2. Forces - TSFX

COLIN HOPKOINS Physics Unit 3 2015 Forces notes 1 of 30

2. Forces Study Design

Apply Newton’s three laws of motion in situations where two or more coplanar forces act along a straight line and in two dimensions;

Types of forces Forces can be divided into two major categories, field forces and contact forces

The relationship between a force and the acceleration it causes was first understood by Isaac Newton (1672 – 1727). Newton summarised all motion by three laws: Newtons First Law

An important consequence of this law was the realisation that an object can be in motion without a force being constantly applied to it. When you throw a ball, you exert a force to accelerate the ball, but once it is moving, no force is necessary to keep it moving. Prior to this realisation it was believed that a constant

force was necessary, and that this force was supplied by that the air pinching in behind the ball. This model, first conceived by Aristotle, proved tenacious, and students still fall into the trap of using it. Newton’s first law is commonly tested on the exam. This is achieved by the inclusion of statements such as “An object is moving with a constant velocity” within questions. Whenever you see the key words constant velocity in a question, you should highlight them. The realisation that the object is travelling at a constant velocity, and hence that the net force on the object is zero, will be essential for solving the problem. Newton’s Second Law

In words, Newton’s Second Law states that a force on an object causes the object to accelerate (change its velocity). The amount of acceleration that occurs depends on the size of the force and the mass of the object. Large forces cause large accelerations. Objects with large mass accelerate less when they experience the same force as a small mass. The acceleration of the object is in the same direction as the net force on the object.

Newtons 1st law of motion If an object has zero net force acting on it, it will remain at rest, or continue moving with an unchanged velocity.

Newtons 2nd law of motion This law relates to the sum total of the forces on the body ( ΣF ) the body's mass (m) and the acceleration produced (a)

ΣF = ma.

a = F

m

Note ΣF must have the same direction as 'a'.

Forces that act at a distance are called FIELD FORCES, (gravitational, electrical or magnetic)

Forces created by travelling bodies are called CONTACT FORCES.


Newton’s Third Law This law is the most commonly misunderstood. You need to appreciate that these action/reaction forces act on DIFFERENT OBJECTS and so you do not add them to find a resultant force. For example, consider a book resting on a table top as shown in the diagram below. There are two forces acting on the book: Gravity is pulling the book

downward and the tabletop is pushing the book upwards. These forces are the same size, and are in opposite directions but THEY ARE NOT a Newton’s thirds law pair, because they both act on the same object.

The best way of avoiding making a mistake using Newton’s third law is to use the following statement.

FA on B = - FB on A

In the example of the book on the table the force Table on Book is a Newton third law pair with the force Book on Table. Notice the first force is on the book and the second force is on the table. They do not act on the same object. Similarly the weight force, which is the gravitational attraction of the earth on the book, is a Newton third law pair with the gravitational force of the book on the earth. The gravitational effect of the book on the earth is not apparent because the earth is so massive that no acceleration is noticeable. Drawing Force Diagrams You will often be asked to draw diagrams illustrating forces. There are several considerations when drawing force diagrams:

The arrows that represent the forces should point in the direction of applied force. The length of the arrow represents the strength of the force, so some effort should be made to draw the arrows to scale.

An arrow representing a field force should begin at the centre of the object. An arrow representing a contact force should begin at the point on contact where the force

is applied. All forces should be labelled.

Some sample force diagrams of common situations are drawn below. Mass on a string Mass in free flight m

Newtons 3rd law of motion For every action force acting on one object, there is an equal but opposite reaction force acting on the other object.

Velocity v = 0, so T = mg Velocity v = constant upwards, so T = mg Velocity v = constant downwards, so T = mg Accelerating Upwards, T - mg = ma. Acceleration Downwards, mg - T = ma.

F = mg = ma

N

W

T

mg mg


Mass pulled along a plane Smooth (No Friction) Rough (Friction) N a N a T Fr T mg mg T = ma, N + mg = 0 T - F = ma, N + mg = 0 Bodies with parallel forces acting Bodies with non-parallel forces acting

The vectors need to be resolved in order to solve for the acceleration.

Inclined planes Another example of forces acting at angles to each other is an object on an incline plane. There are only three different types of examples of a body on an incline plane without a driving force. A body accelerating The component of the weight force acting down the plane is larger then the frictional forces. (This is also true if there are no frictional forces). For these situations you would take down the plane to be positive, the reason for this is that the acceleration is down the plane.

m m

m m m

a a a

F1 F1 F1 F2

F2 F2

F1 + F2 = ma F1 + F2 = ma F2 – F1 = ma

Forces perpendicular to the plane Fnet = mgcos - N = 0

Forces parallel to the plane

Fnet = mgsin - F = ma Thus the acceleration is down the plane. If there is not friction then the acceleration is gsin

mg

N

F

m m m

F1

F2

F2

F1 F1

F2

F1 + F2 = ma F1 + F2 = ma F1 + F2 = ma

a a a


A body travelling at constant speed This can be the when an object is not changing its speed whilst travelling down an incline or when the object is at rest on the incline plane.

