In Fig. 12-25, suppose the length Lof the uniform bar is $\mathbf{3}\mathbf{.00}\mathbf{\text{\hspace{0.17em}m}}$and its weight is$\mathbf{200}\mathbf{\text{\hspace{0.17em}N}}$Also, let the block’s weight $\mathit{W}\mathbf{=}\mathbf{}\mathbf{300}\mathbf{\text{\hspace{0.17em}N}}$and the angle$\mathit{\theta }\mathbf{=}\mathbf{}\mathbf{30}\mathbf{.0}\mathbf{°}$ . The wire can withstand a maximum tension of$\mathbf{500}\mathbf{\text{\hspace{0.17em}N}}$.(a)What is the maximum possible distance xbefore the wire breaks? With the block placed at this maximum x, what are the(b) horizontal and (c) vertical components of the force on the bar from the hinge at A?

i) Length of the bar, $L=3.00\text{\hspace{0.17em}m}$.

ii) Weight of bar,$w=200\text{\hspace{0.17em}N}$ .

iii) Weight of block, $W=300\text{\hspace{0.17em}N}$.

iv) The angle between the wire and the bar,$\theta ={30}^{\circ }$ .

v) Maximum tension without breaking the wire is, data-custom-editor="chemistry" $500\text{\hspace{0.17em}N}$.

Youcan use the condition for net torque at equilibrium to find the maximum possible distance x where the block can be placed before the wire breaks. Then using the condition for net force acting on the system at equilibrium,you can find the force components of the hinge on the bar

The calculation for the maximum possible distance X_max where the block can be placed without breaking the wire:

At equilibrium,

${\tau }_{net}=0$

From the figure, we can write for the net torque about the hinge as,

$\begin{array}{c}{T}_{\max }L\sin \theta -W\left(X_{\max }\right)-w\left(\frac{L}{2}\right)=0\\ W\left(X_{\max }\right)=T_{\max }L\sin \theta -w\left(\frac{L}{2}\right)\end{array}$

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

$\begin{array}{c}300\left(X_{\max }\right)=500\left(3.00\right)\sin {30}^{\circ }-200\left(\frac{3.00}{2}\right)\\ X_{\max }=\frac{450}{300}\\ =1.5\text{\hspace{0.17em}m}\end{array}$

Thus, the maximum possible distance x where a block can be placed before breaking of the wire is $1.50\text{\hspace{0.17em}m}$.

The calculation for thehorizontal component of the force on the bar from the hinge at A:

For equilibrium,the net horizontal force acting on the system is zero.

$\begin{array}{c}{F}_{xnet}=0\\ F_x-T_{\max }\cos \theta =0\\ F_x=T_{\max }\cos \theta \end{array}$

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

$\begin{array}{c}{F}_x=500\cos {30}^{\circ }\\ =433\text{\hspace{0.17em}N}\end{array}$

Thus, the horizontal component of the force on the bar from the hinge at A is $433\text{\hspace{0.17em}N}$.

The calculation for the vertical component of the force on the bar from the hinge at A:

At equilibrium,

$\begin{array}{c}{F}_{ynet}=0\\ F_y+T_{\max }\sin {30}^{\circ }-W-w=0\\ F_y=W+w-T_{\max }\sin {30}^{\circ }\end{array}$

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

$\begin{array}{c}{F}_y=300+200-500\sin {30}^{\circ }\\ =250\text{\hspace{0.17em}N}\end{array}$

Thus, the vertical component of the force on the bar from the hinge at A is$250\text{\hspace{0.17em}N}$ .