Define inertia and explain three different directions of forces with diagrams. Sketch the Free Body Diagram for a loaded beam with pin and roller supports.

Here are the solutions to the questions:

Q1(a) (i):
The Reluctance of a body at rest to move or a moving body to stop is called Inertia.
The correct term is $\boxed{\text{Inertia}}$.

Q1(a) (ii):
Three different directions of forces are:

• Vertical force (e.g., weight acting downwards, normal force acting upwards).
• Horizontal force (e.g., pushing or pulling force).
• Inclined force (e.g., a force applied at an angle to the horizontal or vertical).

Sketch:

\begin{tikzpicture}[scale=0.8] \draw[->, thick] (0,0) -- (0,2) node[above] {Vertical}; \draw[->, thick] (0,0) -- (2,0) node[right] {Horizontal}; \draw[->, thick] (0,0) -- (1.5,1.5) node[above right] {Inclined}; \fill (0,0) circle (1.5pt); \end{tikzpicture}

Q1(b):
Figure Q1b shows a loaded beam. Assuming it is a simply supported beam with a pin support at the left end and a roller support at the right end, its Free Body Diagram (FBD) would be:

\begin{tikzpicture}[scale=0.8] % Beam \draw[thick] (0,0) -- (10,0); % Pin support at A \draw[fill=black] (0,0) circle (2pt); \draw (0,-0.2) -- (-0.5,-0.7) -- (0.5,-0.7) -- cycle; \draw[dashed] (-0.7,-0.7) -- (0.7,-0.7); \node at (-0.5, 0.5) {$R_{Ay}$}; \draw[->, thick] (0,0) -- (0,0.8); \node at (0.5, -0.5) {$R_{Ax}$}; \draw[->, thick] (0,0) -- (0.8,0); % Roller support at B (assuming it's at the end of the beam for simplicity, as no specific location is given for supports in the original sketch) \draw[fill=black] (10,0) circle (2pt); \draw (9.5,-0.2) circle (0.2); \draw (10.5,-0.2) circle (0.2); \draw[dashed] (9.3,-0.7) -- (10.7,-0.7); \node at (10.5, 0.5) {$R_{By}$}; \draw[->, thick] (10,0) -- (10,0.8); % UDL w1 \draw[decorate,decoration={snake,amplitude=2pt,segment length=10pt}] (1,0.5) -- (4,0.5); \foreach \x in {1.2, 1.7, ..., 3.8} \draw[->] (\x,0.5) -- (\x,0); \node at (2.5, 0.8) {$w_1 (\text{kN/m})$}; % Point load Rw \draw[->, thick] (5.5,1) -- (5.5,0); \node at (5.5, 1.2) {$R_W (\text{kN})$}; % UDL w2 \draw[decorate,decoration={snake,amplitude=2pt,segment length=10pt}] (7,0.5) -- (9,0.5); \foreach \x in {7.2, 7.7, ..., 8.8} \draw[->] (\x,0.5) -- (\x,0); \node at (8, 0.8) {$w_2 (\text{kN/m})$}; % Lengths (approximate based on original sketch) \draw[<->] (0,-1) -- (4,-1) node[midway, below] {$L_1$}; \draw[<->] (4,-1) -- (5.5,-1) node[midway, below] {$L_2$}; \draw[<->] (5.5,-1) -- (7,-1) node[midway, below] {$L_3$}; \draw[<->] (7,-1) -- (10,-1) node[midway, below] {$L_4$}; \end{tikzpicture}

Q2(a):
Momentum is a Vector Quantity with a S.I Unit of kNm.
This statement is FALSE.

• Momentum is indeed a Vector Quantity.
• However, the S.I. unit of momentum is kilogram-meter per second ($\text{kg} \cdot \text{m/s}$) or Newton-second ($\text{N} \cdot \text{s}$).
• $\text{kNm}$ (kiloNewton-meter) is the unit for moment (torque) or energy (kiloJoule).

Q2(b):
Figure Q2b shows a simply supported beam. We need to determine the reaction forces.
Let $R_A$ be the vertical reaction at the pin support (left end) and $R_B$ be the vertical reaction at the roller support (at $8 \, \text{m}$ from the left end).

First, convert the uniformly distributed load (UDL) into an equivalent point load:
The UDL is $40 \, \text{kN/m}$ over $4 \, \text{m}$.
Equivalent point load $F_{UDL} = 40 \, \text{kN/m} \times 4 \, \text{m} = 160 \, \text{kN}$.
This load acts at the centroid of the UDL, which is $4 \, \text{m} / 2 = 2 \, \text{m}$ from the left support.

Now, let's apply the equilibrium equations: