2020 A/L Physics Past Paper · Question 39 (English Medium)

A sagging metal wire — Young's modulus

A metal wire of length 2 m2\ \text{m} and cross-sectional area 5 mm25\ \text{mm}^2 is stretched taut between two points AA and BB which are 2 m2\ \text{m} apart in the same horizontal plane. A mass of 2.4 kg2.4\ \text{kg} is then hung from the midpoint of the wire as shown in the figure. In this situation the midpoint of the wire is found to sag 2.0 cm2.0\ \text{cm} from its original position, and the total extension of the wire is found to be 0.04 cm0.04\ \text{cm}. What is the approximate value of Young's modulus of the metal?

  1. (1) 2×1011 N m−22 \times 10^{11}\ \text{N m}^{-2}
  2. (2) 3×1011 N m−23 \times 10^{11}\ \text{N m}^{-2}
  3. (3) 4×1011 N m−24 \times 10^{11}\ \text{N m}^{-2}
  4. (4) 6×1011 N m−26 \times 10^{11}\ \text{N m}^{-2}
  5. (5) 12×1011 N m−212 \times 10^{11}\ \text{N m}^{-2}
Answer and explanation

Answer: (4) 6×1011 N m−26 \times 10^{11}\ \text{N m}^{-2}

Given information

We know: l=2 ml = 2\ \text{m}, A=5 mm2=5×10−6 m2A = 5\ \text{mm}^2 = 5 \times 10^{-6}\ \text{m}^2; mass m=2.4m = 2.4 kg (weight 2424 N, g=10 m s−2g = 10\ \text{m s}^{-2}); sag of the midpoint y=2.0y = 2.0 cm; total extension e=0.04 cm=4×10−4e = 0.04\ \text{cm} = 4 \times 10^{-4} m. The wire is initially taut (straight, with no tension stated) between AA and BB.

What the question asks

We must find: Young's modulus YY of the metal.

Relevant theory

Every object continues to be at rest or move with uniform velocity unless acted upon by a resultant force.

The tensile force acting on a unit area of cross section of the wire is known as the tensile stress.

The extension per unit length of the wire is known as the tensile strain.

The ratio of tensile stress to tensile strain is called the Young's modulus of the material of the wire.

Required equations

2Tsin⁡θ=mg2T\sin\theta = mg

sin⁡θ≈yl/2\sin\theta \approx \frac{y}{l/2}

Y=T/Ae/lY = \frac{T/A}{e/l}

Applying it to this question

Plan: (1) the slope of each half from the sag.

(2) since the midpoint is at rest, the tension TT from the balance of forces.

(3) stress T/AT/A.

(4) strain e/le/l.

(5) divide.

The tension is the same throughout the wire (TT), since the wire is light.

Reasoning and calculation

0.020.02

sin⁡θ≈2.0 cm100 cm\sin\theta \approx \frac{2.0\ \text{cm}}{100\ \text{cm}}

600600

T=242×0.02T = \frac{24}{2 \times 0.02}

1.2×1081.2 \times 10^8

TA=6005×10−6\frac{T}{A} = \frac{600}{5 \times 10^{-6}}

2×10−42 \times 10^{-4}

el=4×10−42\frac{e}{l} = \frac{4 \times 10^{-4}}{2}

6×10116 \times 10^{11}

Y=1.2×1082×10−4Y = \frac{1.2 \times 10^8}{2 \times 10^{-4}}

Checks

Unit check: N m−2(unitless)=N m−2\dfrac{\text{N m}^{-2}}{\text{(unitless)}} = \text{N m}^{-2} ✓.

Physical plausibility: a 2424 N weight needs a 600600 N tension — since the wire is nearly horizontal, a small vertical component needs a large tension ✓.

Y=6×1011Y = 6 \times 10^{11} is larger than steel's 2.0×10112.0 \times 10^{11}, but still within the range of metals (an approximate value is asked for) ✓.

Number check: from the sag geometry, the extension 2(1+0.022−1)≈4×10−42\left(\sqrt{1 + 0.02^2} - 1\right) \approx 4 \times 10^{-4} m =0.04= 0.04 cm — matches the given data ✓.

(m∗10/(2∗y)/A)/(e/l)(m * 10 / (2 * y) / A) / (e / l)

Conclusion

Y=6×1011 N m−2Y = 6 \times 10^{11}\ \text{N m}^{-2}

T=600T = 600 N, stress 1.2×108 N m−21.2 \times 10^8\ \text{N m}^{-2}, strain 2×10−42 \times 10^{-4}: Y=6×1011 N m−2Y = 6 \times 10^{11}\ \text{N m}^{-2} — the answer is (4).

Taking the tension as equal to the hanging weight (2424 N).

The tensions are nearly horizontal; it is their vertical components that bear the weight: 2Tsin⁡θ=mg2T\sin\theta = mg.

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