Static & Kinetic Friction, Tension, Normal Force, Inclined Plane & Pulley System Problems - Physics

Angle of Incline

• The angle of the incline plays a crucial role in analyzing forces acting on objects.

Weight Force Components

• The weight force (Fg) can be broken into two components:

  • Fg parallel: Acts down the incline

  • Fg perpendicular: Acts perpendicular to the incline

Calculating Components

• These components can be calculated using trigonometric functions:

  $Fg_{\text{parallel}} = mg \times \text{sin}(\theta)$

  $Fg_{\text{perpendicular}} = mg \times \text{cos}(\theta)$

Friction on the Incline

• Frictional force (F_friction) that opposes motion down the incline:

  • Normal force (N) equals the perpendicular component of the weight:

  $N = Fg_{\text{perpendicular}} = mg \times \text{cos}(\theta)$

  • Frictional force can be expressed as:

  $F_{\text{friction}} = \text{μ}N = \text{μ}(mg \times \text{cos}(\theta))$

Force Balance on Incline

• When the block is at rest, the forces must balance:

  $mg \times \text{sin}(\theta) = \text{μ}(mg \times \text{cos}(\theta))$

Block Sliding Down the Incline

• For a block sliding down, the net force equation is set up:

  $F_{net} = mg \times \text{sin}(\theta) - \text{μ}(mg \times \text{cos}(\theta))$

  • Using Newton's second law:

  $ma = mg \times \text{sin}(\theta) - \text{μ}(mg \times \text{cos}(\theta))$

Example Calculation

• For a block with mass 10 kg, incline of 30 degrees, and coefficient of kinetic friction 0.2:

  1. Calculate weight force:

  $Fg = mg = 10 \times 9.8 = 98 \text{N}$ 2. Calculate components of the weight force:

  $Fg_{\text{parallel}} = 98 \times \text{sin}(30^\text{o}) = 49 \text{N}$

  $Fg_{\text{perpendicular}} = 98 \times \text{cos}(30^\text{o}) ightarrow 98 \times \frac{\text{√3}}{2} ightarrow 84.87 \text{N}$ 3. Calculate frictional force:

  $F_{\text{friction}} = 0.2 \times 84.87 \text{N} ightarrow 16.97 \text{N}$ 4. Find net force:

  $F_{net} = 49 \text{N} - 16.97 \text{N} ightarrow 32.03 \text{N}$ 5. Calculate acceleration:

  $ma = F_{net} ightarrow a = \frac{32.03 \text{N}}{10 \text{kg}} ightarrow 3.20 \text{m/s}²$

Equations to Remember

1. Normal force on an incline:

   $N = mg \times \text{cos}(\theta)$

2. Gravitational force down the incline:

   $Fg = mg \times \text{sin}(\theta)$

Frictionless Incline Example

• For a frictionless incline of 30 degrees and a mass of 20 kg:

  • Calculate the acceleration:

    • Only gravitational force accelerates the block down the incline:

      $a = g \times \text{sin}(\theta) = 10 \times 0.5 = 5 \text{m/s}²$

Final Velocity on an Incline

• If the length of incline = 100 m:

  • Initial speed is zero, find final speed using:

    $V_{final}^{2} = V_{initial}^{2} + 2ad$

    $V_{final} = \text{√}(1000) ightarrow 31.6 \text{m/s}$

Stopping Distance on Horizontal Surface

• Assuming initial speed of 31.6 m/s with coefficient of kinetic friction of 0.20:

  1. Calculate the stopping distance:

  $V_{final}^{2} = V_{initial}^{2} + 2ad$

  • Where:

    $a = -2 \text{m/s}²$ (deceleration due to friction)

  • Solve for D:

    $D = \frac{31.6^{2}}{4} = 249.6 \text{m}$

Determine Sliding vs. Rest on Incline with Friction

1. Calculate weight force (Fg) and frictional forces (FS and FK).

2. Compare states:

   • If FG > FS, block will slide down.

   • If FS >= FG, block remains at rest.

Example with Static and Kinetic Friction

• Consider a block with mass 10 kg on a 30-degree incline, μS = 30, μK = 10.

1. Calculate FG and FS to determine which force is larger and the block's state of motion.

Summary of Tension and Net Forces in Pulley Problems

• In pulley systems, calculate:

  • Net force:

    • Taking the weight forces and any frictional forces into account.

  • Tension across ropes based on the net forces acting on each block.

  • For two different mass blocks, determine the resultant forces to find the acceleration and tension values.