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Question
A rod of length l and negligible mass is suspended at its two ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths (Figure). The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 mm2, respectively.
(YAl = 70 × 109 Nm−2 and Ysteel = 200 × 109 Nm–2)
- Mass m should be suspended close to wire A to have equal stresses in both the wires.
- Mass m should be suspended close to B to have equal stresses in both the wires.
- Mass m should be suspended at the middle of the wires to have equal stresses in both the wires.
- Mass m should be suspended close to wire A to have equal strain in both wires.
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Solution
b and d
Explanation:
Let the mass is placed at x from the end B.
Let TA and TB be the tensions in wire A and wire B respectively.
For the rotational equilibrium of the system,
∑τ = 0 ......(Total torque = 0)
⇒ `T_Bx = T_A (l - x)` = 0
⇒ `T_B/T_A = (l - x)/x` ......(i)
Stress in wire `A = S_A = T_A/a_A`
Stress in wire `B = S_B = T_B/a_B`
Where aA and aB are cross-sectional areas of wire A and B respectively.
By question aB = 2aA
Now, for equal stress SA = SB
⇒ `T_A/a_A = T_B/a_B`
⇒ `T_B/T_A = a_B/a_A` = 2
⇒ `(l - x)/x` = 2
⇒ `l/x - 1` = 2
⇒ `x = l/3`
⇒ `l - x = l - l/3 = (2l)/3`
Hence, mass m should be placed closer to B.
For equal train, (strain)A = (strain)B
⇒ `(Y_A)/S_A = Y_B/S_B` ......(Where YA and YB are Young moduli)
⇒ `Y_(steel)/(T_A/a_A) = Y_(Al)/(T_B/a_B)`
⇒ `Y_(steel)/Y_(Al) = T_A/T_B xx a_B/a_A = (x/(l - x))((2a_A)/a_A)`
⇒ `(200 xx 10^9)/(70 xx 10^9) = (2x)/(l - x)`
⇒ `20/7 = (2x)/(l - x)`
⇒ `10/7 = x/(l - x)`
⇒ `10l - 10x = 7x`
⇒ `17x = 10l`
⇒ `x = (10l)/17`
`l - x = l - (10l)/17 = (7l)/17`
Hence, mass m should be placed closer to wire A.
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