The resultant of two forces 3p and 2p is R.If the first force is doubled,then the resultant is also doubled.The angle between the two forces is
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Resultant,R =√(a^2+b^2+2abcosx)
As, r = √{(3p)^2+(2p)^2+2×3p×2p×cosx}
or, r = √(9p^2+4p^2+12p^2cosx)
or, r = √(13p^2+12p^2cosx)
According to question;
2r = √{(6p)^2+(2p)^2+2×6p×2p×cosx}
or, 2r =√(36p^2+4p^2+24p^2cosx)
or, 2r = √(40p^2+24p^2cosx)
or, 2{√(13p^2+12p^2cosx)}
= √(40p^2+24p^2cosx)
or, 4(13p^2+12p^2cosx) = (40p^2+24p^2cosx)
or, 52p^2 + 48p^2cosx= 40p^2 + 24p^2cosx
or, 12 p^2 = - 24 p^2cosx
or, cosx = - 1/2
or, x = cos^-1(-1/2)
or, x = 120°
So, angle between the forces is 120°.
As, r = √{(3p)^2+(2p)^2+2×3p×2p×cosx}
or, r = √(9p^2+4p^2+12p^2cosx)
or, r = √(13p^2+12p^2cosx)
According to question;
2r = √{(6p)^2+(2p)^2+2×6p×2p×cosx}
or, 2r =√(36p^2+4p^2+24p^2cosx)
or, 2r = √(40p^2+24p^2cosx)
or, 2{√(13p^2+12p^2cosx)}
= √(40p^2+24p^2cosx)
or, 4(13p^2+12p^2cosx) = (40p^2+24p^2cosx)
or, 52p^2 + 48p^2cosx= 40p^2 + 24p^2cosx
or, 12 p^2 = - 24 p^2cosx
or, cosx = - 1/2
or, x = cos^-1(-1/2)
or, x = 120°
So, angle between the forces is 120°.
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Here is the answer.
Resultant,R =√(a^2+b^2+2abcosx)
As, r = √{(3p)^2+(2p)^2+2×3p×2p×cosx}
or, r = √(9p^2+4p^2+12p^2cosx)
or, r = √(13p^2+12p^2cosx)
According to question;
2r = √{(6p)^2+(2p)^2+2×6p×2p×cosx}
or, 2r =√(36p^2+4p^2+24p^2cosx)
or, 2r = √(40p^2+24p^2cosx)
or, 2{√(13p^2+12p^2cosx)}
= √(40p^2+24p^2cosx)
or, 4(13p^2+12p^2cosx) = (40p^2+24p^2cosx)
or, 52p^2 + 48p^2cosx= 40p^2 + 24p^2cosx
or, 12 p^2 = - 24 p^2cosx
or, cosx = - 1/2
or, x = cos^-1(-1/2)
or, x = 120°
So, angle between the forces is 120