In Fig. 12-82, a uniform beam of length $\mathbf{12}\mathbf{.0}\mathbf{\text{\hspace{0.17em}m}}$ is supported by a horizontal cable and a hinge at angle $\mathbf{\text{θ}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{50.0}\mathbf{°}$. The tension in the cable is $\mathbf{400}\mathbf{\text{\hspace{0.17em}N}}$ .

In unit-vector notation, what are

(a) the gravitational force on the beam and

(b) the force on the beam from the hinge?

The length of the uniform beam,$\text{L}=12.0\text{\hspace{0.17em}m}$

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

The tension in the cable, $\text{T}=400\text{\hspace{0.17em}N}$

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.

$\begin{array}{l}\mathbf{\Sigma }{\overrightarrow{\mathbf{F}}}_{\mathbf{net}}\mathbf{=}\mathbf{0}\\ \mathbf{\Sigma }{\overrightarrow{\mathbf{\tau }}}_{\mathbf{net}}\mathbf{=}\mathbf{0}\end{array}$

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

$\varnothing =90-\mathrm{\theta }$

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

$\begin{array}{c}{\mathrm{\Sigma \tau }}_{\mathrm{net}}=0\\ \mathrm{W}\left(\frac{\mathrm{L}}{2}\sin \mathrm{\theta }\right)-{\mathrm{T}}_{\mathrm{c}}(\mathrm{Lsin}\varnothing )=0\\ \mathrm{W}\left(\frac{\mathrm{L}}{2}\sin \mathrm{\theta }\right)={\mathrm{T}}_{\mathrm{c}}(\mathrm{Lsin}\varnothing )\\ \mathrm{W}=\frac{{\mathrm{T}}_{\mathrm{c}}(\mathrm{Lsin}\varnothing )}{\left(\frac{\mathrm{L}}{2}\mathrm{sin\theta }\right)}\\ \mathrm{W}=\frac{{\mathrm{T}}_{\mathrm{c}}[\mathrm{Lsinsin}(90-\mathrm{\theta })]}{\left(\frac{\mathrm{L}}{2}\sin \mathrm{\theta }\right)}\\ \mathrm{W}=671\text{\hspace{0.17em}N}\end{array}$

The gravitational beam on the beam,${\text{F}}_{\text{w}}=671\text{\hspace{0.17em}N}$.

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

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

Hence,

$\begin{array}{c}{\mathrm{\Sigma F}}_{\mathrm{y}\mathrm{net}}=0\\ {\mathrm{F}}_{\mathrm{h}\mathrm{y}}-\mathrm{W}=0\\ {\mathrm{F}}_{\mathrm{h}\mathrm{y}}=\mathrm{W}=671\text{\hspace{0.17em}N}\end{array}$

For x-axis as

$\begin{array}{c}{\mathrm{\Sigma F}}_{\mathrm{x}\mathrm{net}}=0\\ {\mathrm{F}}_{\mathrm{h}\mathrm{x}}-{\mathrm{T}}_{\mathrm{c}}=0\\ {\mathrm{F}}_{\mathrm{h}\mathrm{x}}={\mathrm{T}}_{\mathrm{c}}\\ {\mathrm{F}}_{\mathrm{h}\mathrm{x}}=400\text{\hspace{0.17em}N}\end{array}$

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