PLAIN AND CIVIL

EXAMPLE 5.0.1: MOMENT OF COPLANAR FORCES

Determine the moment of force F about point A

With force F lying on plane xy, we can solve this problem using either the vector method (r x F) or with the scalar method (Fd). With either case, it is longer if we use the force directly, that is, it is more difficult to determine the perpendicular distance from point A to the line of action of force F. Thus, the force needs to be converted to its components.

USING THE VECTOR METHOD

Force F in vector form is computed as follows:

$\mathbf{F}=\left ( -\frac{4}{5} \right )200i+\left ( \frac{3}{5} \right )200i=-160i+120j$

Next is to identify the distance of these components from point A. Remember the transmissibility of the force which makes it travel to its line of action. Force is considered a sliding vector and from the illustration, we can identify the exact distance using three points - B, C, and D. Let us use point B for this method.

Distances from point A to point B shows -4inches horizontally and +6inches vertically. In vector form

$r_{AB}=-4i+6j$

By taking the cross product of these two, we obtain the moment of the force
about point A.

$M_{A}=rxF=\left | \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ -4& 6& 0\\ -160& 120& 0 \end{matrix} \right |$

$M_{A}=rxF=0i+0j+\left [ \left ( -4x120 \right ) -\left ( 6x-160 \right )\right ]k=480\mathbf{k} lb.in$ ANSWER

USING SCALAR METHOD

The easier method for coplanar moments is the scalar method though trickier. Check the rotation to have the proper sign when solving the resultant.

1. Point B as force origin:

By using the sign convention (clockwise-positive, counterclockwise-negative), the moment shows

$M_{A}=-F_{1}d_{1}+F_{2}d_{2}$ $M_{A}=-160\left ( 6 \right )+120\left ( 4 \right )=-480lb.in(counterclockwise)$ ANSWER

2. Point C as force origin:

Using point C as the origin, notice that F2 runs through point A. This will not yield any perpendicular distance d2, thus d2 = 0. So,

$M_{A}=-F_{1}d_{1}+0$

$M_{A}=-160lb\left ( 3 in\right )=-480lb.in$

3. Point D as force origin:

Similar with point C, F1 will run through point A if the components are slid down to point D. d1 for this case will also be zero, thus

$M_{A}=0-120lb\left ( 4in \right )=-480lb.in$ ANSWER

Back to 5. MOMENT AND ITS PRINCIPLE

5. MOMENT AND ITS PRINCIPLE

Another parameter used to complete the formulas for equilibrium is Moment. Moment is defined as the turning effect brought about by a force with reference to a certain point. By formula, this is defined as the product of the force by its moment arm, whereby moment arm is the shortest distance between the point of reference from the line of action of the force.

$Moment=Force\left ( distance \right )_{perpendicular}$

THE PRINCIPLE OF MOMENTS

The principle of moments is also known as Varignon's theorem, which states

"The moment of a force about a point is equal to the sum of the moments of its components about that point."

By this, the adopted signs for each component are the rotating motions instead of the direction in the Cartesian plane. The most adopted conventional sign convention for moments are

clockwise - positive moment
counterclockwise - negative moment

Simply stated the principle of moment of force F about point O (where perpendicular distance from O to line of action R is d) is as follows:

$M_{O}^{R}=\sum M_{O}=\sum \left ( Fxd \right )$

COPLANAR APPLICATIONS

Determining moments from vertical and horizontal forces is easy because the distances can be computed directly. However, it is not the case for inclined forces. It is often easier to solve for the moment of an inclined force by breaking it down first into components and taking the moment instead of directly solving the moment from the force itself. This is due to the fact that distances can be easily solved using the vertical and horizontal axes.

Moments in two dimensional planes can be computed by Vector solution using cross product and Scalar solution using resultant of moments.

EXAMPLE 1: COPLANAR MOMENTS
EXAMPLE 2: COPLANAR MOMENTS
EXAMPLE 3: COPLANAR MOMENTS

EXAMPLE 5.0.4. MOMENT OF COPLANAR FORCES

EXAMPLE 5.0.5: RESULTANT MOMENT OF COPLANAR FORCES
EXAMPLE 5.0.6: APPLICATION OF MOMENT OF COPLANAR FORCES

SPATIAL APPLICATIONS

We can simplify the determination of moment of a force in space by taking the i-, j-, and k- components of the force and the position vector of the point in consideration. These procedures have already been discussed in spatial forces.

