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Chapter 7: Problem 874

A uniform plank of young's modulus \(\mathrm{Y}\) is moved over a smooth horizontal surface by a constant horizontal force \(\mathrm{F}\), The area of cross-section of the plank is \(\mathrm{A}\). What is the compressive strain on its plank in the direction by the force ? (A) (F / AY) (B) \((2 \mathrm{~F} / \mathrm{AY})\) (C) \((1 / 2)(\mathrm{F} / \mathrm{AY})\) (D) (3F / AY)

Short Answer

Expert verified
The compressive strain on the plank in the direction of the force is (A) \(\frac{F}{AY}\).

Step by step solution

01

Write down the given information

We are given the following: - Young's Modulus (Y) - A constant horizontal force (F) - Cross sectional area of the plank (A) We need to find the compressive strain on the plank in the direction of the force.
02

Recall the definition of strain and the formula for Young's Modulus

Strain is the change in length per unit length. The formula for strain is: \[ Strain=\frac{\Delta L}{L} \] Young's modulus (Y) is defined as the ratio of stress to strain. The formula for Young's modulus is: \[ Y = \frac{Stress}{Strain} \]
03

Replace stress in the Young's Modulus formula with force and area

Stress is defined as the force per unit area. So, we can replace stress in the Young's modulus formula with force(F) and area(A). \[ Y = \frac{F / A}{Strain} \]
04

Solve the equation for strain

Rearrange the equation to solve for strain. \[ Strain = \frac{F}{AY} \]
05

Identify the correct answer

Now we can compare the obtained result to the given choices: (A) (F / AY) (B) \((2F / AY)\) (C) \((1 / 2)(F / AY)\) (D) (3F / AY) So, the correct answer is (A) (F / AY).

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