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.

(Y_Al = 70 × 10⁹ Nm⁻² and Y_steel = 200 × 10⁹ Nm^–2)

1. Mass m should be suspended close to wire A to have equal stresses in both the wires.
2. Mass m should be suspended close to B to have equal stresses in both the wires.
3. Mass m should be suspended at the middle of the wires to have equal stresses in both the wires.
4. Mass m should be suspended close to wire A to have equal strain in both wires.

Answer 1

b and d

Explanation:

Let the mass is placed at x from the end B.

Let T_A and T_B 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 a_A and a_B are cross-sectional areas of wire A and B respectively.

By question a_B = 2a_A

Now, for equal stress S_A = S_B 

⇒ `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 Y_A and Y_B 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.