2. A push or pull on an object caused by its interaction with another object. Measured in NEWTONS (N)

5. Occurs in a string when the ends are pulled If the string is light the force due to tension is the same throughout the string Occurs in a rod when the ends are pushed If the rod is light, the thrust is the same throughout occur when two surfaces are touching occurs when two surfaces are pressed against one another. A normal reaction always acts at right angles (normal to) the plane of contact occurs when two surfaces are sliding or tending to slide over one another. The friction always acts to try to prevent the motion

6. Example 1Example 2 A large box of mass m kg is being towed upThe box in example 1 is now allowed to a rough slope inclined at 30° to the horizontal, slide down the slope, controlled by the rope. using a rope at 20° to the slope. Draw the forces in this situation. Draw a diagram to show the forces acting on the box. Weight, mg Normal Reaction R Tension, T 20° Friction, F Weight, mg Normal Reaction R Tension, T 20° Friction, F

7. Example 3 The diagram shows four dog leads, OA, OB, OC and OD, each tied to the same post, O. The dogs all pull horizontally on the leads with the forces shown. Find the resultant force on the post. 20 – 30cos45 – 25cos30+ 15cos50 = -13.222N 30sin45 – 25sin30– 15sin50= -2.777N 13.222N 2.777N R

8. resultant force in any direction is zero Example 4 An object of mass 12kg is suspended by two light, inextensible strings, AB and BC, as shown. Find the tensions in the strings. T1T1 T2T2 12g T 1 sin27+ T 2 sin38 = 12g(1) T 1 cos27 = T 2 cos38 Sub into (1) T 2 = 115.6NT 1 = 102.3N

9. Example 5 The diagram shows a block of mass 4kg resting on a smooth plane inclined at 25° to the horizontal. The block is kept in place by a light string inclined at 15° to the plane. Find the tension in the string and the normal reaction of the plane on the block. T 4g R 25° Resolve parallel to plane Tcos15 = 4gsin25 T = 17.15N Resolve perpendicular to plane R + Tsin15 = 4gcos25 R = 4gcos25 – Tsin15 R = 31.09N

10. Example 6 An object of weight W is suspended by light strings AB and AC of lengths 4m and 3m respectively. Points B and C are on the same horizontal level and are 5m apart. A horizontal force, P is applied to A so that the tension in AB is twice that in AC. Find, in terms of W, the value of P and the tension in AC. T 2T W sinB = 3 / 5 cosB = 4 / 5 sinC = 4 / 5 cosC = 3 / 5 Resolve vertically 2TsinB+ TsinC= W Resolve horizontally 2TcosB = TcosC + P

11. Example 7 A smooth ring, C, of mass 3kg is threaded on a light string 64cm long. The ends of the string are fixed to points A and B, 48cm apart on the same horizontal level. A force P, is applied to the ring so that it rests vertically below B. Find the value of P and the tension in the string. 3g T T x 64-x x 2 = 48 2 + (64 – x) 2 x 2 = 2304 + 4096 – 128x + x 2 128x = 6400 x = 50 50cm14cm θ sinθ = 48 / 50 cosC = 14 / 50 Resolve vertically T + Tcosθ= 3g Resolve horizontally P = Tsinθ

12. Friction occurs when one surface slides over another surface which is in contact with it. Friction always acts in the direction opposite to any movement or tendency to move In any situation there is a maximum possible frictional force. If the applied force is greater than this maximum, movement will take place When movement is just about to happen, we have reached a stage called limiting equilibrium The friction force will increase to match the applied force up to this point when it reaches limiting friction

13. The limiting frictional force depends only on the type of surfaces and the normal reaction between them For a given pair of surfaces, the ratio between the limiting friction and the normal reaction is constant This ratio is called the coefficient of friction and is denoted by the symbol µ As the frictional force is always less than or equal to the limiting friction, we can say that F ≤ µR where F is the frictional force, R the normal reaction and µ the coefficient of friction, which is constant for a given pair of surfaces So from this, we can say that when a body is moving or on the point of moving F = µR

14. Example 8 A block of mass 4kg rests on a rough horizontal surface. The coefficient of friction between the block and the surface is 0.35. (a) A horizontal force P is applied to the(b) The applied force P is now inclined at block so that it is just on the point of 20° to the horizontal. The block is just moving. Find the value of P. on the point of moving. Find P. 4g R P μR Resolve vertically R = 4g Resolve horizontally P = μR P = 0.35 x 4g P = 13.72N (a)

15. Example 8 A block of mass 4kg rests on a rough horizontal surface. The coefficient of friction between the block and the surface is 0.35. (a) A horizontal force P is applied to the(b) The applied force P is now inclined at block so that it is just on the point of 20° to the horizontal. The block is just moving. Find the value of P. on the point of moving. Find P. 4g R P 20° μR Resolve vertically R + Psin20 = 4g R = 4g – Psin20 Resolve horizontally Pcos20 = μR Pcos20 = 0.35(4g – Psin20) Pcos20 = 1.4g – 0.35Psin20 P(cos20 + 0.35sin20) = 13.72 P = 13.0N (b)

16. Example 9 A block of mass 5kg rests on a rough plane inclined at 30° to the horizontal. A horizontal force of 70N is applied and the block is on the point of moving up the slope. Find the value of the coefficient of friction between the block and the plane. 30° 5g R 70N μRμR 30° Resolve perpendicular to plane R = 5gcos30 + 70cos60 60° Resolve parallel to plane 70sin60 = 5gsin30 + μR R = 77.43524479N

17. Example 10 A block of weight W rests on a rough plane inclined at an angle α to the horizontal. The value of α is increased until the block is on the point of slipping. Show that the coefficient of friction, µ = tanα. α μR W R α Resolve perpendicular to plane R = Wcosα Resolve parallel to plane μR = Wsinα (1) (2) (2) ÷ (1) μ = tanα