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FORCES What is a force? Intuitively, a force is like a push or a pull which produces or tends to produce motion Forces experienced in Daily Life • Weight • Normal reaction • Friction • Viscous force • Tension • Upthrust • Lift • Electrical force • Magnetic force Weight W • weight is not the same as mass; it is a force • it is the gravitational force exerted by the Earth • it passes through the centre of gravity of the body Normal reaction N • two bodies in contact with each other • perpendicular to the surface of contact Friction • friction is exerted two surfaces slide across one another • direction is along the surface of contact Cause of friction movement F • hollows and humps all over the surface • actual contact area only a fraction 1/10000 of total area • extreme high pressure at contact points causes welding of surfaces • forces are needed to overcome these adhesive forces when trying to slide over the surface Static and kinetic friction • It is harder to move a stationary object than to move the object while it is moving • Static friction is the friction exerted by the ground in order to prevent the object from moving • Kinetic friction is the friction exerted by the ground to oppose the motion of the object while it is moving Limiting friction • Static friction is not constant; it varies in magnitude • Suppose a force P is applied trying to move the object 2 N1 N F • • • • • P 1 N2 N If P is 1 N, F will also be 1 N to prevent object moving If P increased to 2 N, F also increased to 2 N But there is limit to how much F can increase to Maximum possible static friction is called limiting friction P must exceed limiting friction in order to move object Example 1 In Fig 1.1, an object was moving to the right on a rough surface. In Fig 1.2, an object rests in equilibrium on a rough slope. In both cases, draw the friction force acting on each object. Fig 1.1 friction Fig 1.2 Viscous force • When body moves in fluid, it experiences resistance • such resistance is known as viscous force • examples: air resistance and water resistance • viscous force depends on the speed of the body • the greater the speed, the greater the viscous force Terminal velocity release v F W F W V F W VT W gathering speed gathering more speed finally reaches constant terminal velocity Tension • Tension is exerted by a stretched rope, string or spring. • When a body is attached to a string, the tension in the taut string would tend to pull the body. Hooke’s Law F F (l - lo) => F e => F = ke where k is force constant (elastic constant, spring constant or stiffness F constant) Strain energy in a Deformed wire Assume that Hooke’s Law is obeyed. =>For a force-extension graph, it will be a straight line. In general, work done by a force F in extending a wire from x1 to x2 is the area under the force-extension graph. F e x =>Work done in extension or strain energy stored in wire, W = ½ Fe = ½ ke2 Example 2 A vertical wire suspended from one end is stretched by attaching weight of 20 N to the lower end. If the extension is 1 x 10-3 m, what is (a) the force constant; (b) the energy stored in the wire; (c) the gravitational potential energy loss by the weight in dropping a distance of 1 x 10-3 m? F 20 3 2 x10 4 e 10 Assuming Hooke’s law is obeyed, (a) F = ke k = F/e = 20/(1x10-3) Nm-1 = 2 x 104 Nm-1 (b) energy stored in wire, W = ½ Fe = ½ (20)(1x10 -3) = 1 x10-2 J (c)Gravitational potential energy lost by weight = mgh = 2 x 10-2 J By conservation of energy, P.E. lost = Energy stored in wire + heat dissipated when weight at end of wire comes to rest after vibrating. Upthrust upthrust • Upthrust is an upward push on a body when it is immersed in a fluid (gas or liquid) • Upthrust is exerted by the fluid • Upthrust is due to pressure difference of fluid at the top and bottom of immersed portion of the body Example 3 Consider an object partially immersed in a fluid of density . The area of the top surface of the object is A and the immersed depth is h. h h (a) What is the pressure difference across the immersed portion of the object? hg (b) Hence write down the expression for the upthrust acting on the object. hgA (c) What is the volume of fluid displaced by the h A object? (d) Hence write down the expression for the weight of fluid displaced. (e) Comment on your answers to (b) and (d). They are the same. hAg Example 3 shows that Upthrust = weight of fluid displaced This is actually the Achimedes’ Principle Archimedes’ Principle states that the upthrust on a body in a fluid is equal and opposite to the weight of the fluid displaced