected, LetF 1 and F 2
=
=
=
. . . ( i )
Taking moment of forces about thepoint of suspension of mass from the rod, we have
F 1 x = F 2 ( 1.05 − x )
or
=
=
or2.10 − 2 x = x
orx = 0.70 m = 70 c m
(b) Let mass m be suspended from the rod at distance x from the end where wire A is connected. Let andF 1 F 2
=
or
=
=
×
=
As the rod is stationary, so
F 1 x = F 2 ( 1.05 − x )
or
=
=
or10 x = 7.35 − 7 x
orx = 0.4324 m = 43.2 c m
For aluminium wire
(a) Let mass m be suspended from the rod at distance x from theend where wire A is connected, Let
Taking moment of forces about thepoint of suspension of mass from the rod, we have
or
or
or
(b) Let mass m be suspended from the rod at distance x from the end where wire A is connected. Let and
or
As the rod is stationary, so
or
or
or
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