Question

Two steel wires of radii r and 2r are connected together end to end and tied to a wall as shown. The force stretches the combination by 10 mm. How far does the junction point A move, (in mm)

Answer 1

8

Answer 2

The problem involves two steel wires of different radii connected end to end and subjected to a tensile force.

Let the first wire (connected to the wall) have radius $r_1 = r$ and length $L_1 = L$. Its area of cross-section is $A_1 = \pi r_1^2 = \pi r^2$.

Let the second wire (connected to the force $F$) have radius $r_2 = 2r$ and length $L_2 = L$. Its area of cross-section is $A_2 = \pi r_2^2 = \pi (2r)^2 = 4 \pi r^2$.

Both wires are made of steel, so they have the same Young's modulus, let's denote it by $Y$.

When a force $F$ is applied to the combination, both wires are under the same tension $F$. The elongation of a wire under tension is given by the formula $\Delta L = \frac{F L}{A Y}$.

The elongation of the first wire is $\Delta L_1 = \frac{F L_1}{A_1 Y} = \frac{F L}{\pi r^2 Y}$.

The elongation of the second wire is $\Delta L_2 = \frac{F L_2}{A_2 Y} = \frac{F L}{4 \pi r^2 Y}$.

Comparing the elongations, we can see that $\Delta L_2 = \frac{1}{4} \left(\frac{F L}{\pi r^2 Y}\right) = \frac{1}{4} \Delta L_1$.

The total elongation of the combination is the sum of the individual elongations: $\Delta L_{total} = \Delta L_1 + \Delta L_2$. We are given that the total elongation is 10 mm. $\Delta L_{total} = 10 \text{ mm}$.

Substituting the relationship between $\Delta L_1$ and $\Delta L_2$ into the total elongation equation:

$10 \text{ mm} = \Delta L_1 + \frac{1}{4} \Delta L_1$

$10 \text{ mm} = \left(1 + \frac{1}{4}\right) \Delta L_1$

$10 \text{ mm} = \frac{5}{4} \Delta L_1$.

Solving for $\Delta L_1$:

$\Delta L_1 = \frac{4}{5} \times 10 \text{ mm} = 8 \text{ mm}$.

The junction point A is located at the end of the first wire (the one with radius $r$). The other end of the first wire is fixed to the wall. When the force is applied, the first wire elongates by $\Delta L_1$. Since the left end of the first wire is fixed, the junction point A moves by an amount equal to the elongation of the first wire.

Displacement of junction point A = $\Delta L_1$.

Displacement of junction point A = 8 mm.