$\mathbf{r}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$

$\mathbf{F}=F_{x}\mathbf{i}+F_{y}\mathbf{j}+F_{z}\mathbf{k}$

With these two parameters identified, the moment of the force can be solved using the principle of cross product

$\mathbf{M}_{o}=\mathbf{r}x\mathbf{F}$

And also remember Varignon's theorem

$\mathbf{M}_{o}=\mathbf{M}_{x}\mathbf{i}+\mathbf{M}_{y}\mathbf{j}+\mathbf{M}_{z}\mathbf{k}$

It would be easier to compute the moment of a force in space using the determinant form after identifying the components.

EXAMPLE 6: SPATIAL MOMENT
EXAMPLE 7: SPATIAL MOMENT
EXAMPLE 8: RESULTANT MOMENT OF SPATIAL FORCES

EXAMPLE 4.2.2: CROSS PRODUCT: SHORTEST DISTANCE

At a given instant, the position of a plane at A and a train at B are measured relative to a radar antenna at O. Determine the shortest distance the train can interfere the signal of the antenna to the plane. (Shortest distance of point B from OA).

Start by identifying the coordinates of the points A, B, and O. Being the point of origin, point O coordinates can be easily identified (0, 0, 0). For the other points A and B, the components would have to be broken down as per given. Because angles were used to designate the coordinates, trigonometric function procedure is required.

For the plane at Pt. A:

The plane's configuration can be broken down firstly between the z axis and the plane xy.

   $Fz_{plane}=5sin60^{o}=4.33km$

$Fxy_{plane}=5cos60^{o}=2.50km$

While Fz is already the component of the plane on the z axis, the components at the x- and y- axes should still be broken down from Fxy.

$Fx_{plane}=-Fxy_{plane}cos35^{o}=-2.50cos35^{o}=-2.05km$

   $Fy_{plane}=-Fxy_{plane}sin35^{o}=-2.50sin35^{o}=-1.43km$

Both x- and y- components are negative ( going into the plane yz- and going to the left)

For the train at Pt. B:

Start off with the plane xy and axis z just the same for the train.

   $Fz_{train}=-2sin25^{o}=-0.85km$

   $Fxy_{train}=2cos25^{o}=1.81km$

Further break down Fxy into individual components:

   $Fx_{train}=Fxy_{train}sin40^{o}=1.81sin40^{o}=1.16km$

$Fy_{train}=Fxy_{train}cos40^{o}=1.81cos40^{o}=1.39km$

So, coordinates of points A, B, and O are as follows:

Point	i, km	j, km	k, km
A	-2.05	-1.43	4.33
B	1.16	1.39	-0.85
O	0.00	0.00	0.00

For the shortest distance of point B from line OA, take the perpendicular distance d.

From two forces OA and OB, taking the angle $\theta$ between them, we can identify the perpendicular distance, d, trigonometrically as:

   $d=\left | OB \right |sin\theta$

Also recall the geometric definition of cross product as:

$\overrightarrow{OA}X\overrightarrow{OB}=\left | OA \right |\left | OB \right |sin\theta \widehat{n}$

Since, we are solving for the distance, which is just a magnitude we can modify the cross product formula as:

$\left |\overrightarrow{OA}x\overrightarrow{OB} \right |=\left | OA \right |\left | OB \right |sin\theta$

We introduce the distance d into the cross product formula using the trigonometric definition:

$\left |\overrightarrow{OA}x\overrightarrow{OB} \right |=\left | OA \right |d$

Where our working formula for the shortest distance (perpendicular distance) between point B and line OA is:

   $d= \frac{\left |\overrightarrow{OA}x\overrightarrow{OB} \right |}{\left | OA \right |}$

Using this formula, we no longer need to find the value for $sin\theta$ .

