Question

Q. Calculate the resultant force, when four forces of 50 N due east, 40 N due north, 20 N due west and 60 N due south, are acted upon a body:
l) 20$\sqrt{2}$ N, 60$°$, south-west
2) 20$\sqrt{2}$ N 45$°$, south-west
3) 20$\sqrt{2}$ N, 45$°$ north-east
4) 20$\sqrt{2}$ N, 45$°$ south-east

Answer 1

Dear student
$$\begin{gathered} Netforcealongeast=F_X=(50-20)=30N \\ Netforcealongsouth{F}_Y=(60-40)=20N \\ Resul\tan tforce \\ F=\sqrt{{F^2}_X+{F^2}_Y}=\sqrt{{30}^2+{20}^2}=\sqrt{900+400}=10\sqrt{13}N \\ Anglemadebytheresul\tan twiththeeastbe\theta so \\ \tan \theta =\frac{F_Y}{F_X}=\frac{20}{30}=\frac{2}{3} \\ \theta ={\tan }^{-1}\left(\frac{2}{3}\right)=33.69° \\ Regards \end{gathered}$$