Chapter 9, Problem 27P

A uniform rod AB of length 5.0 m and mass M=3.S kg is hinged at A and held in equilibrium by a light cord, as shown in Fig. 9-57.A load W=22N 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.

To determine

A free-body diagram for the rod.

Answer to Problem 27P

Solution:

  

Explanation of Solution

Given:

Weight of the load, $W=22\text{ N}$

Tension in cord, $T=85\text{ N}$

Mass of the rod, $M=3.8\text{ kg}$

Formula used:

Equilibrium condition: the net force and torque would be zero.

  $\begin{array}{l}{\displaystyle \sum F=0}\\ {\displaystyle \sum \tau =0}\end{array}$

Calculation:

  

Conclusion:

Thus, the forces acting on the rod are:

Weight of the rod, weight of hanging object, Tension, vertical and horizontal components of force on the rod exerted by the hinge.

To determine

The vertical and horizontal forces on the rod exerted by the hinge.

Answer to Problem 27P

Solution:

Vertical component of the force exerted by the hinge $=8.64\text{N}$ and the horizontal component of the force exerted by the hinge $=51.2\text{N}$ .

Explanation of Solution

Given:

Weight of the load, $W=22\text{ N}$

Tension in cord, $T=85\text{ N}$

Mass of the rod, $M=3.8\text{ kg}$

Formula used:

Equilibrium condition: the net force and torque would be zero.

  $\begin{array}{l}{\displaystyle \sum F=0}\\ {\displaystyle \sum \tau }=0\end{array}$

Calculation:

  $\begin{array}{l}{\displaystyle \sum F_y=0}\\ F_{ay}+W+mg-T\cos 37°=0\\ F_{ay}+22+3.8(9.8)-85\cos 37°=0\\ F_{ay}=8.64\text{N}\\ {\displaystyle \sum F_x=0}\\ F_{ax}-T\sin 37°=0\\ F_{ax}=85\sin 37°\\ F_{ax}=51.2\text{N}\end{array}$

Conclusion:

Thus the vertical component of the force exerted by the hinge is 8.64N and the horizontal component of the force exerted by the hinge is 51.2N.

To determine

d from the appropriate torque equation.

Answer to Problem 27P

Solution:

  $d=2.44\text{ m}$

Explanation of Solution

Given:

Weight of the load, $W=22\text{ N}$

Tension in cord, $T=85\text{ N}$

Mass of the rod, $M=3.8\text{ kg}$

Formula used:

Equilibrium condition: the net force and torque would be zero.

  $\begin{array}{l}{\displaystyle \sum F=0}\\ {\displaystyle \sum \tau }=0\end{array}$

Calculation:

About A: ${\displaystyle \sum \tau =0}$

  $\begin{array}{l}W(d\sin 53)+mg(2.5\sin 53)+T\cos 37(5\sin 53)+T\sin 37(5\cos 53)=0\\ 22(d\sin 53)=85(5)(\cos 37\sin 53-\sin 37\cos 53)-3.8(9.8)(2.5\sin 53)\\ 22(d\sin 53)=42.793\\ d=2.44\text{m}\end{array}$

Conclusion:

Thus, the distance is $d=2.44\text{ m}$ .