Solving for Fnet using Vector Components --> Question: A 1.00kg object is experiencing a 30.0N force pulling south, a 20.0N force pulling west and a 50.0N force pulling at 60° S of E Answer: (The image below is presented to show my work however it is not clear so I'll type out the answer) First an free body diagram must be made to have a visual of the vectors and their directions. Since there is an angle involved, sin and cos must be used to find the lengths of fx and fy.For our Fnetx we already know one vector which is 20.0N[W], the other one is found by using cos, Fx = 50cos60 = 25.0N[E], Fnetx = 25 -20 = 5.00N[E]. For the Fnety, again we already know one of our vectors which is 30.0N[S], to find the other one Fy = 40sin60 = 43.3N[S] so the Fnety = 30 + 43.3 = 58.3N[S]. Then another triangle is made, which I tried showing, using the new information and to find the overall Fnet, the Pythagorean theorem is used, the overall Fnet^2 = 5N[E]^2 + 58.3N[S]^2, Fnet = √5^2 + 58.3^2 = 58.5N, to find the direction, an angle must be chosen (does not matter which angle) θ = tan-1(5/58.3) = 5°. The answer is 58.5N @ 5° E of S.