by the body. Lift What helps birds and aeroplanes maintain its flight? The answer is the upward lift force exerted on their wings when in motion Electric force Electric force is exerted between two electric charges + + like charges repel + - unlike charges attract Magnetic force Magnetic force is exerted between two magnetic materials or between electric currents N N like poles repel N S unlike poles attract Forces experienced in Daily Life • Weight • Normal reaction • Friction • Viscous force • Tension • Upthrust • Lift • Electrical force • Magnetic force Different forces normal reaction upthrust weight weight normal reaction weight Different forces lift tension weight weight Different forces How did this ‘forward force’ come about? normal reaction normal reaction speed air resistance forward force friction weight Different forces How did this ‘thrust’ come about? lift air resistance thrust weight Who exerts on who A force is always exerted by some body on some other body. What makes a car move? Friction exerted by ground on tires What makes a rocket fly? Gases expelled by rocket Test Yourself. Identify the forces normal reaction weight weight FGM weight Fundamental types of force When scientists examined all the forces, they found that many of them are similar in nature. Scientists have identified 4 fundamental types of force: • gravitational force • electromagnetic force • nuclear force • weak force All forces in our daily life can be classified into one of the fundamental types. In the following table, identify the nature of each force: Force Nature Weight (W) of an object gravitational Gravitational Attraction between two oppositely-charged bodies electromagnetic Electromagnetic Attraction by a magnet on a piece of iron electromagnetic Electromagnetic Tension (T) in a string pulling an object electromagnetic Electromagnetic Pushing a person with your hands electromagnetic Electromagnetic Normal reaction (N) by the table on an object resting on it electromagnetic Electromagnetic Friction (F) experienced by an object moving on a rough surface Electromagnetic electromagnetic 2 Addition of Vectors Parallelogram Rule A B A R B Triangle Rule A B B A R Finding resultant force The magnitude of resultant force can be found by • drawing vector diagram to scale • calculation (pythagoras theorem, cosine rule, etc) • resolution Example 4 Two forces are given below: 70º 5N 4N 30º Find the magnitude of the resultant force. Method 1 Drawing vector diagram to scale Scale used is 1 cm : 1 N 5N (5 cm) 4N (4 cm) R (5.8 cm) From the vector diagram, magnitude of resultant R is 5.8 N What is missing in the answer? Method 2: By calculation 5N 30 80 4N 70 x • Using Cosine rule: x2 = 52 + 42-2 (4) (5) cos800 =>x =5.84 N 0 sin • Using Sine rule: sin 80 42.4 0 5.836 4 Method 3 By resolving vectors 4 cos 70° 5 sin 30° 70º 5N 30º 4 sin 70° 5 cos 30° 4N Rx = 5 cos 30° + 4 cos 70° = 5.70 N Ry = 5 sin 30° - 4 sin 70° = -1.26 N Rx 5.70 N Ry 1.26 N Magnitude of resultant R is given by R2 = (5.70)2 + (1.26)2 R = 5.8 N R Example 5 Two horizontal forces act at a point to produce a resultant force of magnitude 40 N in the eastward direction. Given that one of the forces is in the northward direction and has a magnitude of 30 N, find the magnitude and direction of the second force. N F 30 N E 40 N Magnitude of second force F = 302 + 402 = 50 N Angle = tan-1 (40/30) = 53°, direction of F is 53° east of south (or bearing 127°) Centre of gravity and Free body diagram Centre of Gravity • The centre of gravity of a body is the single point at which the entire weight of the body can be considered to act. Free body diagram (Important) • is a diagram showing all the forces acting on a particular object • is an important tool for solving problems Example 6 An object A of weight w rests on top of another object B of weight W placed on the ground, as shown. Draw separate free body diagrams showing forces acting on (a) A only (b) B only, and (c) A and B together. Answer N1 N2 w N3 W N3 W+w N1 = normal reaction exerted by B on A N2 = normal reaction exerted by A on B N3 = normal reaction exerted by ground on B N1 is numerically equal to N2 (action / reaction pair) Common forces in free body diagrams Force Appli cable when weight W Object has a mass tension T Object is attached to a string normal reaction N friction F Direction of force vertically downwards through centre of gravity along the string pulling towards the centre of the string Object is in con tact with a surface normal to and away from the contact surface Object tries to move across a rough surface along the rough surface Exampl