Answer:
[tex]F_2 = 29.54 N[/tex]
Explanation:
As we know that the combination is maintained at rest position
So we will take net torque on the system to be ZERO
so we know that
[tex]\tau = \vec r \times \vec F[/tex]
here we will have
[tex]\vec r_1 \times F_1 = \vec r_2 \times F_2[/tex]
so we have
[tex]13 \times 50 = 22 \times F_2[/tex]
so we have
[tex]F_2 = \frac{13 \times 50}{22}[/tex]
[tex]F_2 = 29.54 N[/tex]