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17 Sep 2011 17:27:49 IST

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Mathematics

The sum of the magnitudes of two forces actinf at a point is 18 units and the magnitude of their resultant is 12units. The resultant makes and angle of 90 degree with the force of the smaller magnitude. The magnitude of the individual forces are_____?

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sudhindra shenoy

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18 Sep 2011 22:56:28 IST

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It is given by the formula R= sq root(P^2+Q^2+2PQcos)

cos 900=-1 P+Q=18 R=12 substitute the values

144=(P+Q)^2-4PQ

144=324-4PQ PQ=45 P^2+Q^2=234

take P=18-Q

sq and simplify u get Q=15 or 3 P=15or 3

Deepak

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20 Sep 2011 19:00:53 IST

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actually the angle is 90.... and how do you consider the force of smaller magnitude,

sudhindra shenoy

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20 Sep 2011 23:08:12 IST

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oops i had taken cos 900 instead of cos90

but u r s question is invalid i hope so i am not sure since P+Q=18 & R^2 =P^2+Q^2

r=12 144=(18-Q)^2+Q^2 u will get discriminant negative i.e roots are complex Hence i think the q does not exists .If it is correct ask me again .

Deepak

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3 Oct 2011 12:57:51 IST

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But this is the question from my study material for IIT JEE from Brilliant tutorials. i may post the answer and will seek the explanation.

sudhindra shenoy

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9 Oct 2011 22:37:08 IST

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@ deepak ur q. is correct

in such cases if apply the formula R=sq rt A2+B2+2ABcostheta u will get d discriminat -ve

in all such cases u hav to take the resultant as d altitude of right angled triangle and on the hypotenuse sum of the forces.

now the problem goes like this

R^2=(18-A)^2-A^2

solving this A=5N, B=13N

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