Inclined planes

Motion in Gravity

Inclined Plane Overview

• The concept of motion in gravity includes understanding how objects behave when on an inclined plane, which is a flat surface tilted at an angle.

Frictionless Surface

• Vector Addition: The forces acting on an object can be resolved into components using vector addition.

• Component of Weight (W): Weight can be divided into two components when on an inclined plane:

  • Perpendicular to the slope: W_ot = mg imes ext{cos}( heta) where (m) is mass and ( heta) is the angle.

  • Parallel to the slope: $W_ ext{parallel} = mg imes ext{sin}( heta)$.

Friction on Surface

• When friction is considered, the dynamics become more complex. The forces must still be addressed using vector addition to find resultant forces.

• Components of Weight and Friction: Friction opposes the motion and its force can be calculated, typically expressed as:

  • Static friction when stationary can be negligible.

  • Kinetic (dynamic) friction is significant when the object is in motion.

Constant Velocity Down the Incline

• When an object moves down an inclined plane at constant velocity:

  • The net force (ΣE) acting on the object is zero: $ΣE = 0$.

  • The force of gravity is balanced by the force of friction and the normal force, allowing a constant speed.

• Forces Involved:

  • Weight (W)

  • Normal force (FN)

  • Applied force (F)

Basic Concept of Inclined Planes

• When analyzing inclined planes, key parameters include:

  • The angle of the slope (e.g., 30°): The steeper the angle, the greater the component of gravitational force acting down the plane.

• Calculating Forces:

  • The net force ( ext{net}) is found out from:

  • $F_ ext{net} = mg imes ext{sin}( heta)$

  • Where (g) is the acceleration due to gravity (approximately $9.81 ext{ m/s}^2$).

Estimating the Angle of Slope to Begin Sliding

• A problem can arise when determining how steep an angle needs to be before an object starts sliding:

  • Given a frictional force of (F_f = 29N) and the relation of weight to angle can be expressed and solved.

• Sliding Condition: The angle of incline (ϑ) at which sliding begins can be found using:

  • $F_f = mg imes ext{sin}(ϑ)$

Worked Example - Inclined Planes

Example: Heavier Skier

Problem Statement: A skier of mass 85.0 kg travels down an icy slope inclined at 20.0° with negligible friction and g = 9.80 m/s².

1. Calculating Weight:

   • $W = mg = 85.0 imes 9.80 = 833 ext{ N}$

2. Weight Components:

   • Perpendicular to slope: W_ot = 833 ext{N} imes ext{cos}(20.0°) = 783.88 ext{N}

   • Parallel to slope: $W_ ext{parallel} = 833 ext{N} imes ext{sin}(20.0°) = 284.5 ext{N}$

3. Normal Force Calculation:

   • Normal Force (FN) is equal to the perpendicular component of the weight because there's no vertical acceleration:

   • $F_N = 783.88 ext{ N}$

4. Acceleration Calculation:

   • Using Newton’s second law, $F = ma$,

   • The acceleration down the slope:

   • $a = rac{F_ ext{parallel}}{m} = rac{284.5 ext{N}}{85.0 ext{kg}} = 3.34 ext{ m/s}^2$

Experimental Calculation of Forces on an Object on an Inclined Plane

Setup:

• An object with a mass of 1.50 kg on a plane inclined at 25°.

• Friction Considerations:

  • Static friction negligible when at rest.

  • Dynamic friction when in motion is 10.0 N.

Calculations:

1. Object at Rest:

   • Calculate force on probe:

   • At rest: $F = mg imes ext{sin}(25°) = 1.50 imes 9.81 imes ext{sin}(25°)$

2. Object Moving at Constant Velocity (2 m/s):

   • The force registered is equal to the weight component parallel to incline, minus dynamic friction (10.0 N).

   • $F = mg imes ext{sin}(25°) + F_f$

3. Object Accelerating (2 m/s²):

   • Additional force caused by acceleration:$F = mg imes ext{sin}(25°) + F_f + ma$

Problem with a Block Sliding Down an Inclined Plane

Parameters: Block of mass 12.0 kg, inclined at 38.0° with 24.4 N of friction.

1. Force Analysis:

   • Draw a vector diagram showing:

     • Forces acting on the block (weight, normal, and friction forces).

2. Components:

   • Normal force and perpendicular weight component calculated as:

   • $F_N = mg imes ext{cos}(38.0)$

3. Acceleration Calculation:

   • $a = rac{( ext{Weight Parallel}) - F_f}{m}$

4. Pendulum Scenario:

   • A pendulum bob attached to a mass of 8 kg swings in a vertical circular arc.

   • Key aspects to analyze:

     • Speed at lowest position (7.67 m/s), tension in the string, and acceleration of the pendulum.

5. Period Calculation:

   • Link tension and period of rotation to angle:

   • $ext{cos}( heta) = rac{gT^2}{4 ext{π}^2L}$

   • Where (L) represents the length of the string.