e Force exerted by surface (only) Total force R exerted by surface on moving object consists of two components - normal reaction N - frictional force F R is also known as the contact force N R Motion F A Non lecture Note Example An object of weight W, resting on a rough surface, is connected to a suspended object of weight w by a string over a smooth pulley. Draw and label the forces acting on each object. Normal reaction tension tension friction w W Turning effect of a force Consider a water wheel which is free to rotate about its centre. Water flowing to the right exerts force on lower blades. This force causes the wheel to rotate about its centre. We say that the force has a turning effect. Turning effect of a force is also known as its moment. Amount of moment depends on force and distance away. Moment of a force The moment of a force about an axis is defined as the product of the force and the perpendicular distance between the axis and the line of action of the forces. The moment of a force is also known as the torque. A Moments about A = F l (clockwise) l F Moment of a force Example: Not in lecture notes A l 300 Mtd 2 300 Mtd 1 F A A l F cos 300 F F sin 300 l 300 F Moments about A = F l sin 30 0 (clockwise) = 1/2 F l Example 7 Find the moments of the following forces about point A. 5m 4m A 40 20 N 3m 30 N 40 N 60 Moment of 30 N about A = 30 × 4 = 120 N m (anticlockwise) Moment of 40 N about A = 40 × 3 sin 60 = 104 N m (clockwise) Moment of 20 N about A = 20 × 5 cos 40 = 77 N m (clockwise) Torque of a couple The torque of a couple is equal to the product of one of its forces the perpendicular between Couple = pairand of equal and oppositedistance forces whose lines the lines of forces. of action action of dothe nottwo coincide F A x d F Taking moment about any arbitrary point, say A, total anticlockwise moment = F × (d+x) - F × x = Fd Example 8 Calculate the torque acting on the rod 2.0 m long in Figs 9.1 and 9.2. 10 N 10 N Fig. 9.1 Fig. 9.2 2.0 m 30º 2.0 cos 30º 10 N 10 N Fig. 9.2: 9.1: Torque = F d distance Perpendicular = 10 × 2.0 between = 20 N10 m N forces = 2.0 cos 30º Torque = 10 × 2.0 cos 30º = 17 N m 5 System in equilibrium A system is in equilibrium when there is no resultant force and no resultant torque. Second First condition: condition: Resultant Resultant force torque is zero is zero forces would form a closed triangle or polygon • total clockwise moment = total anticlockwise moment sum of components resolved in any direction is zero • if there are only 3 forces, would intersect at translational a common point • they system is said to be in equilibrium • system in rotational equilibrium is eitherisatsaid restto orbe moving with constant velocity • is atconstant rest or rotating with constant angular velocity has linear momentum • has constant angular momentum Example 9 A horizontal force F is exerted on the pendulum of weight W, causing the pendulum to be suspended at an angle to the vertical, as shown. Find F in terms of W and . T T W F F W From the vector diagram, tan = F / W F = W tan Example 10 A body of weight 200 N is suspended by two cords, A and B, as shown in the diagram. Find the tension in each cord. 60º cord A cord B TB TA W TB W 60º TA From the vector diagram, tan 60º = W / TA TA = W / tan 60º = 200 / tan 60º = 115 N sin 60º = W / TB TB = W / sin 60º = 200 / sin 60º = 231 N Example 11 A uniform rod is supported with the fulcrum exactly at the centre of the rod. Two masses were placed on the rod and the system is in equilibrium. Find m. 0.45 m 2.0 kg 2.0 × g 0.30 m N m W mg Taking moments about the fulcrum, clockwise moments = anticlockwise moments m g × 0.30 = 2.0 × g × 0.45 m = 3.0 kg Example 12 A uniform rod XY of weight 20 N is freely hinged to a wall at X. It is held horizontal by a string attached at Y at an angle of 20º to the rod, as shown. string X 20º Y Find (a) the tension in the string, (b) the magnitude of the force exerted by the hinge. Example 12 (continued) string R R X T 20º 20 N Y 70º T 29 N 20 N (a) Let R be of the rod (b) bethe thelength force exerted by and the hinge T be tension in theusing stringcosine rule, From thethe vector diagram, 2 - 2(20)(29) Taking about X, R2moments = 202 + 29 cos 70º anticlockwise R = 29 N moments = clockwise moments T sin 20º × = 20 × ( / 2) T = 29 N Example 13 A heavy uniform beam of length is supported by two vertical cords as shown. cord A TA (3/10) cord B TB (7/10) weight Taking momentstension about the centre in cord A of gravity, Find the ratio moments = anticlockwise moments clockwise tension in cord B TA × (2/10) = TB × (5/10) ratio TA / TB = 5/2 The End