Answer to Question #233596 in Physics for jopay

Question

 The following forces act on an object resting on a level and frictionless surface: 10 N to the north; 20 N to the east; 10 N at an angle 40^o south of east; and 20 N at an angle 50^o west of south. Find the magnitude and the direction of the resultant force acting on an object.

Answer 1

Let's first find $x$- and $y$-components of resultant force:

$F_x=10\ N\cdot cos90^{\circ}+20\ N\cdot cos0^{\circ}+10\ N\cdot cos320^{\circ}+20\ N\cdot cos230^{\circ}=14.8\ N,$

$F_y=10\ N\cdot sin90^{\circ}+20\ N\cdot sin0^{\circ}+10\ N\cdot sin320^{\circ}+20\ N\cdot sin230^{\circ}=-11.75\ N.$

We can find the magnitude of the resultant force acting on an object from the Pythagorean theorem:

$$F=\sqrt{F_x^2+F_y^2}=\sqrt{(14.8\ N)^2+(-11.75\ N)^2}=18.9\ N.$$

We can find the direction of the resultant force acting on an object from the geometry:

$$\theta=sin^{-1}(\dfrac{F_y}{F})=sin^{-1}(\dfrac{-11.75\ N}{18.9\ N})=38^{\circ}\ S\ of\ E.$$