4-120
Solution:-(1)4-120The free body diagram of the force systemResultant folce along the x-axis{:[SigmaF_(x)=(F_(R))_(x)],[(F_(R))_(x)=(900N)((3)/(5))-(400N)((4)/(5))],[(F_(R))_(x)=220N]:}along the y-axis{:[sumF_(y)=(F_(R))_(y)],[(F_(R))_(y)=(600 N)+(400 N)((3)/(5))-(400 N)-(900 N)((4)/(5))]:}{:[(F_(R))_(y)=-280N],[(F_(R))_(y)=280N(darr)]:}(2)magnitude of the resultant forces{:[F_(R)=sqrt((F_(R))^(2)x+(F_(R))^(2)y)],[=sqrt((220)^(2)+(-280)^(2))],[=sqrt126800],[F_(R)=356.08N]:}Angle of the resultart force with ho ... See the full answer