91Ó°ÊÓ

In equilibrium, the net force in the x and y directions should be zero, and the torque about any point is zero.For this problem, you can calculate the tension in the wire using the condition for the zero torque about the hinge.

The weight of the shop sign is \({W_1} = 215\;{\rm{N}}\).

The weight of the beam is \({W_2} = 155\;{\rm{N}}\).

The angle of the wire is \(\theta = {35.0^ \circ }\).

The length of the beam is \({L_1} = 1.70\;{\rm{m}}\).

The supported wire is attached at a distance of \({L_2} = 1.35\;{\rm{m}}\) from the hinge.

You can assume that the mass of the beam is in the middle of the beam.

Let \({F_{\rm{x}}}\) and \({F_{\rm{y}}}\) be the horizontal and vertical components of the forces exerted by the hinge on the beam.

The torque about the hinge is

\(\begin{array}{c}T\sin \theta \times {L_2} - {W_1}{L_1} - {W_2}\frac{{{L_1}}}{2} = 0\\T\sin \theta \times {L_2} = \left( {{W_1} + \frac{{{W_2}}}{2}} \right){L_1}\\T \times \sin {35.0^ \circ } \times 1.35\;{\rm{m}} = \left( {215\;{\rm{N}} + \frac{{155\;{\rm{N}}}}{2}} \right)1.70\;{\rm{m}}\\T = 642.2\;{\rm{N}}\end{array}\).

Hence, the tension in the supporting wire is 642.2 N.

Now, the condition for the horizontal forces at equilibrium is

\(\begin{array}{c}{F_{\rm{x}}} = T\cos \theta \\ = \left( {642.2\;{\rm{N}}} \right) \times \cos {35.0^ \circ }\\ = 526.1\;{\rm{N}}\end{array}\).

Now, the condition for the vertical forces at equilibrium is

\(\begin{array}{c}{F_{\rm{y}}} + T\sin \theta = {W_1} + {W_2}\\{F_{\rm{y}}} + \left[ {\left( {642.2\;{\rm{N}}} \right) \times \sin {{35.0}^ \circ }} \right] = 215\;{\rm{N}} + 155\;{\rm{N}}\\{F_{\rm{y}}} = 1.6\;{\rm{N}}\end{array}\).

Hence, the horizontal and vertical components of the forces by the hinge are \(526.1\;{\rm{N}}\) and \(1.6\;{\rm{N}}\), respectively.