 # Engineering Statics: Open and Interactive

## Section4.3Scalar Components

We saw in Subsection 3.3.2 that vectors can be expressed as the product of a scalar component and a unit vector.
For example, a $$\N{100}$$ force acting down can be represented by $$F_y\, \jhat\text{,}$$ where $$F_y$$ is the scalar component and $$F_y=\N{-100}\text{.}$$ This describes a vector $$\vec{F}$$ which has a magnitude of $$\N{100}$$ and acts in the $$-\jhat$$ direction, i.e. $$\downarrow\text{.}$$ The unit vector $$\jhat$$ along with the sign $$(+/-)$$ of $$F_y$$ determines the direction, while the absolute value of $$F_y$$ determines the vector’s magnitude.
Moments in two dimensions are either clockwise or counter-clockwise, or alternately they point into or out of the page. This means that a single scalar value is sufficient to completely specify such a moment if we have established which direction is positive. The choice is arbitrary, but the default sign convention is based on the right-handed Cartesian coordinate system, as illustrated in Figure 4.1.3.
When using the standard convention, counter-clockwise moments are positive and clockwise moments are negative. Simply append a positive sign to the magnitude for counter-clockwise moments or a negative sign for clockwise moments to create a scalar component. You are free to use the opposite convention, but this should be explicitly stated.

### Example4.3.1.Sign Conventions.

For each scalar component, determine the direction of the corresponding moment vector.
1. $$\displaystyle M_1 = \Nm{30}$$
2. $$\displaystyle M_2 = \kNm{-400}$$
3. $$\displaystyle M_3 = \Nm{25} \circlearrowright$$
4. $$\displaystyle M_4 = \ftlb{-100} \circlearrowright$$
1. CCW
2. CW
3. CW
4. CCW
Solution.
1. CCW. Use the default sign convention, i.e. CCW is positive.
2. CW. Negative value means the moment acts opposite to positive direction.
3. CW. The arrow overrides default sign convention, so now CW is positive direction.
4. CCW. Negative CW is CCW.
Scalar components are most useful when combining several clockwise and counter-clockwise moments. The resulting algebraic sum of the scalar components will be either positive, negative, or zero, and this sign indicates the direction of the resultant moment.

Use scalar moments to determine the magnitude of the resultant of three moments:
$$\vec{M_1} = \kNm{25} \circlearrowright\text{,}$$ $$\vec{M_2} =\kNm{40} \circlearrowleft\text{,}$$ and $$\vec{M_3} = \kNm{30} \circlearrowright$$
$$|\vec{M}| = \kNm{15}$$
Solution.
Manually attaching the signs according to the standard sign convention (CCW +) gives the scalar moments:
\begin{gather*} M_1 = \kNm{-25}\\ M_2 =\kNm{+40}\\ M_3 = \kNm{-30}\text{.} \end{gather*}
Adding these moments gives the resultant scalar moment.
\begin{align*} M \amp= M_1 + M_2 + M_3\\ \amp = (\kNm{-25}) + (\kNm{40}) + (\kNm{ - 30})\\ \amp = \kNm{-15}\text{.} \end{align*}
The corresponding magnitude of $$\vec{M}$$ is
In three dimensions, moments, like forces, can be resolved into components in the $$x\text{,}$$ $$y\text{,}$$ and $$z$$ directions.