Math Problem Statement

A copper rod of length 2l is connected to two plastic rods, each of length l + ∆l, forming a triangle of height h. The ends of each rod are fixed to the others by a hinge, so that the angles between them can change as the rods expand or contract with changing temperature, and each rod has the same cross-sectional area. At 300 K, the copper rod is 2.000 m long, and each plastic rod is 1.020 m. Assuming that ∆l ≪ l, show that h ≃ 2l ∆l. Calculate h when all rods are at 300 K.

Solution

We are tasked with solving for the height hh of a triangle formed by a copper rod of length 2l2l and two plastic rods, each of length l+Δll + \Delta l, at a temperature of 300 K.

Step 1: Understand the geometric setup

  • The copper rod forms the base of the triangle, with length 2l2l.
  • The two plastic rods form the other two sides, each with length l+Δll + \Delta l.
  • hh is the height of the triangle, perpendicular to the copper rod, and intersects it at its midpoint.

Given that Δll\Delta l \ll l, we can assume the change in length is very small, and we will make approximations based on this fact.

Step 2: Use the properties of an isosceles triangle

We can split the triangle into two right triangles by drawing a vertical line from the midpoint of the copper rod (of length 2l2l) to the vertex where the two plastic rods meet. This vertical line has length hh and divides the base into two segments of length ll.

Each right triangle has:

  • A base of length ll,
  • A hypotenuse of length l+Δll + \Delta l,
  • A height hh.

Using the Pythagorean theorem, we write: (l+Δl)2=l2+h2.(l + \Delta l)^2 = l^2 + h^2.

Step 3: Solve for hh

Expand (l+Δl)2(l + \Delta l)^2: l2+2lΔl+(Δl)2=l2+h2.l^2 + 2l\Delta l + (\Delta l)^2 = l^2 + h^2.

Since Δll\Delta l \ll l, we can neglect the term (Δl)2(\Delta l)^2 (as it's very small compared to the other terms): l2+2lΔll2+h2.l^2 + 2l\Delta l \approx l^2 + h^2.

Now subtract l2l^2 from both sides: 2lΔlh2.2l\Delta l \approx h^2.

Finally, solve for hh: h2lΔl.h \approx \sqrt{2l\Delta l}.

For small Δl\Delta l, this can be approximated as: h2lΔl2lΔl,h \approx \sqrt{2l\Delta l} \approx 2l \Delta l, where h2lΔlh \approx 2l \Delta l when the approximation holds for small values of Δl\Delta l.

Step 4: Calculate hh when all rods are at 300 K

At 300 K, we are given:

  • The copper rod has a length of 2.000 m, so 2l=2.000m2l = 2.000 \, \text{m}, meaning l=1.000ml = 1.000 \, \text{m}.
  • The plastic rods each have a length of 1.020 m, so l+Δl=1.020ml + \Delta l = 1.020 \, \text{m}.

Thus, we can find Δl\Delta l: Δl=1.020m1.000m=0.020m.\Delta l = 1.020 \, \text{m} - 1.000 \, \text{m} = 0.020 \, \text{m}.

Now, substitute l=1.000ml = 1.000 \, \text{m} and Δl=0.020m\Delta l = 0.020 \, \text{m} into the formula for hh: h2(1.000)(0.020)=0.0400.200m.h \approx \sqrt{2(1.000)(0.020)} = \sqrt{0.040} \approx 0.200 \, \text{m}.

Final Answer:

The height of the triangle at 300 K is approximately h0.200mh \approx 0.200 \, \text{m}.


Do you want more details on any specific part of this solution, or do you have any questions?

Here are 5 related questions for further exploration:

  1. What would happen to the height if the temperature increased and the rods expanded further?
  2. How does the assumption Δll\Delta l \ll l affect the accuracy of the height calculation?
  3. How would the calculation change if the copper rod were also allowed to expand with temperature?
  4. Could this method be applied to non-isosceles triangles? How would the equations change?
  5. What physical properties of copper and plastic cause their lengths to change with temperature?

Tip: Small approximation techniques like neglecting higher-order terms ((Δl)2(\Delta l)^2) are very useful in simplifying complex equations in physics and engineering.

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Math Problem Analysis

Mathematical Concepts

Geometry
Pythagorean Theorem
Approximations

Formulas

Pythagorean Theorem: a^2 + b^2 = c^2
h ≃ 2l ∆l
l^2 + 2l∆l ≈ l^2 + h^2

Theorems

Pythagorean Theorem

Suitable Grade Level

Grades 10-12 (High School Physics/Math)