The length of the wire is 3.0 m, the change in length (elongation) of the wire is 2.1 mm, the young’s modulus of the copper wire is 120×109 Pa and the force or weight at the end is 120 N.

Write the ex

Write the expression to calculate the power output of the pump.

P=mgyt

Here, P is the power, m is the mass of the water, g is the acceleration due to gravity, y is the depth of the well.

Write the expression for the force and young’s modulus of the wire.

FA=YΔLL (I)

Here, F is the force at the end, A is the cross-sectional area of the wire, Y is the young’s modulus, L is the length of the wire and ΔL is the change in length.

Write the expression to calculate the cross-sectional area of the wire.

A=π(d2)2

Here, d is the diameter of the wire.

Use the above expression in (I) to rewrite.

Fπ(d2)2=YΔLL

Rewrite the above equation in terms of d.

d=2FYπ(LΔL)

Substitute 3.0 m for L, 2.1 mm for ΔL, 120×109 Pa for Y and 120 N for F in the above equation to calculate d.

d=2(120 N120×109 Pa(π))(3.0 m2.1 mm(10−3 m1 mm))d=0.0013 m(103 mm1 m)=1.3 mm

Write the expression for the tensile stress.

FA=YΔLL

Here, FA is the tensile stress.

Substitute 3.0 m for L, 2.1 mm for ΔL, 120×109 Pa for Y in the above equation to calculate FA.

FA=(120×109 Pa)2.1 mm(10−3 m1 mm)3.0 m=8.4×107 N/m2

Conclusion:

Therefore, the diameter and tensile stress in the wire is respectively 1.3 mm and 8.4×107 N/m2.

The tensile strength is 400 MPa.

Write the expression for the tensile stress.

WA=F′

Here, F′ is the tensile strength and W is the maximum weight.

Rewrite the above equation using A=π(d2)2 in terms of W.

Wπ(d2)2=F′W=πFd24

Substitute 400 MPa for F′ and 1.3 mm for d in the above equation to calculate W.

W=π(400 MPa)(106 Pa1 MPa)(1.3 mm(10−3 m1 mm))24=531 N

Conclusion:

Therefore, the maximum weight in order to hang the wire is 531 N.