1iron40 cmsteel70 cmHenry has a rod that is made outof 2 parts. An iron and steelsection. The iron section is 40cmlong (Y₁=9.1 x 10¹⁰ N/m²). Theiron piece is attached to a steelpiece that is 70 cm long (Y,=7.0 x10¹0⁰ N/m²). The rod has a uniformcross sectional diameter of 3.40cm. Determine the totalcompression (in mm)when henryuses a compressive force on therod of magnitude F=6000N. Lookat the image provided to get abetter understanding.

Solution: Given: 1 m = 103 mm L1 = length of iron section = 40 cm = 0.40 m L2 = length of steel…

iron -140 cm steel 70 cm Henry has a rod that is made out of 2 parts. An iron and steel section. The iron section is 40cm long (Y₁-9.1 x 10¹⁰ N/m²). The iron piece is attached to a steel piece that is 70 cm long (Y₂=7.0 x 10¹0 N/m²). The rod has a uniform cross sectional diameter of 3.40 cm. Determine the total compression (in mm)when henry uses a compressive force on the rod of magnitude F=6000N. Look at the image provided to get a better understanding.

iron -140 cm steel 70 cm Henry has a rod that is made out of 2 parts. An iron and steel section. The iron section is 40cm long (Y₁-9.1 x 10¹⁰ N/m²). The iron piece is attached to a steel piece that is 70 cm long (Y₂=7.0 x 10¹0 N/m²). The rod has a uniform cross sectional diameter of 3.40 cm. Determine the total compression (in mm)when henry uses a compressive force on the rod of magnitude F=6000N. Look at the image provided to get a better understanding.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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