Question

The 4-lb force is applied at point A of the crank assembly. Determine the moment about point O using the geometric method.

Use the geometric method. Meaning, use Mox=+/-fydz +/- fzdy, Moy=+/-fxdz +/- fzdx , Moz=+/-fxdy +/-fydx

The 4-lb force is applied at point A of the crank assembly. Determine the moment about point O using the geometric method. Use the geometric method. Meaning, use Mox=+/-fydz +/- fzdy, Moy=+/-fxdz +/- fzdx , Moz=+/-fxdy +/-fydx

Answer 1

Concepts and reason

This question is based on the concept of the moment of the force.

Moment of the force: In a system of forces, the vector product of the distance between the line of action of force from a certain point and the force is known as moment of force. That certain point is the point about which the moment has been considered.

First of all, obtain the vector form of the force $F$ using the orientation of the line of action of the force. Obtain the position vector of the point A, assuming O as the origin. Perform the vector product of the force and position vector of the point A in order to calculate the moment of the force.

Fundamentals

A force F can be expressed in a vector form as,

${\bf{\vec F}} = F\left( {\cos {\theta _x}{\bf{i}} + \cos {\theta _y}{\bf{j}} + \cos {\theta _z}{\bf{k}}} \right)$

Here, ${\theta _x}$ , ${\theta _y}$ and ${\theta _z}$ is the angle made by the force with axes x, y, and z respectively, and F is the magnitude of the applied force.

The moment about O is calculated as,

${{\bf{\vec M}}_O} = {{\bf{\vec R}}_{OA}}{\bf{ \times \vec F}}$

Here, ${{\bf{\vec R}}_{RO}}$ is the perpendicular position vector to the force from O and ${\bf{F}}$ is the force vector.

The 3D diagram of arrangement is given as,

Consider the coordinates of the point O.

$A = \left( {0{\rm{ in}},1.5{\rm{ in}},0.75{\rm{ in}}} \right)$

Consider the vector form of the force F.

$\vec F = F\left( { - \cos {\theta _x}{\bf{i}} + \cos {\theta _y}{\bf{j}} + \cos {\theta _z}{\bf{k}}} \right)$

Here, ${\theta _x}$ is angle made by the force with the $x$ -axis, ${\theta _y}$ is angle made with the y-axis and ${\theta _z}$ is angle made by the force with the z-axis, i, j, k are the unit vector along x, y and z direction.

Substitute $30^\circ$ for ${\theta _x}$ , $60^\circ$ for ${\theta _y}$ , $90^\circ$ for ${\theta _z}$ and $4{\rm{ lb}}$ for $F$ .

$\begin{array}{c}\\\vec F = \left( {{\rm{4 lb}}} \right)\left( { - \cos 30^\circ {\bf{i}} + \cos 60^\circ {\bf{j}} + \cos 90^\circ {\bf{k}}} \right)\\\\ = \left( { - 3.468{\bf{i}} + 2{\bf{j}}} \right)\\\end{array}$

Assume O as origin and obtain the position vector for the point A.

${\vec R_{OA}} = \left( {1.5{\bf{j}} + 0.75{\bf{k}}} \right){\rm{ in}}$

Consider the moment of the force about point O.

${\vec M_O} = {\vec R_{OA}}{\bf{ \times }}\vec F$

Here, ${\vec M_O}$ is moment of the force about the point O.

‎

Substitute $\left( { - 3.468{\bf{i}} + 2{\bf{j}}} \right){\rm{ lb}}$ for $\vec F$ and $\left( {1.5{\bf{j}} + 0.75{\bf{k}}} \right){\rm{ in}}$ for ${\vec R_{OA}}$ .

$\begin{array}{c}\\{{\vec M}_O} = \left( {1.5{\bf{j}} + 0.75{\bf{k}}} \right){\rm{ lb}}{\bf{ \times }}\left( { - 3.468{\bf{i}} + 2{\bf{j}}} \right){\rm{ in}}\\\\{\rm{ = }}\left( { - 1.5{\bf{i}} - {\rm{2}}{\rm{.6}}{\bf{j}} + {\rm{5}}{\rm{.2}}{\bf{k}}} \right){\rm{ lb}} \cdot {\rm{in}}\\\end{array}$

Ans:

The magnitude of the moment about O is $\left( { - 1.5{\bf{i}} - {\rm{2}}{\rm{.6}}{\bf{j}} + {\rm{5}}{\rm{.2}}{\bf{k}}} \right){\rm{ lb}} \cdot {\rm{in}}$ .