In Fig. 12-47, a nonuniform bar is suspended at rest in a horizontal position by two massless cords. One cord makes the angle$\mathit{\theta }\mathbf{=}\mathbf{}\mathbf{36}\mathbf{.}\mathbf{9}\mathbf{°}$with the vertical; the other makes the angle$\mathit{\varphi }\mathbf{=}\mathbf{53}\mathbf{.1}\mathbf{°}$with the vertical. If the length$\mathit{L}$of the bar is$\mathbf{6}\mathbf{.}\mathbf{10}\mathbf{}\mathit{m}$, compute the distance$\mathit{x}$from the left end of the bar to its center of mass.

Length of the bar

$\begin{array}{l}\varphi =53.1°\\ \theta =36.9°\end{array}$

For the condition of equilibrium to be satisfied, the net forces and the net torques acting on the pivot point are zero as per Newton's third law. Thus, we consider the forces along the horizontal and vertical directions with their distances from the line of force to get the net torque about these forces. Now, we get a final expression to the net torque acting at that point as zero for the condition of balancing of forces.

Formula:

The net force acting on a body as per Newton’s third law$\mathbf{\sum }{\mathit{F}}_{\mathbf{n}\mathbf{e}\mathbf{t}}\mathbf{=}\mathbf{0}$,

The net torque acting on a pivot point is zero,$\mathbf{\sum }{\mathit{\tau }}_{\mathbf{n}\mathbf{e}\mathbf{t}}\mathbf{=}\mathbf{0}$

Let$T_l$be the tension force of the left-hand cord ,$Tr$be the tension force of the right-hand cord, and m be the mass of the bar.

Now, for the condition of equilibrium,

Condition-1: The vertical forces acting on the body is equated to zero.

$Tlcos\theta +Trcos\varphi -mg=0$

Condition-2: The horizontal forces acting on the body is equated to zero.

$-Tlsin\theta +Trsin\varphi =0$

Condition-3: The net torque acting at the point is zero.

$mgx-TrLcos\varphi =0$

The origin was chosen to be at the left end of the bar for purposes of calculating the torque. The unknown quantities are$Tl$,$Tr$and.$x$

The second equation yields:

$Tl=T_{r}sin\varphi /sin\theta$

Now, this is substituted into the first and solved for$Tr$the result is given as follows:

$Tr=\frac{mgsin\theta }{sin\varphi cos\theta +cos\varphi sin\theta }$

Now, substituting the above value in the third equation, we get that

$\begin{array}{c}x=L\frac{sin\theta cos\varphi }{sin\varphi cos\theta +cos\varphi sin\theta }\\ =L\frac{sin\theta cos\varphi }{sin[\theta +\varphi ]}\left(\because sin(A+B)=sin{\rm A}cosB+cosAsinB\right)\end{array}$

For the special case ofthis problem$\theta +\varphi =90°$and$sin[\theta +\varphi ]=1$.Thus, the value of x becomes:

$\begin{array}{c}x=Lsin\theta cos\varphi \\ =(6.10m)sin36.9°cos53.1°\\ =2.20m\end{array}$

The distance$x$ from the left end of the bar to its center of mass is $2.20m$.