Question

A force of 8.0 N is along x direction, another force of 6.0 N is along +y direction. If both forces are acting on a point object located at the origin, what is their resultant force on the object?

Answer 1

Answer:

Resultant force, R = 10 N

Explanation:

It is given that,

Force acting along +x direction, $F_x=8\ N$

Force acting along +y direction, $F_y=6\ N$

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,

$R=\sqrt{F_x^2+F_y^2+F_xF_y\ cos\theta}$

Here $\theta=90^{\circ}$

$R=\sqrt{F_x^2+F_y^2}$

$R=\sqrt{8^2+6^2}$

R = 10 N

So, the resultant force on the object is 10 N. Hence, this is the required solution.