Question

forces of 3.0 N (west) and 4.0 N (33° S of E) act on an object. What is the net force of the object? Calculate both magnitude and direction

Answer 1

The net force is 2.2 N at 80.9 degrees south of east

Explanation:

In order to find the net force, we have to resolve each force along the x-y direction, and then add the components in each direction.

Taking east as positive x-direction and north as positive y-direction, we have:

- First force:

$F_{1x} = -3.0 N\\F_{1y} = 0$

- Second force:

$F_{2x} = (4.0)(cos (-33^{\circ})=3.35 N\\F_{2y} = (4.0)(sin (-33^{\circ})=-2.18 N$

Therefore, the components of the net force are

$F_x = F_{1x}+F_{2x}=-3.0+3.35=+0.35 N\\F_y = F_{1y}+F_{2y}=0+(-2.18)=-2.18 N$

And so, the magnitude of the net force is

$F=\sqrt{F_x^2+F_y^2}=\sqrt{(0.35)^2+(-2.18)^2}=2.2 N$

And the direction is

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{-2.18}{0.35})=-80.9^{\circ}$

which means 80.9 degrees south of east.

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