91Ó°ÊÓ

Given that the uniform horizontal beam supporting the sign is 1.50 m long, has a mass of 16.0 kg, and is hinged to the wall. The sign itself is uniform with a mass of 28.0 kg and overall length of 1.20 m. The two wires supporting the sign are each 32.0 cm long, are 90.0 cm apart, and are equally spaced from the middle of the sign. The cable supporting the beam is 2.00 m long.

Mass of beam ${\mathit{m}}_{\mathbf{1}}\mathbf{=}\mathbf{16}\mathbf{\text{kg}}$

Mass of sign,${\mathit{m}}_{\mathbf{2}}\mathbf{=}\mathbf{28}\mathbf{\text{kg}}$

Since the wires are symmetrically placed on either side, therefore, their tensions are equal. So Tension is${\mathit{T}}_{\mathbf{w}}\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{1}}\mathbf{g}}{\mathbf{2}}\mathbf{=}\mathbf{137}\mathbf{ }\mathbf{\text{N}}$

Torque$\mathit{Ï„}\mathbf{=}\mathit{F}\mathit{l}$

Where Fis force exerted and l is distance.

Since the cable is 2 m long and the beam is 1.5 m long,

$\begin{array}{rcl}\cos \mathrm{θ}& =& \frac{1.5 \text{m}}{2 \text{m}}\\ \mathrm{θ}& =& 41.4°\end{array}$

Let tension in the cable is $T_c$.

Applying the condition for equilibrium,

$∑\mathrm{τ}=0$gives

$$\begin{gathered} T_c\left(\sin 41.4°\right)\left(1.50 \text{m}\right)-w_{beam}\left(0.75 \text{m}\right)-T_W\left(1.50 \text{m}\right)-T_w\left(0.6 \text{m}\right)=0 \\ {T}_c=\frac{\left(1.6 \text{m}\right)\left(9.8 {\text{m/s}}^2\right)\left(0.75 \text{m}\right)+\left(137 \text{N}\right)\left(1.5 \text{m}+0.6 \text{m}\right)}{\left(1.5 \text{m}\right)\left(\sin 41.4°\right)} \\ =408.6 \text{N} \end{gathered}$$

The minimum tension is 408.6 N.

Let horizontal component of the force that the cliff face exerts on the climber’s feet is F_H and vertical component is F_V .

Applying equilibrium condition

Net horizontal and vertical force is zero.

$∑F_y=0$

$∑F_y=0$implies $F_v+T_c\sin 41.4°-w_{beam}-2T_w=0$

$\begin{array}{rcl}{F}_v& =& 2T_w+w_{beam}-T_c\sin 41.4°\\ & =& 2\left(137 \text{N}\right)+\left(16 \text{kg}\right)\left(9.8 {\text{m/s}}^2\right)-\left(408.6 \text{N}\right)\left(\sin 41.4°\right)\\ & =& 161 \text{N}\end{array}$

Hence, vertical component is 161 N.