Hi there!
We can use a summation of torques to solve.
Recall the following:
[tex]\Sigma \tau = r \times F\\\\[/tex]
For a system to be static:
[tex]\Sigma \tau = 0 \\\\\Sigma F = 0[/tex]
In this case, the counterclockwise torques must sum up to the clockwise torques. We can use the pivot as our fulcrum.
Clockwise torques:
1. Weight of man (F = Mg = 70 · 9.8 = 686 N)
2. Weight of beam (F = Mg = 20 · 9.8 = 196 N)
Counterclockwise torque:
3. VERTICAL comp. of tension (F = Ty = Tsin(φ))
Do a summation of torques. The 'r' value is the distance from the force's line of action to the pivot.
Ex: For the beam, its center of mass (assuming uniform density) is at 1.5 m, or its center.
[tex]\tau_{cc} = \tau_{ccw}\\\\rF_1 + rF_2 = rF_3\\\\(1)(686) + 1.5(196) = 3(Tsin(30))}}\\\\686 + 294 = 3T(0.5)\\\\980 = 1.5T\\\\T = \boxed{653.33 N}[/tex]
Now, we can use a summation of forces to determine the vertical and horizontal components of the pivot's force.
[tex]\Sigma F_y = 0[/tex]
Sum the forces in the vertical direction. Let 'V' represent the pivot's vertical force.
[tex]0 = T_y + V - W_m - W_b\\\\0 = Tsin(30) + V - 686 - 196\\\\882 = 326.67 + V\\\\V = 555.34 N[/tex]
Now, sum the horizontal forces. 'H' is the pivot's horizontal force.
[tex]\Sigma F_x = 0 \\\\0 = T_x - H \\\\H = Tcos(30)\\\\H = 653.33cos(30) = 565.80 N[/tex]
These are the components, so use Pythagorean Theorem to find the total pivot force.
[tex]F_p = \sqrt{V^2 + H^2} = \boxed{792.80 N}[/tex]
The force will point towards the UPPER LEFT (diagonal). We can solve for the exact angle:
[tex]tan\theta = V/H \\\\tan{-1}(555.34/565.80) = \boxed{44.47^o}[/tex]
