S08P2AChap10Sol

# S08P2AChap10Sol - Chapter 10 Solution Problem 1 A 28-kg...

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Chapter 10 Solution Problem 1. A 28-kg child sits at one end of a 3.5-m-long seesaw. Where should her 65-kg father sit so the center of mass will be at the center of the seesaw? Solution Take the x -axis along the seesaw in the direction of the father, with origin at the center. The center of mass of the child and her father is at the origin, so x cm = 0 = m c x c + m f x f , where the masses are given, and x c = ! (3.5 m) = 2 (half the length of the seesaw in the negative x direction). Thus, x f = ! m c x c = m f = ( 28 = 65 )( 1 . 75 m ) = 75 . 4 cm from the center. Problem 3. Four trucks, with masses indicated in Fig. 10-23, are on a rectangular barge of mass 35 Mg whose center of mass is at its center. The trucks’ individual centers of mass are located 25 m apart on the barge’s long dimension and 10 m apart on the short dimension, as shown. Where is the center of mass of the entire system? Express in relation to the truck at lower left. FIGURE 10-23 Top view of four trucks on a barge, with truck masses given. Dots mark individual trucks’ centers of mass (Problem 3). Solution As explained in the text (see Figs. 10-8 and 9), in order to find the center of mass of this system, the trucks and the barge can be treated as point masses located at their centers of mass. With x-y axes as shown superimposed on Fig. 10-23, x cm = ( 18 Mg )( 0 ) + ( 23 Mg )( 25 m ) + ( 19 Mg 25 m ) + ( 11 Mg 0 ) + ( 35 Mg )( 12 . 5 m ) ( 18 + 23 + 19 + 11 + 35 ) Mg = 14 . 0 m , y cm = ( 19 Mg + 11 Mg )( 10 m ) + ( 35 Mg 5 m ) 106 Mg = 4 . 48 m

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230 CHAPTER 10 Problem 5. Three equal masses lie at the corners of an equilateral triangle of side l . Where is the center of mass? Solution Take x-y coordinates with origin at the center of one side as shown. From the symmetry (for every mass at x , there is an equal mass at ! x ) x cm = 0. Since y = 0 for the two masses on the x -axis, and y = l sin60 ° = l = 2 for the other mass, Equation 10-2 gives y cm = m ( l = 2 ) = 3 m = l = = 0 . 289 l . Problem 5 Solution. Problem 9. Find the center of mass of a pentagon of side a with one triangle missing, as shown in Fig. 10-24. Hint: See Example 10-3, and treat the pentagon as a group of triangles. Solution Choose coordinates as shown. From symmetry, x cm = If the fifth isosceles triangle (with the same assumed uniform density) were present, the center of mass of the whole pentagon would be at the origin, so 0 = ( my 5 + 4 cm ) = 5 m , where y cm gives the position of the center of mass of the figure we want to find, and y 5 is the position of the center of mass of the fifth triangle. Of course, the mass of the figure is four times the mass of the triangle. In Example 10-3, the center of mass of an isosceles triangle is calculated, so y 5 = ! 2 3 l , and from the geometry of a pentagon, tan36 ° = 1 2 a = l . Therefore, y cm = ! 1 4 y 5 = 1 6 l = 1 12 a cot 36 ° = 0 . 115 a . FIGURE 10-24 Problem 9 Solution.
CHAPTER 10 231 Problem 10. A solid cube of side a has a density that varies linearly from zero at the bottom to ! 0

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## This note was uploaded on 07/25/2009 for the course PHYS 2A taught by Professor Hicks during the Winter '07 term at UCSD.

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S08P2AChap10Sol - Chapter 10 Solution Problem 1 A 28-kg...

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