Define normal force (N), including the direction of the normal force.

Step 1: Plotting the graph
Record data in the table, assuming typical values (e.g., mass per block = 15 g, tray + spring balance mass ≈ 50 g for illustration). Total mass determines normal force $N = mg$. Maximum static friction $f_\text{max}$ is the spring balance reading when tray just starts moving.

Example table (substitute your measurements):

| Number of blocks | Mass of blocks (g) | Total mass (g) | Normal force $N$ (g wt) | Max static friction $f_\text{max}$ (g wt) |
|------------------|--------------------|----------------|--------------------------|-------------------------------------------|
| 1 | 15 | 65 | 65 | 8 |
| 2 | 30 | 80 | 80 | 10 |

Plot $f_\text{max}$ (y-axis) vs $N$ (x-axis): straight line through origin.
$f_max = \mu N$
Graph slope $=$ coefficient of static friction $\mu$.

Step 2: Interpretation of graph (Q1)
The graph is a straight line passing through the origin, showing $f_max \propto N$. Slope $\mu \approx 0.12$ (from example data).
Normal force acts perpendicular to surface (vertically upward on tray base).

Step 3: Why does friction force increase? (Q2)
As blocks added, total mass increases $\to N$ increases $\to f_max = \mu N$ increases (direct proportion).

Step 4: Why blocks on tray and friction when pulling tray? (Q3)
Blocks increase $N$ without changing contact surface. Pulling tray slowly measures $f_\text{max}$ at limiting friction (tray just slips).

Step 5: Why mass of spring balance/tray in $N$? (Q4/Q5)
Total $N = (m_tray + m_blocks + m_spring)g$. Omit $\to$ underestimate $N$, incorrect $\mu$.

Final answers:
$\mu \approx 0.12 \quad \text{(dimensionless; from your slope)}$
Graph: linear, $f_max = \mu N$, $\mu =$ slope.