Question

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?

Short answer

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

Step-by-step solution

Understanding the given information

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, $\text{T}=400\text{\hspace{0.17em}N}$

Concept and formula used in the given question

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.

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

(a) Calculation for the gravitational force on the beam

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, $\overrightarrow{{\mathrm{F}}_{\mathrm{w}}}=-(671\mathrm{N})\widehat{\mathrm{j}}$.

(b) Calculation for the force on the beam from the hinge 

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, $\overrightarrow{{\mathrm{F}}_{\mathrm{h}}}=\left(400\mathrm{N}\right)\widehat{\mathrm{i}}+(671\mathrm{N})\widehat{\mathrm{j}}$.