Answer with Explanation:
We are given that
[tex]F_1=6.9 N[/tex]
[tex]F_2=4.5 N[/tex]
We have to find the direction and magnitude of the third force acting on the object.
Resultant force,F=[tex]\sqrt{F^2_1+F^2_2}[/tex]
[tex]F=\sqrt{(6.9)^2+(4.5)^2}=8.24 N[/tex]
The object moves with constant velocity .Therefore, net force on object is equal to zero
So,Third force,[tex]F_3=F=8.24 N[/tex]
Direction,[tex]\theta=tan^{-1}(\frac{F_2}{F_1})[/tex]
[tex]\theta=tan^{-1}(\frac{4.5}{6.9})=33.02^{\circ}[/tex]
Angle lies in second quadrant because the direction of third force is opposite to the direction of the resultant force of F1 and F2.
Therefore,[tex]\theta=\pi-\theta=180-33.02=146.98^{\circ}[/tex]
