Pushes and Pulls

1. Pushes and Pulls for IJSO training course

2. Content • What are forces? • Measurement of a force • Daily life examples of forces • Useful mathematics: Vectors • Newton’s laws of motion • Free body diagram • Mass, weight and gravity • Density vs. mass • Turning effect of a force

3. 1. What are forces? • Force, simply put, is a push or pull that an object exerts on another. • We cannot see the force itself but we can observe what it can do: • It can produce a change in the motion of a body. The body may change in speed or direction. • It can change the shape of an object. A force is the cause of velocity change or deformation.

4. 2. Measurement of a force • Force is measured in units called Newton (N). We can measure a force using a spring balance (彈簧秤). The SI unit of force: N (Newton) (Wikimedia commons)

5. Many materials including springs extend evenly when stretched by forces, provided that the force is not too large. This is known as Hooke’s law (虎克定律). • A spring balance uses the extension of a spring to measure force. The extension is proportional to the force acting on it as shown below.

6. 3. Daily examples of force Weight • The weight of an object is the gravitational force acting on it. Weight

7. Normal force • A book put on a table does not fall because its weight is balanced by another force, the normal force, from the table. • Normal: perpendicular to the table surface. normal force normal force force by the hand weight

8. Tension • Tension (張力) in a stretched string tends to shorten it back to the original length. • Once the string breaks or loosens, the tension disappears immediately. • Since tension acts inward to shorten the string, we usually draw two “face-to-face” arrows to represent it. Draw “face-to-face” arrows to represent tension

9. Example These two forces counterbalance each other (suppose the weight of the hook is negligible). tension 10 N “face to face” arrows representing tension tension 10 N The tension balances the weight, therefore the mass does not fall down. 1-kg mass weight 10 N

10. Friction • Friction(摩擦力) arises whenever an object slides or tends to slide over another object. • It always acts in a direction opposite to the motion. • Cause: No surface is perfectly smooth. When two surfaces are in contact, the tiny bumps catch each other. motion friction Friction drags motion.

11. Friction can be useful • We are not able to walk on a road without friction, which pushes us forward. • In rock-climbing, people need to wear shoes with studs. The studs can be firmly pressed against rock to increase the friction so that the climber will not slide easily. backward push of foot on road forward push of road on foot

12. The tread patterns on tyres also prevents the car from slipping on slippery roads. Moreover, road surfaces are rough so as to prevents slipping of tyres. Tread pattern on a car tyre (Wikimedia commons) Tread pattern on a mountain bicycle tyre (Wikimedia commons)

13. Disadvantages of friction • There are some disadvantages of friction. For example, in the movable parts of machines, energy is wasted as sound and heat to overcome friction. Friction will also cause the wear in gears. • Friction can be reduced by the following ways. • bearings (Wikimedia commons)

14. The streamlined shape cuts down the air-friction on the racing car. • using lubricating oil • using air cushion • streamlining 1. Propellers2. Air3. Fan4. Flexible skirt BHC SR-N4 The world's largest car and passenger carrying hovercraft (All pictures are from Wikimedia commons)

15. 4. Useful mathematics: Vectors • A scalar(標量) is a quantity that can be completely described by a magnitude (size). • Examples: distance, speed, mass, time, volume, temperature, charge, density, energy. • It is not sensible to talk about the direction of a scalar: the temperature is 30oC to the east(?). • A vector (向量) is a quantity that needs both magnitude and direction to describe it. • Examples: displacement, velocity, acceleration, force. A vector has a direction.

16. Example: displacement • A mouse moves 4 cm northward and then 3 cm eastward. • What is the distance travelled? • Answer = 4 cm + 3 cm = 7 cm • What is the displacement of the mouse? • Answer = 5 cm towards N36.9oE 3 cm 4 cm 5 cm How to find the angle?

17. Example: velocity • A bird is flying 4 m/s northward. There suddenly appears a wind of 3 m/s blowing towards the east. • What is the velocity of the bird? • Answer = 5 m/stowards N36.9oE • What is the speed of the bird? • Answer = 5 m/s • Note 1: No need to specify the direction. • Note 2: the answer is not simply= 3 m/s + 4 m/s = 7 m/s 3 m/s 4 m/s 5 m/s

18. Example: force • You push a cart with 4 N towards north. Your friend helps but he pushes it with 3 N towards the east. • What is the resultant force? • Answer = 5 Ntowards N36.9oE • What is the magnitude of the force? • Answer = 5 N • Note: A magnitude does not have a direction. 3 N 4 N 5 N A magnitude does not have a direction.

