Answer:
34.8 N at 2.5 degrees from the positive x-axis
Step-by-step explanation:
From the given information:
The force F makes an angle A with the positive x axis can be expressed in terms horizontal and vertical components.
[tex]F_x = F cos A\\\\ F_y = Fsin A[/tex]
Given that
[tex]F_1 = 50 \ N[/tex]
[tex]\theta_1 = 30 ^0 \ \ \ (x-axis)[/tex]
[tex]F_{1x} = F_1 \times Cos A_1[/tex]
= 50 × cos 30
= 43.3 N
[tex]F_{1y} = F_1 \times Sin \ A_1[/tex]
= 50 × sin 30
= 25 N
Similarly;
[tex]F_2 = 25 \ N[/tex]
[tex]\theta_2 = 250 ^0 \ \ \ ( x-axis)[/tex]
[tex]F_{2x} = F_2 \times cos \ A_2[/tex]
= 25 × cos 250
= - 8.55 N
[tex]F_{2y} = F_2 \times A_2[/tex]
= 25 × sin 250
= -23.5 N
The total net force;
[tex]F_{net} = F_{(net)}_x + F_{(net)}_y[/tex]
[tex]F_{net} = (F_{1x} + F_{2x} ) i + (F_{1y} + F_{2y} )j[/tex]
[tex]F_{net} = (43.3 - 8.55) i + (25-23.5 ) j[/tex]
[tex]F_{net} =34.75 i + 1.5 j[/tex]
[tex]|F_{net} | = \sqrt{F_{net}_x^2 + F_{net}_y^2}[/tex]
[tex]|F_{net} | = \sqrt{34.75^2 + 1.5^2}[/tex]
[tex]|F_{net} | = 34.8 \ N[/tex]
Finally, the direction of the angle for the net force is:
[tex]tan \theta = \dfrac{F_{net_y}}{F_{net_x}}[/tex]
[tex]\theta = tan^{-1} \Big (\dfrac{F_{net_y}}{F_{net_x}} \Big)[/tex]
[tex]\theta = tan^{-1} \Big (\dfrac{1.5}{34.75}} \Big)[/tex]
[tex]\theta = tan^{-1}( 0.043165)[/tex]
[tex]\theta \simeq 2.5^0\ with \ positive \ x-axis[/tex]
