(III) A uniform rod AB of length 5.0 m and mass \({\bf{M = 3}}{\bf{.8}}\;{\bf{kg}}\) is hinged at A and held in equilibrium by a light cord, as shown in Fig. 9–67. A load \({\bf{W = 22}}\;{\bf{N}}\) hangs from the rod at a distance d so that the tension in the cord is 85 N. (a) Draw a free-body diagram for the rod. (b) Determine the vertical and horizontal forces on the rod exerted by the hinge. (c) Determine d from the appropriate torque equation.

For a system to be in static equilibrium, the value of the net torque and net force working on the system must be equivalent to zero.

Given data:

The length of the rod is\(l = 5.0\;{\rm{m}}\).

The mass of the rod is \(M = 3.8\;{\rm{kg}}\).

The load is \(W = 22\;{\rm{N}}\).

The tension in the rope is \({F_{{\rm{rope}}}} = 85\;{\rm{N}}\).

The angle is \(\phi = 37^\circ \).

The angle is \(\theta = 53^\circ \).

The free body diagram of the rod can be drawn as:

Here, \({F_{\rm{H}}}\) is the horizontal force, \({F_{\rm{V}}}\) is the vertical force, \(W\) is the load, and \(Mg\) is the force due to the weight of the rod.

Apply the force equilibrium condition along the horizontal direction.

\(\begin{array}{c}\sum {F_x} = 0\\{F_{{\rm{rope}}}}\sin \phi - {F_{\rm{H}}} = 0\\\left( {85\;{\rm{N}}} \right)\sin \left( {37^\circ } \right) - {F_{\rm{H}}} = 0\\{F_{\rm{H}}} = 51.1\;{\rm{N}} \approx {\rm{51}}\;{\rm{N}}\end{array}\)

Thus, the magnitude of the net horizontal force on the rod exerted by the hinge is \({\rm{51}}\;{\rm{N}}\).

Apply the force equilibrium condition along the vertical direction.

\(\begin{array}{c}\sum {F_{\rm{y}}} = 0\\{F_{{\rm{rope}}}}\cos \phi + {F_{\rm{V}}} - Mg - W = 0\\\left( {85\;{\rm{N}}} \right)\cos \left( {37^\circ } \right) + {F_{\rm{V}}} - \left( {3.8\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right) - \left( {22\;{\rm{N}}} \right) = 0\\{F_{\rm{V}}} = - 8.6\;{\rm{N}}\end{array}\)

Here, a negative sign indicates the vertical force points downward.

Thus, the magnitude of the net vertical force on the rod exerted by the hinge is \(8.6\;{\rm{N}}\).

Now, take the torques about the hinge point, with the clockwise torques as positive.

\(\begin{array}{c}\sum \tau = 0\\Wd\sin \theta + Mg\frac{l}{2}\sin \theta - {F_{{\rm{rope}}}}l\sin \left( {\theta - \phi } \right) = 0\\\left[ \begin{array}{l}\left( {22\;{\rm{N}}} \right)d\sin \left( {53^\circ } \right) + \left( {3.8\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\left( {\frac{{5\;{\rm{m}}}}{2}} \right)\sin \left( {53^\circ } \right)\\ - \left( {85\;{\rm{N}}} \right)\left( {5\;{\rm{m}}} \right)\sin \left( {53^\circ - 37^\circ } \right)\end{array} \right] = 0\\d = 2.4\;{\rm{m}}\end{array}\)

Thus, the value of the distance \(d\) is \(2.4\;{\rm{m}}\).