19. c b a b c a Addition and resolution F F • Two usual ways to denote a vector • Boldface • Adding an arrow • Vectors can be added by using the tip-to-tail or the parallelogram method. • If vectors a and b add up to become c, we can write c = a + b. Tip-to-tail method Parallelogram method

20. Two vectors can add up to form a single vector, a vector can also be resolved into two vectors. • In physics, we usually resolved a vector into two perpendicular components. • Below, a force F is resolved into two components, Fx and Fy.

21. 5. Newton’s laws of motion • Isaac Newton developed three laws of motion, which give accurate description on the motion of cars, aircraft, planet, etc. • The laws are important but simple. They are just the answers to three simple questions. • Consider a cue and a ball.

22. Newton’s 3 laws of motion answer 3 questions: • If the cue does not hit the ball, what will happen to the ball? • Newton’s first law • If the cue hits the ball, what will happen to the ball? • Newton’s second law • If the cue hits the ball, what will happen to the cue? • Newton’s third law

23. The first law • Also called “The law of inertia” (慣性定律) • A body continues in a state of rest or uniform motion in a straight line unless acted upon by some net force. • Galileo discovered this. • If the cue does not hit the ball, the ball will remain at rest.

24. The second law • The acceleration of an object is directly proportional to, and in the same direction as, the unbalanced force acting on it, and inversely proportional to the mass of the object. • In the form of equation, the second law can be written as F = ma • F is the acting force • m is the mass of the object • a is the acceleration (a vector) of the object • If the cue hits the ball, the ball will accelerate. Second law: F= ma

25. But .. what is acceleration? N B (1 km, 3 km) • Consider an object moving from A to B in 2 hours with a uniform velocity. What is the velocity? A (3 km, 1 km) Final displacement from O = OB O E Initial displacement from O = OA Change in displacement = OB – OA = AB Change in displacement AB = Velocity = Time required 2 hours

26. AB = (Note: This AB does not have an arrow. It indicates a length, which is a scalar.) N B (1 km, 3 km) Speed = AB / 7200 s = 0.39 m/s (Note: speed is also a scalar.) A (3 km, 1 km) O E Velocity = 0.39 m/s towards NW. Change in displacement Velocity = Time required

27. N v2 • Consider a bird. At time t = 0 s, it was moving 5 m/s towards SE. Its velocity gradually changed such that at t = 2 s, its velocity became 5 m/s towards NE. • Calculate the acceleration. vc = v2 - v1 E v1 Change in velocity = vc Change in velocity vc = Acceleration = Time required 2 s

28. N vc = (Note: This vc does not have an arrow. It indicates a magnitude.) v2 vc = v2 - v1 Magnitdue of acceleration = vc / 2 s = 3.54 m/s2 E v1 Acceleration = 3.54 m/s2 towards N. Change in velocity Acceleration = Time required

29. Equations of motion in 1D • In the 1D, there are only two directions, left and right, up and down, back and forth, etc. • For these simple cases, once we have chosen a positive direction, we can use + and - signs to indicate direction. We can also use a symbol without boldface to denote a vector. • Example: If we choose downward positive, the velocity v = -5 m/s describes an upward motion of speed 5 m/s.

30. Uniform acceleration • Let • t = the time for which the body accelerates • a = acceleration (which is assumed constant) • u = the velocity at time t = 0, the initial velocity • v = the velocity after time t, the final velocity • s = the displacement travelled in time t • We can prove that

31. Velocity-time graph Displacement-time graph v s parabola slope = a u 0 0 t t

32. Back … to the second law: F = ma • Mass is a measure of the inertia, the tendency of an object to maintain its state of motion. The SI unit of mass is kg (kilogram). • 1 Newton (N) is defined as the net force that gives an acceleration of 1 m/s2 to a mass of 1 kg. • The same formula can be applied to the weight of a body of mass m such that W = mg. • W: the weight of the body. It is a force, in units of N. • g: gravitational acceleration = 9.8 m/s2 downward, irrespective of m. W= mg

33. Force of man accelerates the cart. The same force accelerates two carts half as much. Twice as much force produces acceleration twice as much.

