Two steel wires are connected together, end to end, and attached to a wall as sh

Question

Two steel wires are connected together, end to end, and attached to a wall as shown below. The two wires have the same length and elastic modulus, but the ratio of the radius of the first wire to the radius of the second wire is 5:4. As the wires are initially the same length, the midpoint of the combination coincides with the connection point. An applied force then stretches the combination by 1.850 mm while the two wires stay connected together. After the wires are stretched, what is delta, the distance from the midpoint to the connection point?

Explanation / Answer

By using Young's modulus we know the stretch of each wire is proportional to the cross sectional area of each wire or the squares of their radii. The smaller wire will stretch more by a factor of 5^2/4^2 = 25/16

x + 25x/16 = 1.850

x = 0.722 mm

The big wire stretches 0.722 mm and

the smaller wire stretches =1.85 -0.722 = 1.128 mm

The midpoint of the wires moves half the total stretch =1.85/2 = 0.925 mm

The connection point moves the stretch of the bigger wire because according to the question it is the wire connected to the immovable wall.

So the distance from the midpoint to the connection point = 0.925 - 0.722 = 0.203 mm

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.