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:

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

$\theta =53^{\circ }$

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

$\Sigma F_{x}=0$
$\Sigma F_{y}=0$
$\Sigma \tau =0$

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	$\tau =(2)(200)\sin {143^{\circ }}=240Nm$
$\Sigma \tau$		0

Solving for $F_{T}$,

$F_{T}={240 \over 0.6}=400N$

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
$F$	$-F_{x}$	$F_{y}$
$\Sigma {\vec {F}}$	0	0

Solving the two equations we find

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