Answer:
Resultant force, R = 10 N
Explanation:
It is given that,
Force acting along +x direction, [tex]F_x=8\ N[/tex]
Force acting along +y direction, [tex]F_y=6\ N[/tex]
Both the forces are acting on a point object located at the origin. Let the resultant force of the object is given by R. So,
[tex]R=\sqrt{F_x^2+F_y^2+F_xF_y\ cos\theta}[/tex]
Here [tex]\theta=90^{\circ}[/tex]
[tex]R=\sqrt{F_x^2+F_y^2}[/tex]
[tex]R=\sqrt{8^2+6^2}[/tex]
R = 10 N
So, the resultant force on the object is 10 N. Hence, this is the required solution.
