Question

The J-shaped member shown in the figure(Figure 1) is supported by a cable DE and a single journal bearing with a square shaft at A. Determine the reaction forces Ay and Az at support A required to keep the system in equilibrium. The cylinder has a weight WB = 5.70 lb , and F = 2.00 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 .

Answer 1

Concepts and reason

Newton’s second law of motion: From Newton’s second law of motion, the summation of force is equal to the product of mass and acceleration.

Resolution of vector: When a vector makes an angle $\theta$ with, say the $x$ axis, then the cosine component of the vector is in the $x$ direction, and the sine component of the vector is in the $y$ direction.

Equilibrium of forces: When the object is in equilibrium condition, then the net forces acting on the object is zero. The forces acting in the positive direction are balanced with the forces acting in the negative direction.

The concept of Newton’s second law of motion, resolution of vector, equilibrium of forces is used to solve this problem. From Newton’s second law of motion, the summation of force is equal to the product of mass and acceleration. When the particle is in equilibrium, then the component of the force along the $x$ , $y$ , and $z$ axis is calculated by taking the acceleration as zero, and equating the summation of the force in a particular direction with zero.

Fundamentals

The resultant displacement is given as,

${\rm{displacement}} = \sqrt {{{\left( {{l_{{\rm{x - axis}}}}} \right)}^2} + {{\left( {{l_{{\rm{y - axis}}}}} \right)}^2} + {{\left( {{l_{{\rm{z - axis}}}}} \right)}^2}}$

Here, ${l_{{\rm{x - axis}}}}$ , ${l_{{\rm{y - axis}}}}$ , and ${l_{{\rm{z - axis}}}}$ is the length from the origin in the x-direction, y-direction, and z-direction respectively.

Take the origin as a reference. Then, take the upward direction as positive, and downward direction as negative. Also, take the forward direction as positive, and backward direction as negative. This sign convention is used throughout the solution.

The diagram for the condition mentioned in the problem is given below:

The force along the $z$ direction is the reaction force along the z axis in the upward direction, the force due to the weight at point $B$ in the downward direction, and the force at point $C$ in the downward direction

The length of the rod $AB$ of the member along the $y$ axis is $l$ and is mentioned as $6.00{\rm{ ft}}$ . Also, the length $BD$ of the member along the $x$ axis is $\left( {1.50{\rm{ ft}} + 1.50{\rm{ ft}}} \right)$ , that is, ${\rm{3}}{\rm{.00 ft}}$ . This means that point $E$ is at a distance of $3.00{\rm{ ft}}$ in the negative $x$ axis, $6{\rm{ ft}}$ in the positive $y$ axis, and $2{\rm{ ft}}$ in the positive $z$ axis. The length of the supporting cable $DE$ from the diagram is calculated as follows:

$DE = \sqrt {{l^2} + {h^2} + {{\left( {w + w} \right)}^2}}$

Substitute $6.00{\rm{ ft}}$ for $l$ , ${\rm{3}}{\rm{.00 ft}}$ for $\left( {w + w} \right)$ , and $2.00{\rm{ ft}}$ for $h$ in the expression of $DE$ , and solve,

$\begin{array}{c}\\DE = \sqrt {{{\left( {6.00{\rm{ ft}}} \right)}^2} + {{\left( {{\rm{2}}{\rm{.00 ft}}} \right)}^2} + {{\left( {{\rm{3}}{\rm{.00 ft}}} \right)}^2}} \\\\ = \sqrt {36.00{\rm{ f}}{{\rm{t}}^2} + 4.00{\rm{ f}}{{\rm{t}}^2} + 9.00{\rm{ f}}{{\rm{t}}^2}} \\\\ = \sqrt {49.00{\rm{ f}}{{\rm{t}}^2}} \\\\ = 7.00{\rm{ ft}}\\\end{array}$

Equate the forces along the $x$ direction at point $E$ ,

$\sum {{F_x} = 0}$

Substitute ${F_{DE}}\sin \theta$ for $\sum {{F_x}}$ in the expression $\sum {{F_x} = 0}$ , and solve,

${F_{DE}}\sin \theta = 0$

From the diagram, the value of ${F_{BC}}$ is ${F_{DE}}\sin \theta$ . Then, the above expression ${F_{DE}}\sin \theta = 0$ can be written as,

${F_{BC}} = 0$

Similarly, consider the equilibrium of force along the $y$ direction. The only reaction force, ${A_y}$ , is acting along the y-axis. Equate the forces along the $y$ direction. This gives,

$\sum {{F_y} = 0}$

Substitute ${A_y}$ for the summation of forces $\sum {{F_y}}$ in the expression $\sum {{F_y} = 0}$ , and solve,

${A_y} = 0$

Consider the equilibrium of force along the $z$ direction. The forces along the $z$ direction are the reaction force along the z axis in the upward direction, the force due to the weight at point $B$ in the downward direction, and the force at point $C$ in the downward direction. Equate the forces along the $z$ axis. This gives,

$\sum {{F_z} = 0}$

Substitute ${A_z} - {W_B} - F$ for $\sum {{F_z}}$ in the expression $\sum {{F_z} = 0}$ , and solve,

${A_z} - {W_B} - F = 0$

Further, substitute $5.70{\rm{ lb}}$ for ${W_B}$ , $2.00{\rm{ lb}}$ for $F$ in the expression ${A_z} - {W_B} - F = 0$ , and solve,

$\begin{array}{c}\\{A_z} - 5.70{\rm{ lb}} - 2.00{\rm{ lb}} = 0\\\\{A_z} = 7.70{\rm{ lb}}\\\end{array}$

Ans:

The reaction force ${A_y}$ is $0{\rm{ lb}}$ .