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Two forces whose magnitudes are in the r...

Two forces whose magnitudes are in the ratio f 3:5 give a resultant of 35 N. If the angle of inclination is `60^(@),` Thin magnitude of each force,

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We are given: `(F _(1))/( F _(2)) = (3)/(5)`
`F _(1) = 0. 6 F _(), R = 3 5 N , theta = 60^(@)` We are to calculate : `F _(1) and F _(2)`
Using relation: `R = sqrt ( F _(1) ^(2) + F _(2) ^(2) + 2 F _(1) F _(2) cos theta)`
We have, `35 = sqrt (( 0. 0 6F _(2)) ^(2) + (F _(2)) ^(2) + ( 0. 6 F _(2)) ( F _(2)) cos 60)`
`35 = sqrt ( 36 F _(2) ^(2) + F _(2) ^(2) + 2 xx ( 0. 6 F _(2) ^(2) ) xx (1)/(2))`
`35 = sqrt ( 1 . 96 F _(2)^(2)) or (35) ^(2) = 1. 96 F _(2) ^(2)`
`1. 96 F _(2) ^(2) = 1225 implies F_(2) = 25 N`
and `F _(1) = 0. 6 F _(2 ) = 0. 6 xx 25 = 15 N.`
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