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Cool goIITian

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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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find resultant of forces of mag.4,3&6N acting in direction AB,BC,CA of equi triangle

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Comments (4)

Swarna

Cool goIITian

Joined: 25 May 2008

Posts: 71

8 Jun 2008 14:04:09 IST

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DO REPLY!!!!!!!

Swarna

Cool goIITian

Joined: 25 May 2008

Posts: 71

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

Swarna

Cool goIITian

Joined: 25 May 2008

Posts: 71

9 Jun 2008 07:34:28 IST

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do help me out

SUNDEEP ALLAMRAJU

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Joined: 28 Feb 2008

Posts: 1014

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)=27F'=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,F_net²=16+27-2(4)(3rt3)(rt3/2)=7F_net=.

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Comments (4)

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)=27F'=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)=7Fnet=.

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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)=27F'=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,F_net²=16+27-2(4)(3rt3)(rt3/2)=7F_net=.