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HW_solu_09-15-04

HW_solu_09-15-04 - TAM 210 211 Fall 2004 Homework solutions...

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TAM 210 / 211, Fall, 2004 Homework solutions for 09/15 3.88 Given: The applied forces & couples at the positions shown Find: Force and couple resultant of the four forces at O Solution: ( ) ( ) ( ) ( ) AO AO BO BO CF CF DE DE forces r r lb lb lb lb r r r r r r r r F F v v v v v v v v v v 80 100 120 700 + + + = = ( ) ( ) ( ) ( ) lb lb lb DE DE k j i k j i r r ˆ 600 ˆ 200 ˆ 300 6 2 3 ˆ 6 ˆ 2 ˆ 3 700 700 2 2 2 + = + + + = v v ( ) ( ) ( ) ( ) lb lb lb CF CF k i k i r r ˆ 24 ˆ 48 5 1 2 ˆ ˆ 2 120 120 2 2 + = + + = v v ( ) ( ) ( ) ( ) lb lb lb BO BO j i j i r r ˆ 60 ˆ 80 3 4 ˆ 3 ˆ 4 100 100 2 2 = + = v v ( ) ( ) ( ) ( ) lb lb lb AO AO k j k j r r ˆ 48 ˆ 64 3 4 ˆ 3 ˆ 4 80 80 2 2 = + = v v ( ) ( ) ( ) ( ) [ ] lb r k j j i k i k j i F ˆ 48 ˆ 64 ˆ 60 ˆ 80 ˆ 24 ˆ 48 5 ˆ 600 ˆ 200 ˆ 300 + + + + + = v ( ) lb r k j i F ˆ 606 ˆ 324 ˆ 113 + = v = = + × = = couples forces i N i i N i i OA couples forces O O r 1 1 , , C F r M M v v v v v The last two forces calculated above have lines of action that intersect with the origin. Hence the moments due to those forces about the origin is 0. The other two forces have lines of action that intersect with the x-axis: i r i r ˆ 5 . 7 ; ˆ 5 . 6 ft ft OD OC = = v v Of the two couples, one is in the x-direction, and the other is in the y-direction. So we write: ( ) ( ) ( ) j i k i i k j i i M ˆ ˆ 200 ˆ 24 ˆ 48 5 ˆ 5 . 6 ˆ 600 ˆ 200 ˆ 300 ˆ 5 . 7 , + + + × + + × = lb ft lb ft lb ft O r v ( ) lb ft O r = k j i M ˆ 1500

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