Question

The J-shaped member shown in the figure is supported by a cable DE and a single journal bearing with a square shaft at A. Determine the reaction forces A_y and A_z at support A required to keep the system in equilibrium. The cylinder has a weight W_B = 5.30 lb, and F = 1.20 lb is a vertical force applied to the member at C. The dimensions of the member are w = 1.50 ft, l = 6.00 ft, and h = 2.00 ft. Find Ay and Az. (I have found Ay=0 and Az=6.5 lb).

Now, for the same J-shaped member, determine M_A_x, M_A_y, and M_A_z, the couple moments at the support about the x, y, and z axes, respectively, required to keep the system in equilibrium. The cylinder weighs W_B = 5.30 lb; a vertical force F = 1.20 lb acts at C; and the member's dimensions are w = 1.50 ft, l = 6.00 ft, and h = 2.00 ft. Find M_A_x, M_A_y and M_A_z.

Answer 1

Concepts and reason

The free body diagram shows all the forces acting on the body and also the direction of forces acting. The arrow represents each forces acting on a particular direction.

In two dimensional Cartesian systems, the force which point in the direction of upward or rightward must be taken as positive and the force which point in the direction of downward or leftward must be taken as negative.

Fundamentals

Resolving of forces:

The resolving of force into two components is shown in figure (1).

The horizontal force $\left( {{F_H}} \right)$ is calculated as shown in figure (1).

${F_H} = F\cos \theta$

Here, the force is F and angle of force with the horizontal is $\theta$ .

The vertical force $\left( {{F_V}} \right)$ is calculated as shown in figure (1).

${F_V} = F\sin \theta$

Equilibrium equations of the system:

There are six conditions in total to satisfy the equilibrium of a body in three dimensional Cartesian coordinate system

1. Force equilibrium along x-axis

$\sum {{F_x} = 0}$

2. Force equilibrium along y-axis

$\sum {{F_y} = 0}$

3. Force equilibrium along z-axis

$\sum {{F_z} = 0}$

4. Moment equilibrium along x-axis.

$\sum {{M_x} = 0}$

5. Moment equilibrium along y-axis.

$\sum {{M_y} = 0}$

6. Moment equilibrium along z-axis.

$\sum {{M_z} = 0}$

The free body diagram of the system is given figure (1).

Write the force equilibrium equation along x-axis to find the reaction force at A along x direction.

$\begin{array}{l}\\\sum {{F_x} = 0} \\\\{F_{DE}}\sin \theta = 0\\\\{F_{DE}} = 0\\\end{array}$

Write the force equilibrium equation along y-axis to find the reaction force at A along y direction.

$\begin{array}{l}\\\sum {{F_y} = 0} \\\\{A_y} = 0\\\end{array}$

Write the force equilibrium equation along z axis to find the reaction force at A along z direction.

$\begin{array}{l}\\\sum {{F_z} = 0} \\\\{A_z} - W - F = 0\\\end{array}$

Substitute $5.3{\rm{ lb}}$ for $W$ and $1.2{\rm{ lb}}$ for $F$ .

$\begin{array}{l}\\{A_z} - 5.3{\rm{ lb}} - 1.2{\rm{ lb}} = 0\\\\{A_z} = 6.5{\rm{ lb}}\\\end{array}$

Write the moment equilibrium equation along x-axis to find the moment at point A along x direction.

$\begin{array}{l}\\\sum {{M_x} = 0} \\\\{M_{Ax}} - W \times l - F \times l = 0\\\end{array}$

Substitute $5.3{\rm{ lb}}$ for $W$ , $6{\rm{ ft}}$ for $l$ , and $1.2{\rm{ lb}}$ for $F$ .

$\begin{array}{l}\\{M_{Ax}} - 5.3 \times 6 - 1.2 \times 6 = 0\\\\{M_{Ax}} = 39{\rm{ lb}}{\rm{.ft}}\\\end{array}$

Write the moment equilibrium equation along y axis to find the moment about point A along y direction.

$\begin{array}{l}\\\sum {{M_y} = 0} \\\\{M_{Ay}} - F \times w = 0\\\end{array}$

Substitute $1.5{\rm{ ft}}$ for $w$

$\begin{array}{l}\\{M_{Ay}} - 1.2 \times 1.5 = 0\\\\{M_{Ay}} = 1.8{\rm{ lb}}{\rm{.ft}}\\\end{array}$

Write the moment equilibrium equation along z axis to find the moment at point A along z direction.

$\begin{array}{l}\\\sum {{M_z} = 0} \\\\{M_{Az}} = 0\\\end{array}$

Ans:

The support reaction along y direction at point A $\left( {{A_y}} \right)$ is $0{\rm{ lb}}$ .