Answer:
1. The magnitude of the net force is 1.33 N.
2. The direction is 14.8° with respect to the vertical.
Explanation:
1. The magnitude of the net force is given by:
[tex] |F| = \sqrt{F_{x}^{2} + F_{y}^{2}} [/tex]
Where:
[tex]F_{x}[/tex]: is the sum of the forces acting in the x-direction
[tex]F_{y}[/tex]: is the sum of the forces acting in the y-direction
Let's find the forces acting in the x-direction and in the y-direction.
In the x-direction:
[tex] \Sigma F_{x} = -F_{a}cos(\theta) [/tex]
Where:
Fa: is the force of air resistance = -0.40 N. The negative sign is because this force is in the negative x-direction.
θ: is the angle = 32°
[tex] \Sigma F_{x} = -0.40 N*cos(32) = -0.34 N [/tex]
In the y-direction:
[tex] \Sigma F_{y} = F_{g} + F_{a}sin(\theta) [/tex]
Where:
[tex]F_{g}[/tex]: is the gravitational force = -1.5 N
[tex] \Sigma F_{y} = -1.5 N + 0.40 N*sin(32) = -1.29 N [/tex]
Hence, the magnitude of the net force is:
[tex] |F| = \sqrt{(-0.34 N)^{2} + (-1.29)^{2}} = 1.33 N [/tex]
2. The direction of the net force is:
[tex] tan(\alpha) = \frac{F_{x}}{F_{y}} = \frac{-0.34}{-1.29} [/tex]
[tex] \alpha = tan^{-1}(\frac{-0.34}{-1.29}) = 14.8 [/tex]
The angle is 14.8° with respect to the vertical.
I hope it helps you!
