Explanation The equation of the line from ( 2 , 0 ) to (0,6) is y = − 3 x + 6 , so x = 2 − 1 3 y The length of the horizontal cut at height y is then 2 ( 2 − 1 3 y ) = 4 − 2 3 y ... To determine b) Use the symmetry principles to show that M y = 0 and find the centre of mass

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ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 9.4, Problem 11E

To determine

a)

Horizontal cut at height $y$ has length $4 - \frac{2}{3} y$ and to compute $M_{x} (w i t h p = 1)$

To determine

b)

Use the symmetry principles to show that $M_{y} = 0$ and find the centre of mass

Expert Solution & Answer

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See solution

Chapter 9.4, Problem 10E

Chapter 9.4, Problem 12E

Chapter 9 Solutions

CALCULUS (CLOTH)

Ch. 9.1 - Prob. 1PQ Ch. 9.1 - Prob. 2PQ Ch. 9.1 - Prob. 3PQ Ch. 9.1 - Prob. 1E Ch. 9.1 - Prob. 2E Ch. 9.1 - Prob. 3E Ch. 9.1 - Prob. 4E Ch. 9.1 - Prob. 5E Ch. 9.1 - Prob. 6E Ch. 9.1 - Prob. 7E

Ch. 9.1 - Prob. 8E Ch. 9.1 - Prob. 9E Ch. 9.1 - Prob. 10E Ch. 9.1 - Prob. 11E Ch. 9.1 - Prob. 12E Ch. 9.1 - Prob. 13E Ch. 9.1 - Prob. 14E Ch. 9.1 - Prob. 15E Ch. 9.1 - Prob. 16E Ch. 9.1 - Prob. 17E Ch. 9.1 - Prob. 18E Ch. 9.1 - Prob. 19E Ch. 9.1 - Prob. 20E Ch. 9.1 - Prob. 21E Ch. 9.1 - Prob. 22E Ch. 9.1 - Prob. 23E Ch. 9.1 - Prob. 24E Ch. 9.1 - Prob. 25E Ch. 9.1 - Prob. 26E Ch. 9.1 - Prob. 27E Ch. 9.1 - Prob. 28E Ch. 9.1 - Prob. 29E Ch. 9.1 - Prob. 30E Ch. 9.1 - Prob. 31E Ch. 9.1 - Prob. 32E Ch. 9.2 - Prob. 1PQ Ch. 9.2 - Prob. 2PQ Ch. 9.2 - Prob. 3PQ Ch. 9.2 - Prob. 4PQ Ch. 9.2 - Prob. 5PQ Ch. 9.2 - Prob. 1E Ch. 9.2 - Prob. 2E Ch. 9.2 - Prob. 3E Ch. 9.2 - Prob. 4E Ch. 9.2 - Prob. 5E Ch. 9.2 - Prob. 6E Ch. 9.2 - Prob. 7E Ch. 9.2 - Prob. 8E Ch. 9.2 - Prob. 9E Ch. 9.2 - Prob. 10E Ch. 9.2 - Prob. 11E Ch. 9.2 - Prob. 12E Ch. 9.2 - Prob. 13E Ch. 9.2 - Prob. 14E Ch. 9.2 - Prob. 15E Ch. 9.2 - Prob. 16E Ch. 9.2 - Prob. 17E Ch. 9.2 - Prob. 18E Ch. 9.2 - Prob. 19E Ch. 9.2 - Prob. 20E Ch. 9.2 - Prob. 21E Ch. 9.2 - Prob. 22E Ch. 9.2 - Prob. 23E Ch. 9.2 - Prob. 24E Ch. 9.2 - Prob. 25E Ch. 9.2 - Prob. 26E Ch. 9.2 - Prob. 27E Ch. 9.2 - Prob. 28E Ch. 9.2 - Prob. 29E Ch. 9.2 - Prob. 30E Ch. 9.2 - Prob. 31E Ch. 9.2 - Prob. 32E Ch. 9.2 - Prob. 33E Ch. 9.2 - Prob. 34E Ch. 9.2 - Prob. 35E Ch. 9.2 - Prob. 36E Ch. 9.2 - Prob. 37E Ch. 9.2 - Prob. 38E Ch. 9.2 - Prob. 39E Ch. 9.2 - Prob. 40E Ch. 9.2 - Prob. 41E Ch. 9.2 - Prob. 42E Ch. 9.2 - Prob. 43E Ch. 9.2 - Prob. 44E Ch. 9.2 - Prob. 45E Ch. 9.2 - Prob. 46E Ch. 9.2 - Prob. 47E Ch. 9.2 - Prob. 48E Ch. 9.2 - Prob. 49E Ch. 9.2 - Prob. 50E Ch. 9.2 - Prob. 51E Ch. 9.2 - Prob. 52E Ch. 