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A force of 3N acts on 90degree to a force of 4N.find the magnitude and direction of the resultant R. ​

User Dumoko
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1 Answer

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R=A2+B2+2ABcosβ−−−−−−−−−−−−−−−−√R=A2+B2+2ABcosβ

A=4NA=4N , B=3NB=3N , β=90°β=90° , cosβ=0cosβ=0

R=A2+B2−−−−−−−√R=A2+B2

R=42+32−−−−−−√=25−−√=5NR=42+32=25=5N

tanα=Bsin90°A+Bcos90°=34tanα=Bsin90°A+Bcos90°=34

α=37°α=37°

Therefore the resultant of the two forces has a magnitude of 5N5N and is at an angle of 37°37° with respect to

User Gtaborga
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