Static Equilibrium EX 4

The rod is being studied. Three forces act on the rod: the tension in the wire $F_T$; the load $F_L$ and the support at the hinge $F$. The wire is 60 cm long and it appears to be horizontal. It is attached in the middle of the rod. Thus, the wire, the wall and the rod form a right angle triangle with hypotenuse 1 m and one side of 60 cm. From this information, we can compute the angle between the rod and the wall:

$\cos \theta =\frac{0.6}{1}=0.6$

$\theta =53^{\circ }$

The diagram below shows the forces and the lever arm for each of the forces. Note that the load acts 2.00 m from the pivot and the wire pulls at a distance 1 m from the pivot. The torque exerted by the support is equal to zero because it acts at the pivot.

The rod is in equilibrium and so

$\mathrm{\Sigma }{F}_x=0$
$\mathrm{\Sigma }{F}_y=0$
$\mathrm{\Sigma }\tau =0$

We will first compute the torques:

Force	Direction	Torque
$F_T$	CW	$\tau =-(1)F_T\sin 143^{\circ }=-0.6F_T$
$F_L$	CCW	$\tau =(2)(200)\sin 143^{\circ }=240Nm$
$\mathrm{\Sigma }\tau$		0

Solving for $F_T$,

$F_T=\frac{240}{0.6}=400N$

Then we will determine the components of the support at the pivot:

Forces	x-component	y-component
$F_T$	400 N	0
$F_L$	0	-200 N
$F$	$-F_x$	$F_y$
$\mathrm{\Sigma }\overrightarrow{F}$	0	0

Solving the two equations we find

$\overrightarrow{F}=400\hat{i}+200\hat{j}$