NEET Class neet Answered
Solve it sir
Asked by brijk456 | 01 Aug, 2019, 09:53: AM
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 Thiyagarajan K | 01 Aug, 2019, 05:25: PM
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