Answer:
F= 8.07N : Magnitude of the third force acting on the object
β =33.88° , above the negative axis of the x, magnitude of the third force acting on the object
Explanation:
Conceptual analysis
Newton's first law says that an object is in equilibrium if it remains at rest or if it moves at constant speed, then, the sum of forces must be equal to zero:
Nomenclature
Fx: Component of the third force in x direction.
Fy: Component of the third force in x direction.
Problem development
∑Fx=0
Fx+6,7N=0
Fx= -6.7N
∑Fy=0
Fy-4.5N=0
Fy= +4.5N
Magnitude of the third force (F)
[tex]F=\sqrt{(F_{x} )^{2}+(F_{y} )^{2} } =\sqrt{(6.7)^{2} +{(4.5)^{2}}[/tex]
F= 8.07N
Direction of the third force (β)
[tex]\beta = tan^{-1} (\frac{F_{y} }{F_{x} } )= tan^{-1} (\frac{4.5}{6.7 })[/tex]
β =33.88° , above the negative axis of the x
Look at the graph of F in the attached graph
