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Solving Vector Problems (DP IB Physics: SL)

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Katie M

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Solving Vector Problems

  • In physics, vectors appear in many different topic areas
    • Specifically, vectors are often combined and resolved to solve problems when considering motion, forces, and momentum

Forces on an Inclined Plane

  • Objects on an inclined plane is a common scenario in which vectors need to be resolved
    • An inclined plane, or a slope, is a flat surface tilted at an angle, θ

  • Instead of thinking of the component of the forces as horizontal and vertical, it is easier to think of them as parallel or perpendicular to the slope
  • The weight of the object is vertically downwards and the normal (or reaction) force, R is always vertically up from the object
  • The weight W is a vector and can be split into the following components:
    • W cos (θ) perpendicular to the slope
    • W sin (θ) parallel to the slope

  • If there is no friction, the force W sin (θ) causes the object to move down the slope
  • If the object is not moving perpendicular to the slope, the normal force will be R = W cos (θ)

Vectors On an Inclined Plane, downloadable AS & A Level Physics revision notes

The weight vector of an object on an inclined plane can be split into its components parallel and perpendicular to the slope

Worked example

A helicopter provides a lift of 250 kN when the blades are tilted at 15º from the vertical.Resolving Forces Worked Example, downloadable AS & A Level Physics revision notesCalculate the horizontal and vertical components of the lift force.

Step 1: Draw a vector triangle of the resolved forces

4.1.2 Resolving Forces Worked Example Answer

Step 2: Calculate the vertical component of the lift force

Vertical = 250 × cos(15) = 242 kN

Step 3: Calculate the horizontal component of the lift force

Horizontal = 250 × sin(15) = 64.7 kN

Worked example

A person is exploring a new part of town, from their starting point they walk 100 m in the direction 30.0º South of West. They then walk 200 m in the direction 40.0º degrees South of East and finally they walk 150 m directly East.Vector-Problem-Worked-Example, downloadable IB Physics revision notesCalculate the magnitude of their displacement from their original position.

In order to calculate the answer, the vectors of displacement must be resolved into their x-components and y-components and then combined. In this case, this effectively means the x-direction is East-West and the y-direction is North-South

Step 1: Consider positive and negative directions for reference

    • Since East is likely to be larger consider it the positive displacement and West as negative
    • Similarly, consider South as positive and North as negative

Step 2: Resolve the first displacement (100 m magnitude) into its components

Vector-Problem-Worked-Example-Step2, downloadable IB Physics revision notes
    • The horizontal component can be resolved from:

cos(30°) × 100 = 86.6 m

    • This is in a Western (negative horizontal) direction

    • The vertical component can be resolved from:

sin(30°) × 100 = 50.0 m

    • This is in a Southern (positive vertical) direction

Step 3: Resolve the second displacement (200 m magnitude) into its components

Vector-Problem-Worked-Example-Step3, downloadable IB Physics revision notes

    • The horizontal component can be resolved from:

cos(40°) × 200 = 153 m

    • This is in an Eastern (positive horizontal) direction

    • The vertical component can be resolved from:

sin(40°) × 200 = 129 m

    • This is in a Southern (positive vertical) direction

Step 4: Resolve the third displacement (150 m magnitude) into its components

    • The horizontal component is already resolved into

150 m

    • This is in an Eastern (positive horizontal) direction

    • There is no vertical component for this vector

Step 5: Combine the horizontal (East-West) components

Vector-Problem-Worked-Example-Step5, downloadable IB Physics revision notes

153 + 150 - 86.6 = 166 m

    • This is in an Eastern (positive horizontal) direction

Step 6: Combine the vertical (North-South) components

Vector-Problem-Worked-Example-Step6, downloadable IB Physics revision notes

50.0 + 129 = 170 m

    • This is in a Southern (positive vertical) direction

Step 7: Using Pythagoras theorem to find the resultant hypotenuse vector

√(1662 + 1792) = 244 m

Equilibrium

  • Coplanar forces can be represented by vector triangles
  • Forces are in equilibrium if an object is either
    • At rest
    • Moving at constant velocity

  • In equilibrium, coplanar forces are represented by closed vector triangles
    • The vectors, when joined together, form a closed path

  • The most common forces on objects are
    • Weight
    • Normal reaction force
    • Tension (from cords and strings)
    • Friction

  • The forces on a body in equilibrium are demonstrated below:

Vector triangle in equilibrium, downloadable AS & A Level Physics revision notes

Three forces on an object in equilibrium form a closed vector triangle

Worked example

A weight hangs in equilibrium from a cable at point X. The tensions in the cables are T1 and T2 as shown.WE - Forces in equilibrium question image 1, downloadable AS & A Level Physics revision notesWhich diagram correctly represents the forces acting at point X?WE - Forces in equilibrium question image 2, downloadable AS & A Level Physics revision notes

Equilibrium Worked Example (3), downloadable AS & A Level Physics revision notes

Examiner Tip

If you're unsure as to which component of the force is cos θ or sin θ, just remember that the cos θ is always the adjacent side of the right-angled triangle AKA, making a 'cos sandwich'Resolving Vectors Exam Tip, downloadable AS & A Level Physics revision notes

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Katie M

Author: Katie M

Expertise: Physics

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.