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Get Full Access to University Physics - 13 Edition - Chapter 11 - Problem 37e
Get Full Access to University Physics - 13 Edition - Chapter 11 - Problem 37e

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# A copper cube measures 6.00 cm on each side. The bottom

ISBN: 9780321675460 31

## Solution for problem 37E Chapter 11

University Physics | 13th Edition

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University Physics | 13th Edition

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Problem 37E

A copper cube measures 6.00 cm on each side. The bottom face is held in place by very strong glue to a flat horizontal surface, while a horizontal force ?F ?is applied to the upper face parallel to one of the edges. (Consult Table 1.1.) (a) Show that the glue exerts a force ?F ?on the bottom face that is equal but opposite to the force on the top face. (b) How large must ?F ?be to cause the cube to deform by 0.250 mm? (c) If the same experiment were performed on a lead cube of the same size as the copper one, by what distance would it deform for the same force as in part (b)?

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Solution 37E This question is based on the concept of shear modulus. Mathematically, shear modulus is expressed as Shear modulus = Shear stress Shear strain Shear strain on the other hand is force per unit area and shear strain is the ratio of the deformation to transverse length. Given that, length of each side of the cube h = 6.00 cm h = 0.06 m (a) The bottom of the cube is fixed to the horizontal surface with a very strong glue. Thus the bottom will not move upon the application of the horizontal force F. If the horizontal force F overcomes the attachment force force between the cube and the floor, the cube as a whole will move. But if the cube has to remain fixed at its position, it will also exert an equal and opposite direction force to that of the horizontal force F. This can be understood from Newton’s third law of motion. 3 (b) Given deformation, x = 0.250 mm = 0.250 × 10 m Now, the transverse length of the cube = 0.06 m The area of the surface of a cube = (0.06) m2 2 From the table, the shear modulus of copper S = 4.4 × 10 10 Pa Shear stress Therefore, Shear modulus = Shear strain 2 4.4 × 10 10= F/(036) 0.250×10 /0.06 5 F = 6.6 × 10 N 5 The required magnitude of F should be 6.6 × 10 N . (c) Shear modulus is lead is S lead= 0.6 × 10 10Pa 5 2 So, 0.6 × 10 10 Pa = (6.6×10 )/(0.06) xlead0.06 x lead= 0.0018 m x = 1.8 mm lead Therefore, the deformation in the case of lead would be 1.8 mm.

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##### ISBN: 9780321675460

This textbook survival guide was created for the textbook: University Physics, edition: 13. The full step-by-step solution to problem: 37E from chapter: 11 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. This full solution covers the following key subjects: face, Force, Cube, copper, glue. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. University Physics was written by and is associated to the ISBN: 9780321675460. The answer to “A copper cube measures 6.00 cm on each side. The bottom face is held in place by very strong glue to a flat horizontal surface, while a horizontal force ?F ?is applied to the upper face parallel to one of the edges. (Consult Table 1.1.) (a) Show that the glue exerts a force ?F ?on the bottom face that is equal but opposite to the force on the top face. (b) How large must ?F ?be to cause the cube to deform by 0.250 mm? (c) If the same experiment were performed on a lead cube of the same size as the copper one, by what distance would it deform for the same force as in part (b)?” is broken down into a number of easy to follow steps, and 118 words. Since the solution to 37E from 11 chapter was answered, more than 496 students have viewed the full step-by-step answer.

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