A steel wire of original length 1 m and cross-sectional area 4.00 mm2 is clamped at the two ends so that it lies horizontally and without tension. If a load of 2.16 kg is suspended from the middle point of the wire, what would be its vertical depression ? Y of the steel = 2.0 x1011 n/m2. Take g=10 m/s2.
Answers
Let 'AB' be the length of the wire.
AB= 2l= 1m
Area of cross section, A= 4 mm²= 4×10⁻²cm²
Young's modulus, Y= 2×10¹¹ N/m²
Tension 'T' is built in the steel wire when load is suspended at the mid point of the wire as shown in the fig.
For suspension of load, a pick point is generated in the wire making an angle θ with vertical.
Let the vertical extension be x.
Therefore, cosθ= x/√(x²+l²)
Considering x<<l, then cosθ=x/l
From the figure, increase in length, ΔL=(AC+BC)-AB
But AC= √(l²+x²)= CB
∴ΔL= 2√(l²+x²)- 2l
We know, Young's modulus, Y= F.L/(A.ΔL)----- (1)
From the figure, 2Tcosθ=mg
2T(x/50)= 2.16×10³×980
(1)=> 2×10¹²= (T×100)/[4×10⁻²×{2×√(50²×x²)-100}]
[∵L= 1m= 100 cm]
=> [2×(2×10¹²)×(4×10⁻²)×{2×√(50²+x²)-100}x}/ (100×50)= 2.16×10³×98
Solving the above equation, we find the value of x is 1.5 cm