Question

(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.

Short answer

(a) The FBD of the rod is shown as:

(b)The magnitudes of the net horizontal force and vertical force on the rod exerted by the hinge are \({\rm{51}}\;{\rm{N}}\) and \(8.6\;{\rm{N}}\), respectively.

(c) The value of distance \(d\) is \(2.4\;{\rm{m}}\).

Step-by-step solution

Meaning of static equilibrium

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 information

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 \(\theta = 53^\circ \).

Draw a free body diagram of the rod

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.

Evaluation of the net horizontal force and vertical force on the rod exerted by the hinge

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 \(8.6\;{\rm{N}}\).

Evaluation of the distance d

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 \(2.4\;{\rm{m}}\).