Question

To move the oil drum, the resultant of the three forces shown must have a magnitude of 500N. Determine the magnitude and direction of the smallest force F that would cause the drum to move

Answer 1

break the forces into two components, horizontal and vertical 
F1(vertical)=250sin90=250N 
F1(horizontal)=250cos90=0N 

F2(vertical)=300sin-25= -126.79N 
F2(horizontal)=300cos-25=271.89N 

resultant force of F1 and F2 
Fr(vertical)=250+-126.79=123.21N 
Fr(horizontal)=271.89N 

colinear forces to have the smallest force F, 
so, the direction of force F = tan^-1(123.21/271.89)=24.38 degree 

set the vertical component of force F as y, then horizontal =(271.89/123.21)y=2.2067y 

Pythagoras' theorem F=square root( Fv^2+Fh^2) 
500=square root((123.21+y)^2+(271.89+2,2067y)^2) 

solve to get y= 83.19N (F vertical) 
2.2067y=183.58N (F horizontal) 

F= square root( Fv^2+Fh^2)=201.55N 

check (123.21+83.19)^2+(271.89+183.58)^2=25005... (slightly more than 500^2, due to the round off)