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The shear modulus of steel is 8.1 × 1010 N/m2. A steel nail of radius 7.5 × 10–4 m projects 0.040 m horizontally outward from a wall. A man hangs a wet raincoat of weight 28.5 N from the end of the nail. Assuming the wall holds its end of the nail, what is the vertical deflection of the other end of the nail?

User Tometoyou
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1 Answer

2 votes

Answer:


\Delta\ y =7.96 * 10^(-6)\ m

Step-by-step explanation:

given,

shear modulus of steel = 8.1 × 10¹⁰ N/m²

radius of steel nail = 7.5 × 10⁻⁴ m

Projection outward = 0.040 m

Weight of wet raincoat = 28.5 N

Vertical deflection of other end of the nail = ?

we know


G = ((F)/(A))/((\Delta y)/(L))


\Delta\ y = (FL)/(AG)

A = π r²

A = π x (7.5 x 10⁻⁴)²

A = 1.767 x 10⁻⁶ m²


\Delta\ y =(28.5 * 0.04)/(1.767* 10^(-6)* 8.1* 10^(10))


\Delta\ y =7.96 * 10^(-6)\ m

Thus, the vertical deflection of the other end of the nail is
\Delta\ y =7.96 * 10^(-6)\ m

User Tonejac
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