91Ó°ÊÓ

The length of the uniform beam,$\text{L}=12.0\text{″¾}$

The inclination angle of the beam at the hinge,$\mathrm{θ}=50.0°$

The tension in the cable, $\text{T}=400\text{ N}$

You draw the free body diagram. The system is at equilibrium. For such a system, the vector sum of the forces acting on it is zero. The vector sum of the external torques acting on the beam about a pivot point is zero. The equations used are given below.

$\begin{array}{l}\mathbf{Î\pounds }{\stackrel{\mathbf{→}}{\mathbf{F}}}_{\mathbf{net}}\mathbf{=}\mathbf{0}\\ \mathbf{Î\pounds }{\stackrel{\mathbf{→}}{\mathbf{Ï„}}}_{\mathbf{net}}\mathbf{=}\mathbf{0}\end{array}$

According to the free body diagram, the uniform beam is in a static equilibrium condition. It makes an angle $\mathrm{θ}$ at the hinge point and angle$\mathrm{ϕ}$with the cable, hence according to the geometry,

$âˆ\dots =90−\mathrm{θ}$

For the static equilibrium condition, the vector sum of the external torque acting on the ladder at about any point is zero.

$\begin{array}{c}{\mathrm{Î\pounds Ï„}}_{\mathrm{net}}=0\\ \mathrm{W}\left(\frac{\mathrm{L}}{2}\sin \mathrm{θ}\right)−{\mathrm{T}}_{\mathrm{c}}(\mathrm{Lsin}âˆ\dots )=0\\ \mathrm{W}\left(\frac{\mathrm{L}}{2}\sin \mathrm{θ}\right)={\mathrm{T}}_{\mathrm{c}}(\mathrm{Lsin}âˆ\dots )\\ \mathrm{W}=\frac{{\mathrm{T}}_{\mathrm{c}}(\mathrm{Lsin}âˆ\dots )}{\left(\frac{\mathrm{L}}{2}\mathrm{²õ¾\pm ²Ôθ}\right)}\\ \mathrm{W}=\frac{{\mathrm{T}}_{\mathrm{c}}[\mathrm{Lsinsin}(90−\mathrm{θ})]}{\left(\frac{\mathrm{L}}{2}\sin \mathrm{θ}\right)}\\ \mathrm{W}=671\text{ N}\end{array}$

The gravitational beam on the beam,${\text{F}}_{\text{w}}=671\text{ N}$.

The gravitational force on the beam in unit vector notation, $\stackrel{→}{{\mathrm{F}}_{\mathrm{w}}}=−(671\text{ }\mathrm{N})\widehat{\mathrm{j}}$.

According to the static equilibrium condition, the sum of the vertical forces acting on the ladder is zero.

Hence,

$\begin{array}{c}{\mathrm{Î\pounds ¹ó}}_{\mathrm{y}\mathrm{net}}=0\\ {\mathrm{F}}_{\mathrm{h}\mathrm{y}}−\mathrm{W}=0\\ {\mathrm{F}}_{\mathrm{h}\mathrm{y}}=\mathrm{W}=671\text{ N}\end{array}$

For x-axis as

$\begin{array}{c}{\mathrm{Î\pounds ¹ó}}_{\mathrm{x}\mathrm{net}}=0\\ {\mathrm{F}}_{\mathrm{h}\mathrm{x}}−{\mathrm{T}}_{\mathrm{c}}=0\\ {\mathrm{F}}_{\mathrm{h}\mathrm{x}}={\mathrm{T}}_{\mathrm{c}}\\ {\mathrm{F}}_{\mathrm{h}\mathrm{x}}=400\text{ N}\end{array}$

The force on the beam from the hinge in unit vector notation, $\stackrel{→}{{\mathrm{F}}_{\mathrm{h}}}=\left(400\text{ }\mathrm{N}\right)\widehat{\mathrm{i}}+(671\text{ }\mathrm{N})\widehat{\mathrm{j}}$.