Key Areas covered

1. Key Areas covered • Resolving the weight of an object on a slope into a component acting down the slope and a component acting normal to the slope. • Systems of balanced forces with forces acting in two dimensions.

2. What we will do today: • Investigate forces involved when an object slides down a slope. • Carry out calculations on the above.

3. Force Acting Down a Slope

4. If an object is placed on a slope, then its weight acts vertically DOWNWARDS. A certain component of this will act down the slope. The weight can be split into two components at right angles to each other. We can say that the component of weight acting down the slope is: F = mg sin θ

5. Reaction Force Component of weight down slope = mg sin θ (commonly described as ‘parallel to the slope’) Component of weight = mg cos θ perpendicular to slope θ mg sin θ mg mg cos θ θ

6. A block of wood of mass 2 kg is placed on a slope which makes an angle of 30º with the horizontal. If a friction force of 4 N acts on the block, what is the unbalanced force down the slope? Solution Weight down slope = mg sin θ = 2 x 9.8 x sin 30º = 9.8 N Fun = Wt. down slope – Friction = 9.8 – 4 = 5.8 N Example

7. Example 2: 2006 Qu: 21 Covered later

8. 2007 Qu: 22

9. 2008 Qu: 22a

10. Questions • Activity sheets: • Page 33 - 35 • Resolution of forces • Qu’s: 6 – 9 • You should now be able to answer all questions in class jotter