I have pasted a picture of the question which will show the actual direction of forces
Let's resolve the force 20 N at 30.0 degrees south of west
20N sin30o will act in the south direction =10 N in the south
20 N cos300 will sct in the west direction =10"\\sqrt3" N in the West
Lets find the resultant of all the forces
We get 30 N acts in the North direction and 30+10"\\sqrt3" N in the West direction
converting this into vector form
"F=-(30+10\\sqrt3)i+30j"
Force =ma
"a=-(2+\\frac{2\\sqrt3}{3})i+2j"
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