HW_solu_09-17-04

HW_solu_09-17-04 - TAM 210 / 211, Fall, 2004 Homework...

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TAM 210 / 211, Fall, 2004 Homework solutions for 09/17 3.120 Given: The applied forces at the positions shown Find: A single force resultant, if possible; if not, the force and couple resultant of the four forces at O Solution: First, let’s find the force and couple resultant at O () ( ) () lb lb lb lb r forces r r k j i F i k j i F F ˆ 100 ˆ 120 ˆ 100 ˆ 50 ˆ 100 12 5 ˆ 12 ˆ 5 130 2 2 + = + + = = v v 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 first force calculated above have a line of action that intersects with the origin. Hence the moment due to that force about the origin is 0. The other two forces have lines of action that intersect with the y-axis: ( ) ( ) lb ft lb ft lb ft O r + = × + × = k i i j k j M ˆ 200 ˆ 200 ˆ 50 ˆ 4 ˆ 100 ˆ 2 , v First, let’s see if a single-force resultant exists: ( ) ( ) 0 200 200 ˆ 200 ˆ 200 ˆ 100 ˆ 120 ˆ 100 , = = + + = k i k j i M F O r r v v It exists; now let’s find it. We know the magnitude and direction is the force resultant above. To find Q, the point of application: r OQ O r F r M v v v × = , () ( ) lb z y x lb z y x lb ft OQ OQ OQ OQ OQ OQ 100 120 100 ˆ ˆ ˆ ˆ 100 ˆ 120 ˆ 100 ˆ ˆ ˆ ˆ 200 ˆ 200 = + + × + + = + k j i k j i k j i k i () () ( ) OQ OQ OQ OQ OQ OQ y x z x z y ft 100 120 ˆ 100 100 ˆ 120 100 ˆ ˆ 200 ˆ 200 + + + = + k j i k i let’s equate one component at a time: x-direction: OQ OQ OQ OQ z ft y z y ft 2 . 1
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HW_solu_09-17-04 - TAM 210 / 211, Fall, 2004 Homework...

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