In Fig12-46, a $\mathbf{50}\mathbf{.0}\mathbf{\text{\hspace{0.17em}kg}}$ uniform square sign, of edge length$\mathit{L}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.00}\mathbf{\text{\hspace{0.17em}m}}$, is hung from a horizontal rod of length ${\mathit{d}}_{\mathbf{h}}\mathbf{=}\mathbf{}\mathbf{3}\mathbf{.00}\mathbf{\text{\hspace{0.17em}m}}$and negligible mass. A cable is attached to the end of the rod and to a point on the wall at distance ${\mathit{d}}_{\mathbf{v}}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.00}\mathbf{}\mathbf{\text{\hspace{0.17em}m}}$above the point where the rod is hinged to the wall.(a) What is the tension in the cable? What are the (b) magnitude and) of the horizontal component of the force on the rod from the wall, and the (c) direction (left or right) of the horizontal component of the force on the rod from the wall, and the (d) magnitude of the vertical component of this force? And (e) direction (up or down) of the vertical component of this force?

i) Mass of the sign,$m=50.0\text{\hspace{0.17em}kg}$.

ii) Length of the rod,$d_h=3.00\text{\hspace{0.17em}m}$ .

iii) Distance from hinge to point where cable attached to the wall,$d_v=4.00\text{\hspace{0.17em}m}$

Using the condition for equilibrium,you can write the torque equation. From this,you can find the tension in the cable. Then using the conditions for the equilibrium for horizontal and vertical forces,you can find the magnitude and direction of the horizontal and vertical components of the force acting on the rod from the wall.

You can find the angle between the cable and the rod at which the sign is attached:

$\tan \theta =\frac{d_v}{d_h}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}\tan \theta =\frac{4.00}{3.00}\\ =1.33\\ \theta ={\tan }^{-1}\left(1.33\right)\\ ={53}^{\circ }\end{array}$

We can find the tension in the cable by using the condition for equilibrium as:

$\begin{array}{c}{\tau }_{net}=0\\ T\sin \theta \left(d_h\right)-\left(\frac{mg}{2}\right)(d_h-L)-\left(\frac{mg}{2}\right)d_h=0\\ T\sin \theta \left(d\right)=\left(\frac{mg}{2}\right)(d_h-L)+\left(\frac{mg}{2}\right)d_h\\ T=\frac{\left(\frac{mg}{2}\right)(d_h-L)+\left(\frac{mg}{2}\right)d_h}{\sin \theta \left(d\right)}\end{array}$

Substitute the values in the above expression, and we get,

localid="1661236210391" $\begin{array}{c}T=\frac{\left(\frac{\left(50.0\right)\left(9.8\right)}{2}\right)(3.00-2.00)+\left(\frac{\left(50.0\right)\left(9.8\right)}{2}\right)\left(3.00\right)}{\left(\sin {53}^{\circ }\right)\left(3.00\right)}\\ =408\text{\hspace{0.17em}N}\end{array}$

Thus, the tension in the cable is, $T=408\text{\hspace{0.17em}N}$.

At equilibrium,

$F_{xnet}=0$

Substitute the terms in the above expression, and we get,

$F_x-T\cos \theta =0$

Substitute the values in the above expression, and we get,

$\begin{array}{c}{F}_x-T\cos {53}^{\circ }=0\\ F_x=T\cos {53}^{\circ }\\ =408\times \cos {53}^{\circ }\\ =245\text{\hspace{0.17em}N}\end{array}$

Thus, the magnitude of the horizontal component of the force on the rod from the wall is, $F_x=245\text{\hspace{0.17em}N}$.

From the figure, we can see that direction of the horizontal component of the force on the rod from the wall is towards the right.

Thus, the direction of the horizontal component of the force on the rod from the wall is towards the right.

At equilibrium,

$F_{ynet}=0$

Substitute the terms in the above expression, and we get,

role="math" localid="1661236093581" $\begin{array}{c}{F}_y+T\sin \theta -mg=0\\ F_y=mg-T\sin \theta \end{array}$

Substitute the values in the above expression, and we get,

Thus, the magnitude of the vertical component of the force on the rod from the wall is,$F_y=163\text{\hspace{0.17em}N}$ .

The direction of the vertical component of the force on the rod from the wall is upward.
Thus, the vertical component of the force on the rod from the wall is in the upward direction.