9.2 - Prob. 53E Ch. 9.2 - Prob. 54E Ch. 9.2 - Prob. 55E Ch. 9.2 - Prob. 56E Ch. 9.2 - Prob. 57E Ch. 9.2 - Prob. 58E Ch. 9.2 - Prob. 59E Ch. 9.2 - Prob. 60E Ch. 9.2 - Prob. 61E Ch. 9.2 - Prob. 62E Ch. 9.2 - Prob. 63E Ch. 9.2 - Prob. 64E Ch. 9.2 - Prob. 65E Ch. 9.3 - Prob. 1PQ Ch. 9.3 - Prob. 2PQ Ch. 9.3 - Prob. 3PQ Ch. 9.3 - Prob. 4PQ Ch. 9.3 - Prob. 5PQ Ch. 9.3 - Prob. 1E Ch. 9.3 - Prob. 2E Ch. 9.3 - Prob. 3E Ch. 9.3 - Prob. 4E Ch. 9.3 - Prob. 5E Ch. 9.3 - Prob. 6E Ch. 9.3 - Prob. 7E Ch. 9.3 - Prob. 8E Ch. 9.3 - Prob. 9E Ch. 9.3 - Prob. 10E Ch. 9.3 - Prob. 11E Ch. 9.3 - Prob. 12E Ch. 9.3 - Prob. 13E Ch. 9.3 - Prob. 14E Ch. 9.3 - Prob. 15E Ch. 9.3 - Prob. 16E Ch. 9.3 - Prob. 17E Ch. 9.3 - Prob. 18E Ch. 9.3 - Prob. 19E Ch. 9.3 - Prob. 20E Ch. 9.3 - Prob. 21E Ch. 9.3 - Prob. 22E Ch. 9.3 - Prob. 23E Ch. 9.3 - Prob. 24E Ch. 9.3 - Prob. 25E Ch. 9.3 - Prob. 26E Ch. 9.3 - Prob. 27E Ch. 9.3 - Prob. 28E Ch. 9.3 - Prob. 29E Ch. 9.3 - Prob. 30E Ch. 9.4 - Prob. 1PQ Ch. 9.4 - Prob. 2PQ Ch. 9.4 - Prob. 3PQ Ch. 9.4 - Prob. 4PQ Ch. 9.4 - Prob. 5PQ Ch. 9.4 - Prob. 6PQ Ch. 9.4 - Prob. 1E Ch. 9.4 - Prob. 2E Ch. 9.4 - Prob. 3E Ch. 9.4 - Prob. 4E Ch. 9.4 - Prob. 5E Ch. 9.4 - Prob. 6E Ch. 9.4 - Prob. 7E Ch. 9.4 - Prob. 8E Ch. 9.4 - Prob. 9E Ch. 9.4 - Prob. 10E Ch. 9.4 - Prob. 11E Ch. 9.4 - Prob. 12E Ch. 9.4 - Prob. 13E Ch. 9.4 - Prob. 14E Ch. 9.4 - Prob. 15E Ch. 9.4 - Prob. 16E Ch. 9.4 - Prob. 17E Ch. 9.4 - Prob. 18E Ch. 9.4 - Prob. 19E Ch. 9.4 - Prob. 20E Ch. 9.4 - Prob. 21E Ch. 9.4 - Prob. 22E Ch. 9.4 - Prob. 23E Ch. 9.4 - Prob. 24E Ch. 9.4 - Prob. 25E Ch. 9.4 - Prob. 26E Ch. 9.4 - Prob. 27E Ch. 9.4 - Prob. 28E Ch. 9.4 - Prob. 29E Ch. 9.4 - Prob. 30E Ch. 9.4 - Prob. 31E Ch. 9.4 - Prob. 32E Ch. 9.4 - Prob. 33E Ch. 9.4 - Prob. 34E Ch. 9.4 - Prob. 35E Ch. 9.4 - Prob. 36E Ch. 9.4 - Prob. 37E Ch. 9.4 - Prob. 38E Ch. 9.4 - Prob. 39E Ch. 9.4 - Prob. 40E Ch. 9.4 - Prob. 41E Ch. 9.4 - Prob. 42E Ch. 9.4 - Prob. 43E Ch. 9.4 - Prob. 44E Ch. 9.4 - Prob. 45E Ch. 9.4 - Prob. 46E Ch. 9.4 - Prob. 47E Ch. 9.4 - Prob. 48E Ch. 9.4 - Prob. 49E Ch. 9.4 - Prob. 50E Ch. 9.4 - Prob. 51E Ch. 9 - Prob. 1CRE Ch. 9 - Prob. 2CRE Ch. 9 - Prob. 3CRE Ch. 9 - Prob. 4CRE Ch. 9 - Prob. 5CRE Ch. 9 - Prob. 6CRE Ch. 9 - Prob. 7CRE Ch. 9 - Prob. 8CRE Ch. 9 - Prob. 9CRE Ch. 9 - Prob. 10CRE Ch. 9 - Prob. 11CRE Ch. 9 - Prob. 12CRE Ch. 9 - Prob. 13CRE Ch. 9 - Prob. 14CRE Ch. 9 - Prob. 15CRE Ch. 9 - Prob. 16CRE Ch. 9 - Prob. 17CRE Ch. 9 - Prob. 18CRE Ch. 9 - Prob. 19CRE Ch. 9 - Prob. 20CRE Ch. 9 - Prob. 21CRE Ch. 9 - Prob. 22CRE Ch. 9 - Prob. 23CRE Ch. 9 - Prob. 24CRE Ch. 9 - Prob. 25CRE Ch. 9 - Prob. 26CRE Ch. 9 - Prob. 27CRE Ch. 9 - Prob. 28CRE

