In Fig. 12-45, a thin horizontal bar ABof negligible weight and length Lis hinged to a vertical wall at Aand supported at B by a thin wire BCthat makes an angle$\mathit{\theta }$ with the horizontal. A block of weight Wcan be moved anywhere along the bar; its position is defined by the distance xfrom the wall to its center of mass. As a function of x, find(a) the tension in the wire, and the (b) horizontal and (c) vertical components of the force on the bar from the hinge at A.

Bar-wire system.

Using the condition for static equilibrium, we can write the torque equation. From this, we will get the tension in the wire. Then using the first and second conditions for the horizontal and vertical components of the forces, we will get thehorizontal and vertical components of force on the bar at hinge A.

At equilibrium, the torque equation can be written as,

$\sum \tau =0$

Computing torque at the hinge as,

$\begin{array}{c}TL\sin \theta -Wx=0\\ T=\frac{Wx}{L\sin \theta }\end{array}$

Thus, the tension in the wire BC attached to the bar is$\frac{Wx}{L\sin \theta }$.

At equilibrium, the net force equation in the x-direction can be given as,

$\sum F_x=0$

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

$F_x-T\cos \theta =0$

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

$\begin{array}{c}{F}_x=\frac{Wx\cos \theta }{L\sin \theta }\\ =\frac{Wx}{L\tan \theta }\end{array}$

Thus, the horizontal component of force on the bar at hinge A is$\frac{Wx}{L\tan \theta }$.

At equilibrium, the net force equation in the y-direction can be given as,

$\sum F_y=0$

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

$F_y-W+T\sin \theta =0$

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

$\begin{array}{c}{F}_y=W-T\sin \theta \\ =W-\frac{Wx}{L\sin \theta }\sin \theta \\ =W(1-\frac{x}{L})\end{array}$

Thus, the vertical component of force on the bar at hinge A is$W(1-\frac{x}{L})$.