Answer:
The values of the forces are
[tex]F_1 = 10.6 \ N[/tex] , [tex]F_2 = 21.33 \ N[/tex]
Explanation:
The diagram for this question is shown on the first uploaded image
From the question we are told that
The magnitude of F is [tex]F = 32 \ N[/tex]
Generally at equilibrium the torque is mathematically evaluated as
[tex]\sum \tau = 0[/tex]
From the diagram we have
[tex]r * F_1 - [\frac{r}{2} ] F_2 + 0 F = 0[/tex]
=> [tex]F_1 = 0.5 F_2[/tex]
Generally at equilibrium the Force is mathematically evaluated as
[tex]\sum F = 0[/tex]
From the diagram
[tex]F - F_ 1 - F_2 = 0[/tex]
substituting values
[tex]32 - (0.5F_2 ) - F_2 = 0[/tex]
[tex]F_2 = 21.33 \ N[/tex]
So
[tex]F_1 = 0.5 * 21.33[/tex]
[tex]F_1 = 10.6 \ N[/tex]