For Force OA:

$\overrightarrow{OA}=\left ( 0-2.05 \right )i+\left ( 0-1.43 \right )j+\left ( 0+4.33 \right )=-2.05i-1.43j+4.33k$

The scalar value for Force OA is already given as 5km. You can check this by using Pythagorean formula with the values for i-, j-, and k-.

   $\left |\overrightarrow{OA} \right |=\sqrt{2.05^{2}+1.43^{2}+4.33^{2}}=5.0km$

For Force OB:

$\overrightarrow{OB} =\left (0+1.16 \right )i+\left ( 0+1.39 \right )j+\left ( 0-0.85 \right )k=1.16i+1.39j-0.85k$

No need to further determine the scalar of force OB.

Cross product of Forces OA and OB:

$\overrightarrow{OA}x\overrightarrow{OB}=\left | \begin{matrix} -2.05 &-1.43 &4.33 \\ 1.16 &1.39 &-0.85 \end{matrix} \right |\begin{matrix} -2.05\\ 1.16 \end{matrix}$

$\overrightarrow{OA}x\overrightarrow{OB}=\left | \begin{matrix} -1.43 &4.33 \\ 1.39&-0.85 \end{matrix} \right |i+\left | \begin{matrix} 4.33 &-2.05 \\ -0.85&1.16 \end{matrix} \right |j+\left | \begin{matrix} -2.05 &-1.43 \\ 1.16&1.39 \end{matrix} \right |k$

$\overrightarrow{OA}x\overrightarrow{OB}=\left [ \left ( -1.43x-0.85 \right )-\left ( 4.33x1.39 \right ) \right ]i+\left [ \left ( 4.33x1.16 \right )-\left ( -2.05x-0.85 \right ) \right ]j+\left [ \left ( -2.05x1.39 \right )-\left ( -1.43x1.16 \right ) \right ]k$

$\overrightarrow{OA}x\overrightarrow{OB}=-4.80i+3.28j-1.19k$

Take the magnitude of the cross product

$\left |\overrightarrow{OA}x\overrightarrow{OB} \right |=\sqrt{4.8^{2}+3.28^{2}+1.19^{2}}=5.93$

From the derived working equation:

$d=\frac{\left |\overrightarrow{OA}x\overrightarrow{OB} \right |}{\left |OA \right |}=\frac{5.93}{5}=1.19km$ ANSWER

Back to 4.2. INTRODUCTION TO CROSS PRODUCT

3.2. COMPONENTS OF SPATIAL FORCES

RESULTANT OF SPATIAL FORCES

Finding the resultant of forces in the two-dimensional plane is easy. The following procedure is followed:

Determine the reference axes
Convert forces into their components
Add or subtract parallel forces according to the sign convention preferred (usually +x going to the right and +y upwards
Take the magnitude of the resultant by using the principle of Pythagorean theorem

$R=\sqrt{\sum F_{x}+\sum F_{y}}$

Solve for the inclination by taking the tangent function of $\sum F_{x}$ and $\sum F_{y}$ .

Solving the resultant of spatial forces follows the same fundamental procedures although there are more steps in converting forces into their components.

Determine the reference axes, this time i, j, k references (similar with x, y, z axes)
Convert forces into i,j,k components. Be mindful of how the forces are drawn in order to properly break them down. This procedure might take several steps, depending on how the forces are given.
Add or subtract parallel forces still following same sign convention: +i going to the right, +j directed upwards, and +k coming out perpendicularly from plane ij (at times, the axes directed outwards from the plane of the paper)
The magnitude of the resultant is taken still using the principle of Pythagorean, only adding the k axis

$R=\sqrt{\sum F_{x}+\sum F_{y}+\sum F_{z}}$

The inclination will be determine using the cosine function of the specific axis and the resultant

$\theta _{x}=cos^{-1}\frac{\sum F_{x}}{R}$

$\theta _{y}=cos^{-1}\frac{\sum F_{y}}{R}$
$\theta _{z}=cos^{-1}\frac{\sum F_{z}}{R}$

E XAMPLE 3.2.1: RESULTANT OF SPATIAL FORCES
EXAMPLE 3.2.2: RESULTANT OF SPATIAL FORCES