Answer:
Second option 6.3 N at 162° counterclockwise from
F1->
Explanation:
Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.
For the address x we have:
[tex]-F_3sin(b) + F_1 = 0[/tex]
For the address and we have:
[tex]-F_3cos(b) + F_2 = 0[/tex]
The forces [tex]F_1[/tex] and [tex]F_2[/tex] are known
[tex]F_1 = 5.7\ N\\\\F_2 = 1.9\ N[/tex]
We have 2 unknowns ([tex]F_3[/tex] and b) and we have 2 equations.
Now we clear [tex]F_3[/tex] from the second equation and introduce it into the first equation.
[tex]F_3 = \frac{F_2}{cos (b)}[/tex]
Then
[tex]-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°[/tex]
Then we find the value of [tex]F_3[/tex]
[tex]F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N[/tex]
Finally the answer is 6.3 N at 162° counterclockwise from
F1->
