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![[Post New]](/templates/default/images/icon_minipost_new.gif) 8 Jun 2008 13:45:00 IST
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find resultant of forces of mag.4,3&6N acting in direction AB,BC,CA of equi triangle
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8 Jun 2008 14:04:09 IST
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DO REPLY!!!!!!!
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8 Jun 2008 17:06:09 IST
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no 1 can solve it.....it shudn't be so difficult 4r 12th students
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9 Jun 2008 07:34:28 IST
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do help me out
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12 Jun 2008 11:24:06 IST
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I think the answer is rt7N.Here's my method.The forces can be assumed to be acting at a point at 1200 to each other.First let us find the resultant of 6N,3N.
F'2=36+9-2(6)(3)(1/2)=27 F'=3rt3 making an angle k with 6N
where k=Tan-1(3sin120/6+3cos120)=Tan-1(3rt3/9)=300.
So,now,we have to find resultant of 4,3rt3 which are at an angle (120+30)=1500 with each other.
So,Fnet2=16+27-2(4)(3rt3)(rt3/2)=7 Fnet= .
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MAKING A MISTAKE IS HUMAN BUT REPEATING IT IS IDIOTIC. |
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