Answer to Question #104402 in Geometry for Jake

Question

When using a pencil eraser, you exert a vertical force of 6.00 N at a distance of 2.00 cm from the hardwood-eraser joint. The pencil is 6.00 mm in diameter and is held at an angle of 20.0° to the horizontal. a. By how much does the wood flex perpendicular to its length? b. How much is it compressed lengthwise?

Answer 1

Given Data

• The diameter of the pencil is: d=6mm=6×10⁻³m
• The vertical force exerted on pencil is: P=6N
• The angle made by the pencil is: θ=20∘
• The original length of the pencil is: l=2cm=2×10⁻²m

The expression to calculate the area is given by,

A=πr²=π(d/2)²

Substitute the values in above equation.

A=π(6×10⁻³2)=2.82×10⁻⁵m2

(a)

The expression to calculate the change in length, force acted perpendicular to area using the formula is given by,

E=σ/ξ=((P_p/A)/(Δl/l)); Δl=P_p/A×l/E=(Pcosθ)/A×l/E

Here, σ is stress, ξis strain and the Eis Young's modulus.

Substitute the values in above equation.

Δl=(6cos20∘)/(2.82×10⁻⁵)×(2×10⁻²)/(200×10⁹)=1.999×10⁻⁸m

Thus, the wood flex perpendicular to its length is 1.999×10⁻⁸m

(b)

The expression to calculate the elongation in pencil using the formula of shear modulus is given by,

G=τ/γ=((P/sinθ)/A)/(Δx/l); Δx=(Psinθ)/A×l/G

Here, τ is shear stress, γ is shear strain.

Substitute the values in above equation.

Δx=(6sin20^∘)/(2.82×10⁻⁵)×(2×10⁻²)/(15×10⁹)=9.7×10⁻⁸m

Thus, the compressed length is 9.7×10⁻⁸m .