34. The third law • For every action, there is an equal and opposite reaction. • When the cue hits the ball, the ball also “hits” the cue. Action: the man pushes on the wall. Reaction: the wall pushes on the man. Action: Earth pulls on the falling man. Reaction: The man pulls on Earth.

35. Example Normal force = mg (upward) • The block does not fall because its weight is balanced by a normal force from the table surface. • Are the weight and the normal force an action-and-reaction pair of force as described by Newton’s third law? • Answer: No! Weight = mg (downward)

36. Explanation • Action and reaction act on different bodies. They cannot cancel each other. • The “partner” of the weight is the gravitational attraction of the block on the Earth. Weight = mg (downward) Gravitational attraction of the block on the Earth = mg (upward)

37. Explanation Normal force = mg (upward) • The “partner” of the normal force acting on the block by the table surface is the force acting on the table by the block surface. • Both have the same magnitude mg. • But they do not cancel each other because they are acting on different bodies. The force acting on the table by the block = mg (downward)

38. 6. Free body diagram • To study the motion of a single object in a system of several bodies, one must isolate the object and draw a simple diagram to indicate all the external forces acting on it. This diagram is called a free body diagram. N Example For an object of mass m at rest on a table surface, there are two external forces acting on it: 1. Its weight W 2. Normal force from the table surface N. Obviously, W = -N, and W = N = mg. W

39. Worked Example 1 • Consider two blocks, A and B, on a smooth surface. • Find • (a) the pushing force on Block B by Block A. • (b) the acceleration of the blocks.  Block A 3 kg Block B 2 kg 10 N

40. Solution: Method 1 Take rightward positive. Let a be the acceleration of the blocks. Let f be the pushing force on Block B by Block A. Consider the free body diagram of Block A normal force from the table surface a 3 kg f (reaction force of the pushing force on Block B) 10 N weight

41. a 3 kg f 10 N Vertical direction: No motion. The weight and the normal force from the table balance each other. Horizontal direction: Applying Newton’s second law (F = ma), we have (with units neglected) (1) 10 - f = 3a

42. Then consider the free body diagram of Block B normal force from the table surface a 2 kg f weight We ignore the vertical direction because the forces are balanced. Consider the horizontal direction. Applying the second law again, we have (2) f = 2a

43. We now have 2 equations in 2 unknowns. (1) 10 - f = 3a (2) f = 2a Solving them, we have f = 4 N a = 2 m/s2 (a) The pushing force on Block B by Block A = 4 N towards the right. (b) The acceleration of the blocks =2 m/s2.

44. Solution: Method 2 • Method 1 is a long method, below is a shorter one. • The whole system is a mass of 5 kg. • We take rightward positive and define the same f and a as those in Method 1. • Applying the second law (F = ma), we have 10 = 5a, hence a = 2 m/s2. • Consider only Block B. The only force acting on it is f. Hence f = 2a = 4 N.

45. Worked Example 2 • Consider a pulley and two balls, A and B. For convenience, take g = 10 m/s2. • Find • (a) the acceleration of Ball A. • (b) the tension in the string. A: 4 kg B: 1 kg

46. Solution Take downward positive. Let tension = T and acceleration of Ball A = a. Consider the free body diagram of Ball A: T A: 4 kg a Weight = 4g We can apply F = ma and get (1) 4g - T = 4a

47. Consider the free body diagram of Ball B: T B: 1 kg a Weight = g We apply F = ma and get (2) g - T = -a Solving Equations (1) and (2), we get a = 6 m/s2 and T = 16 N. (a) The acceleration of Ball A = 6 m/s2 downward. (b) The tension in the string =16 N.

48. Worked Example 3 • Consider a block on an inclined plane. • Label all forces acting on the block and resolve them into components parallel and perpendicular to the plane.

49. Find the acceleration a of the block in terms of g, given that Solution Consider the motion perpendicular to the motion. The forces are balanced, therefore we have

50. Now, consider the motion parallel to the motion. Applying Newton’s second law F = ma, we have