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

Physics

ScanSolve AI Answer

Step-by-step solution

5 steps
  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=mgN = mg. Maximum static friction fmaxf_\text{max} is the spring balance reading when tray just starts moving.

    Example table (substitute your measurements):

    Number of blocksMass of blocks (g)Total mass (g)Normal force NN (g wt)Max static friction fmaxf_\text{max} (g wt)
    11565658
    230808010

    Plot fmaxf_\text{max} (y-axis) vs NN (x-axis): straight line through origin.

    fmax=μNf_\text{max} = \mu N

    Graph slope == coefficient of static friction μ\mu.

  2. Interpretation of graph (Q1)

    The graph is a straight line passing through the origin, showing fmax∝Nf_\text{max} \propto N. Slope μ≈0.12\mu \approx 0.12 (from example data).
    Normal force acts perpendicular to surface (vertically upward on tray base).

  3. Why does friction force increase?

    (Q2)
    As blocks added, total mass increases →N\to N increases →fmax=μN\to f_\text{max} = \mu N increases (direct proportion).

  4. Why blocks on tray and friction when pulling tray?

    (Q3)
    Blocks increase NN without changing contact surface. Pulling tray slowly measures fmaxf_\text{max} at limiting friction (tray just slips).

  5. Why mass of spring balance/tray in NN?

    (Q4/Q5)
    Total N=(mtray+mblocks+mspring)gN = (m_\text{tray} + m_\text{blocks} + m_\text{spring})g. Omit →\to underestimate NN, incorrect μ\mu.

    Final answers:

    μ≈0.12(dimensionless; from your slope)\mu \approx 0.12 \quad \text{(dimensionless; from your slope)}

    Graph: linear, fmax=μNf_\text{max} = \mu N, μ=\mu = slope.

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