Answer
a) The minimum tension that must be in the cable is 409 N.
b) The minimum vertical force that the hinge must support is 161 N.
Work Step by Step
a) Find the minimum tension in the rope.
Let the length of the beam be $x_{b}$ = 1.50m and the mass of the beam be $m_{b}$ = 16.0kg.
Let the mass of the sign be $m_{s}$ = 28.0kg.
Let the sign's length be $l$ = 1.50m
The sign's center of mass is located at it's center. In this case, the sign's center is at $x_{b}$ - $(\frac{1}{2})$$(0.90m)$ = 1.05m from the hinge. Let x_{s} = 1.05m
Let the length of the cable be $x_{c}$ = 2.00m
The two wires supporting the sign are spaced 0.90m apart and spaced equally from the center of the sign.
---Sum the torques:
$\Sigma$$\tau$ = $\tau_{c}$ - $\tau_{b}$ - $\tau_{s}$ = 0
NOTE: Using the right hand rule, the cable's torque is positive, while the beam's torque and the sign's torque are negative.
Find $\theta$:
The adjacent side = $x_{b}$ = $1.50m$ and the hypotenuse = $x_{c}$ = 2.00m. So $arccos(\frac{1.50m}{2.00m})$ = $\theta$ = $41.4^{\circ}$
Torque is defined: $\tau$ = $r\times$$F\times$sin($\theta$) where $r$ is the length of the lever arm, $F$ is the force applied at that distance, and $\theta$ is the angle between the two.
Using this,
$\Sigma$$\tau$ = $x_{b}$$T$$sin(\theta)$ - $\frac{1}{2}$$x_{b}$$m_{b}$$g$$sin(90^{\circ})$ - $x_{s}$$m_{s}$$g$$sin(90^{\circ})$ = $0$
Moving over terms, and since $sin(90^{\circ})$ = $1$,
$x_{b}$$T$$sin(\theta)$ = $\frac{1}{2}$$x_{b}$$m_{b}$$g$ + $x_{s}$$m_{s}$$g$
Therefore,
$T$ = $\frac{\frac{1}{2}x_{b}m_{b}g + x_{s}m_{s}g}{x_{b}sin(\theta)}$
$T$ = $\frac{\frac{1}{2}(1.50m)(16.0kg)(9.81m/s^{2}) + (1.05m)(28.0kg)(9.81m/s^{2})}{(1.50m)(sin(41.4^{\circ}))}$ = $409 N$
b) Find the minimum vertical force the hinge must have to support the sign.
NOTE: In the diagram, $F$ = $T_{y}$
Sum the vertical forces. These include the vertical component of the tension, the weight of the beam, and the weight of the sign.
$\Sigma$$F$ = $T_{y}$ - $m_{b}$$g$ - $m_{s}$$g$
Since $T_{y}$ = $Tsin(\theta)$,
$\Sigma$$F$ = $Tsin(\theta)$ - $m_{b}$$g$ - $m_{s}$$g$
= $(409N)(sin(41.4^{\circ}))$ - $(16.0kg)$$(9.81m/s^{2})$ - $(28.0kg)$$(9.81m/s^{2})$
$\Sigma$$F$ = $-161N$
Therefore, in order for $\Sigma$$F$ to equal $0$, the hinge must support $161N$.
