Opposing shearing forces of 3000 N are applied to the 3.00 cm × 4.00 cm faces of a 3.00 cm × 4.00 cm × 5.00 cm steel block. By what amount does the block become slanted? The shear modulus for steel is 8.0 × 10¹⁰ N/m². A) 1.54 × 10^-6 m B) 5.56 × 10^-7 m C) 9.88 × 10^-7 m D) 3.08 × 10^-6 m E) 6.98 × 10^-⁶ m

Question

Quizplus · Accepted Answer

The shear strain is given by:$$\gamma = \frac{F}{A}\frac{1}{G}$$where:* F is the force applied* A is the area of the face* G is the shear modulusPlugging in the given values, we get:$$\gamma = \frac{3000 	ext{ N}}{(0.03 	ext{ m})(0.04 	ext{ m})}\frac{1}{8.0 	imes 10^{10} 	ext{ N/m}^2} = 1.54 	imes 10^{-6} 	ext{ m}$$The amount the block becomes slanted is given by:$$	heta = \gamma h$$where:* h is the height of the blockPlugging in the given values, we get:$$	heta = (1.54 	imes 10^{-6} 	ext{ m})(0.05 	ext{ m}) = 7.70 	imes 10^{-8} 	ext{ rad}$$

Opposing Shearing Forces of 3000 N Are Applied to the 3.00