Question

Three forces act on a moving object. One force has a magnitude of 83.4 N and is directed due north. Another has a magnitude of 47.2 N and is directed due west. What must be (a) the magnitude and (b) the direction of the third force, such that the object continues to move with a constant velocity? Express your answer as a positive angle south of east.

Answer 1

v=const => ΣF=0
F₁₂=sqrt(F₁²+F₂²)
F₃=F₁₂
F₃(x)= -F₂
F₃(y)= -F₁
tan α= F₃(y)/ F₃(x)=
=F₁/F₂
α is the angle that F₃ makes with +x direction
(south of east)

Answer 2

Components of a force P due North-West acting at an angle A from North will be PcosA and PsinA in North and West directions respectively and should be equal in magnitude to the force acting on the object. Thus PcosA=83.4 and PsinA=47.2. Therefore tanA=47.2/83.4 =0.566 and A=tan inv 0.566 =29.5 deg. Now to neutralize the acting forces, it must act in opposite direction that is SE and will be at an angle of 29.5 deg from South which means 90-29.5=60.5 deg from East. I feel this is the simplest way to find it.