Answer:
The right approach is "128.578 N-m".
Explanation:
Given:
Force,
F = 500 N
then,
[tex]F_y=500 \ Cos30[/tex]
[tex]F_x=500 \ Sin 30[/tex]
Now,
The moment of force will be:
⇒ [tex]\Sigma M_0=500 \ Cos30\times (275)+500 \ Sin30\times (38)[/tex]
[tex]=500\times \frac{\sqrt{3} }{2}\times 275+500\times \frac{1}{2}\times 38[/tex]
[tex]=128578.493[/tex]
or,
[tex]=128.578 \ N-m[/tex]
