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Get Full Access to University Physics - 13 Edition - Chapter 11 - Problem 38e
Get Full Access to University Physics - 13 Edition - Chapter 11 - Problem 38e

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# In lab tests on a 9.25-cm cube of a certain material, a ISBN: 9780321675460 31

## Solution for problem 38E Chapter 11

University Physics | 13th Edition

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Problem 38E

In lab tests on a 9.25-cm cube of a certain material, a force of 1375 N directed at 8.50° to the cube (?Fig. E11.37?) causes the cube to deform through an angle of 1.24°. What is the shear modulus of the material?

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Solution 38E This question is based on the concept of shear modulus. Mathematically, shear modulus is given as Shear modulus = Shear strain.(1) 0 Now, the force 1375 N is applied at an angle of 8.50 . Therefore, we shall have to 0 calculate its horizontal component, which is 1375 N × cos 8.50 = 1360 N . 0 The angular deformation of the cube is 1.24 . For a small angle of deformation, this will be the shear strain. Now, 1.24 = 0.022 rad 2 2 3 2 The area of the cube is = (9.25 cm) = 85.6 cm = 8.56 × 10 m Therefore, from equation (1) 1360/(8.56×10) N/m2 3 Shear modulus = 0.022 rad = 7221 × 10 Pa Shear modulus = 7.22 × 10 Pa 6 6 Therefore, the approximate shear modulus of the material is 7.22 × 10 Pa.

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##### ISBN: 9780321675460

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