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(10%) Problem 9: A uniform density sheet of metal is cut into the shape of an isosceles triangle, which is oriented with the base at

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(10%) Problem 9: A uniform density sheet of metal is cut into the shape of an isosceles triangle, which is oriented with the base at the bottom and a corner at the top. It has a base B = 13 cm, height H= 17 cm, and area mass density o. dy I H B Otheexpertta.com @theexpertta.com - tracking id: 5T88-75-0E-46-BB99-46140. In accordance with Expert TA's Terms of Service. copying this information to any solutions sharing website is strictly forbidden. Doing so may result in termination of your Expert TA Account. A 14% Part (a) The horizontal center of mass of the sheet will be located: O To the right of the center line. O Not enough information to determine. Grade Summary Deductions 0% O To the left of the center line. O On the center line. Potential 100% Submissions Attempts remaining: 3 (33% per attempt) detailed view Submit Hint Feedback I give up! Hints: 1% deduction per hint. Hints remaining: 1 Feedback: 2% deduction per feedback. A 14% Part (b) The vertical center of mass of the sheet will be located: A 14% Part (c) Write a symbolic equation for the total mass of the triangle. A 14% Part (d) Consider a horizontal slice of the triangle that is a distance y from the top of the triangle and has a thickness dy. Write an equation for the area of this slice in terms of the distance y, and the base B and height H of the triangle. A 14% Part (e) Set up an integral to calculate the vertical center of mass of the triangle, assuming it will have the form CJ f(y) where C has all the constants in it and f() is a function of y. What is f())? A 14% Part (f) Integrate to find an equation for the location of the center of mass in the vertical direction. Use the coordinate system specified in the previous parts, with the origin at the top and positive downward. A 14% Part (g) Find the numeric value for the distance between the top of the triangle and the center of mass in cm.The vertical centre of man will be located below the 6 2 M Ixbxh M = ( x b x h ) 6 Area f stop . I dy DOAB = OCD A K b dy 2 2 I Areal 2 by dy CON M h2 dady O COM 2 COM - 2X 14 - 9 - 33 cm

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