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:

  $Fg_{ ext{parallel}} = mg imes ext{sin}( heta)$

  $Fg_{ ext{perpendicular}} = mg imes ext{cos}( heta)$

Friction on the Incline

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

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

  $N = Fg_{ ext{perpendicular}} = mg imes ext{cos}( heta)$

  • Frictional force can be expressed as:

  $F_{ ext{friction}} = ext{μ}N = ext{μ}(mg imes ext{cos}( heta))$

Force Balance on Incline

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

  $mg imes ext{sin}( heta) = ext{μ}(mg imes ext{cos}( heta))$

Block Sliding Down the Incline

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

  $F_{net} = mg imes ext{sin}( heta) - ext{μ}(mg imes ext{cos}( heta))$

  • Using Newton's second law:

  $ma = mg imes ext{sin}( heta) - ext{μ}(mg imes ext{cos}( heta))$

Example Calculation

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

  1. Calculate weight force:

  $Fg = mg = 10 imes 9.8 = 98 ext{N}$ 2. Calculate components of the weight force:

  $Fg_{ ext{parallel}} = 98 imes ext{sin}(30^ ext{o}) = 49 ext{N}$

  Fg_{ ext{perpendicular}} = 98 imes ext{cos}(30^ ext{o}) ightarrow 98 imes rac{ ext{√3}}{2} ightarrow 84.87 ext{N} 3. Calculate frictional force:

  $F_{ ext{friction}} = 0.2 imes 84.87 ext{N} ightarrow 16.97 ext{N}$ 4. Find net force:

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

  ma = F_{net} ightarrow a = rac{32.03 ext{N}}{10 ext{kg}} ightarrow 3.20 ext{m/s}²

Equations to Remember

1. Normal force on an incline:

   $N = mg imes ext{cos}( heta)$

2. Gravitational force down the incline:

   $Fg = mg imes ext{sin}( heta)$

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:

      $a = g imes ext{sin}( heta) = 10 imes 0.5 = 5 ext{m/s}²$

Final Velocity on an Incline

• If the length of incline = 100 m:

  • Initial speed is zero, find final speed using:

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

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

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:

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

  • Solve for D:

    D = rac{31.6^{2}}{4} = 249.6 ext{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.