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Notely Physics 12
Class-12 Physics

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6 - N u m e r i c a l   5

Problem

Young’s Modulus for a particular wood is 1×1010 Nm−21 \times 10^{10} \, \mathrm{Nm}^{-2}. A wooden chair has four legs, each with a length of 42 cm42 \, \mathrm{cm} and a cross-sectional area of 2×10−3 m22 \times 10^{-3} \, \mathrm{m}^{2}. Hamza, with a mass of 100 kg100 \, \mathrm{kg}, stands on the chair.
(a) What is the stress on each leg of the chair when Hamza stands on it?
(b) By how much do the chair legs compress under Hamza’s weight?

Data

  • Length of each leg: L=42 cm=0.42 mL = 42 \, \mathrm{cm} = 0.42 \, \mathrm{m}
  • Cross-sectional area: A=2×10−3 m2A = 2 \times 10^{-3} \, \mathrm{m}^2
  • Mass of Hamza: m=100 kgm = 100 \, \mathrm{kg}
  • Young’s Modulus: Y=1×1010 Nm−2Y = 1 \times 10^{10} \, \mathrm{Nm}^{-2}

Prerequisite Concepts

  1. Tensile Stress: Stress is the force applied per unit area:
Tensile Stress=FA \text{Tensile Stress} = \frac{F}{A}

where FF is the force applied, and AA is the cross-sectional area.

  1. Young’s Modulus: Young’s Modulus relates tensile stress to tensile strain:
Y=Tensile StressTensile Strain Y = \frac{\text{Tensile Stress}}{\text{Tensile Strain}}

where tensile strain is:

Tensile Strain=ΔLL \text{Tensile Strain} = \frac{\Delta L}{L}
  1. Force Due to Weight: The force acting on each leg is determined by Hamza’s weight:
F=mg F = mg

Solution

Step 1: Calculate the Stress on Each Leg

The total force exerted by Hamza is:

F=mg=100 kg×9.8 ms−2=980 NF = mg = 100 \, \mathrm{kg} \times 9.8 \, \mathrm{ms}^{-2} = 980 \, \mathrm{N}

Since the chair has four legs, the force on each leg is:

Fleg=F4=9804=245 NF_{\text{leg}} = \frac{F}{4} = \frac{980}{4} = 245 \, \mathrm{N}

Using the formula for tensile stress:

Tensile Stress=FlegA=2452×10−3\text{Tensile Stress} = \frac{F_{\text{leg}}}{A} = \frac{245}{2 \times 10^{-3}} Tensile Stress=1.225×105 Nm−2\text{Tensile Stress} = 1.225 \times 10^{5} \, \mathrm{Nm}^{-2}

Step 2: Calculate the Compression (Change in Length)

The tensile strain is given by:

Tensile Strain=Tensile StressY\text{Tensile Strain} = \frac{\text{Tensile Stress}}{Y}

Substitute the values:

Tensile Strain=1.225×1051×1010=1.225×10−5\text{Tensile Strain} = \frac{1.225 \times 10^{5}}{1 \times 10^{10}} = 1.225 \times 10^{-5}

The change in length is:

ΔL=Tensile Strain×L=(1.225×10−5)×0.42\Delta L = \text{Tensile Strain} \times L = (1.225 \times 10^{-5}) \times 0.42 ΔL=5.145×10−6 m\Delta L = 5.145 \times 10^{-6} \, \mathrm{m}

Answer

(a) The stress on each leg of the chair is:

Tensile Stress=1.225×105 Nm−2\text{Tensile Stress} = 1.225 \times 10^{5} \, \mathrm{Nm}^{-2}

(b) The compression (change in length) of each chair leg is:

ΔL=5.145×10−6 m or 5.145 μm\Delta L = 5.145 \times 10^{-6} \, \mathrm{m} \, \text{or} \, 5.145 \, \mu\mathrm{m}