FORCES What is a force Intuitively a force
- Slides: 63
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 N 1 N F • • • P 1 N 2 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 Fig 1. 2 fr friction n io t c i
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 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 x 1 to x 2 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 ke 2
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?
Assuming Hooke’s law is obeyed, (a) F = ke k = F/e = 20/(1 x 10 -3) Nm-1 = 2 x 104 Nm-1 (b) energy stored in wire, W = ½ Fe = ½ (20)(1 x 10 -3) = 1 x 10 -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? h g (b) Hence write down the expression for the upthrust acting on the object. h g. A (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. h. A g
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 is exerted between two electric charges + + like charges repel + - unlike charges attract
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 normal reaction weight
Different forces lift tension weight
Different forces How did this ‘forward force’ come about? 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 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: gravitational 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 5 N 4 N 30º Find the magnitude of the resultant force.
Method 1 Drawing vector diagram to scale Scale used is 1 cm : 1 N 5 N (5 cm) 4 N (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 5 N 30 80 4 N 70 x • Using Cosine rule: x 2 = 52 + 42 -2 (4) (5) cos 800 =>x =5. 84 N • Using Sine rule:
Method 3 By resolving vectors 4 cos 70° 5 sin 30° 70º 5 N 30º 4 sin 70° 5 cos 30° 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 R 2 = (5. 70)2 + (1. 26)2 R = 5. 8 N 4 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 30 N F 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 N 1 N 2 w N 3 W+w N 1 = normal reaction exerted by B on A N 2 = normal reaction exerted by A on B N 3 = normal reaction exerted by ground on B N 1 is numerically equal to N 2 (action / reaction pair)
Common forces in free body diagrams
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 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 l A 300 Mtd 1 F 300 F F sin 300 l A Mtd 2 A l F cos 300 0 0 3 n i s l 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. 5 m 4 m A 40 20 N 3 m 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 of dothe nottwo coincide F A x F d 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 Fig. 9. 1 Fig. 9. 2 2. 0 m 30º 2. 0 cos 30º 10 N Fig. 9. 2: 9. 1: Torque = F d distance Perpendicular = 10 × 2. 0 between = 20 N 10 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: Resultant force torque 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 W T 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 = W / tan 60º = 200 / tan 60º = 115 N sin 60º = W / 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 X R 20º T Y 70º T 29 N 20 N (a) Let R (b) �be bethe thelength force exerted of the rod byand the hinge T be the vector tensiondiagram, in the string From using cosine rule, 2 - 2(20)(29) Taking about X, R 2 moments = 202 + 29 cos 70º anticlockwise moments = clockwise moments R = 29 N 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
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