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Area Between The Curve Problem No 1 - Applications Of Definite Integration - Diploma Maths II; Author: Ekeeda;https://www.youtube.com/watch?v=q3ZU0GnGaxA;License: Standard YouTube License, CC-BY

a) Horizontal cut at height y has length 4 − 2 3 y and to compute M x ( w i t h p = 1 )

Solution Summary: The author explains how the triangle lamina is symmetric with respect to the y a x i s, and uses symmetry principles to determine the centre of mass.

Concept explainers

Videos

a)

Horizontal cut at height $y$ has length $4 - \frac{2}{3} y$ and to compute $M_{x} (w i t h p = 1)$

b)

Use the symmetry principles to show that $M_{y} = 0$ and find the centre of mass

Want to see the full answer?

Chapter 9 Solutions

a) Horizontal cut at height y has length 4 − 2 3 y and to compute M x ( w i t h p = 1 )

Solution Summary: The author explains how the triangle lamina is symmetric with respect to the y a x i s, and uses symmetry principles to determine the centre of mass.

Concept explainers

Videos

a) Horizontal cut at height y has length 4−23y and to compute Mx(with p=1)

b) Use the symmetry principles to show that My=0 and find the centre of mass

Want to see the full answer?

Chapter 9 Solutions

a)

Horizontal cut at height $y$ has length $4 - \frac{2}{3} y$ and to compute $M_{x} (w i t h p = 1)$

b)

Use the symmetry principles to show that $M_{y} = 0$ and find the centre of mass