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Solve it sir

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Asked by brijk456 1st August 2019, 9:53 AM
Answered by Expert
Answer:
As shown in figure, Let direction of foce P be along x-axis .
 
Resultant of two forces  R1= (P+Q cosθ) i + (Q sinθ) j
 
Magnitude of resultant, |R1| = [ P2 + Q2 + 2PQ cosθ ]1/2
 
when Q is doubled, resultant R2 = (P+2Q cosθ) i + (2Q sinθ) j
 
Manitde of resultant, |R2| = [ P2 + 4Q2 + 4PQ cosθ ]1/2
 
since |R2| = 2|R1| , we have , P2 + 4Q2 + 4PQ cosθ = 4(P2 + Q2 + 2PQ cosθ)  
 
Hence,  3P2 + 4PQ cosθ = 0  or  cosθ = -3P/(4Q)
Answered by Expert 1st August 2019, 5:25 PM
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