A body decelerating For these situations you would choose up the plane to be positive, this is because this is the direction of acceleration.

Forces perpendicular to the plane

Fnet = mgcos - N = 0 Forces parallel to the plane

Fnet = mgsin - F = 0 Thus the acceleration is zero

mg

N

F

Forces perpendicular to the plane

Fnet = mgcos - N = 0 Forces parallel to the plane

Fnet = F - mgsin = ma Thus the acceleration is up the plane.

mg

F

N


Example 1 1973 Question 8 (1 mark) A car has a maximum acceleration of 3.0 ms-2. What would its maximum acceleration be while towing a car twice its own mass? A train accelerates from rest at one station and travels to another station. The velocity-time for the train is given below. The mass of the train is 5.0 x 105 kg. Assume that a constant frictional resistance of 1.5 x 104 N acts on the train throughout its journey.

Example 2 1976 Question 1 (1 mark) Calculate the distance between the two stations. Example 3 1976 Question 2 (1 mark) Calculate the net force acting on the train during the first 300 seconds. Example 4 1976 Question 3 (1 mark) Calculate the force exerted by the engine on the train during the first 300 seconds. Example 5 1976 Question 4 (1 mark) Calculate the power in kilowatt at which the engine works during the period of constant velocity.


Example 6 1976 Question 9 (1 mark)

A cyclist is accelerating downhill with his brakes off. Which arrow (A – F) best represents the direction of the force of the road on the bicycle? The cyclist now brakes so that his speed is constant. Example 7 1976 Question 10 (1 mark) The mass of the cyclist is m; the road makes an angle θ with the horizontal. The magnitude of the vector sum of all forces acting on the cyclist is A. mg B. mg cos θ C. mg sin θ D. zero Example 8 1976 Question 11 (1 mark) Which arrow (A – F) now best represents the direction of the force of the road on the bicycle? A car of mass 800 kg is towed along a straight road so that its velocity changes uniformly from 10 ms-1 to 20 ms-1 in a distance of 200 m. The frictional force is constant at 500N. Example 9 1977 Question 1 (1 mark) Calculate the acceleration of the car. Example 10 1977 Question 2 (1 mark) What is the magnitude of the net force on the car during this 200 m? Example 11 1977 Question 3 (1 mark) What is the magnitude of the force exerted on the car by the towing vehicle?


Example 12 1977 Question 4 (1 mark) When the speed of 20 ms-1 is reached, the towing force is adjusted so that the car now moves at constant velocity. What is now the magnitude of the towing force?

Two masses A and B are accelerated together along a smooth surface by a force of 48 N, as shown above. The acceleration of A and B is 4.0 ms-2. The mass of A is 4.0 kg Example 13 1979 Question 8 (1 mark) What is the mass of B? Example 14 1979 Question 9 (1 mark) What is the magnitude of the force exerted by A on B? Example 15 1979 Question 10 (1 mark) What is the magnitude of the force exerted by B on A? The bodies are now accelerated together along a smooth surface in the opposite direction by a force of 48 N, as shown below.

Example 16 1979 Question 11 (1 mark) What now is the magnitude of the force exerted by B on A?


A 1.0 kg mass is suspended from a spring balance which is attached to the roof of a lift. The balance is graduated in newton and reads 10 N when the lift is stationary. Example 17 1980 Question 14 (1 mark) What is the reading of the spring balance when the lift moves up with an acceleration of 2.0 ms-2? Example 18 1980 Question 15 (1 mark) What is the reading of the spring balance when the lift moves up with an upward constant velocity of 2.0 ms-1? Example 19 1981 Question 5 (1 mark) A man of mass m is suspended from a parachute of mass M and descends at a constant speed. What is the net force acting on the man? A mg B mg + Mg C mg - Mg D zero Example 20 1981 Question 6 (1 mark) A girl starts to run in a northerly direction across a floor. The frictional force of the floor on her shoes is A in a northerly direction: it is the friction between the floor and her shoes which allows her to

travel in that direction. B in a southerly direction: friction always acts to oppose motion. C virtually zero, because the girl is running and not sliding across the floor. D zero, because action and reaction forces are equal and opposite.

For a block sliding down the frictionless inclined plane of height h metre and length L metre, it can be shown that the acceleration, a ms-2 is given by

ha L i.e. ha = k L where k is a constant

Example 21 1981 Question 7 (1 mark) What is the numerical value of the constant k? A 1 B 5 C 10 D 20 E 100


A rocket is drifting sideways from P to Q in outer space. It is not subject to any outside forces.

When the rocket reaches Q, its engine is fired to produce a constant thrust at right angles to PQ. The engine is turned off again when it reaches R. Example 22 1983 Question 4 (1 mark) Which of the following (A, B, C, D, E, or F) best represents the path of the rocket?

Example 23 1984 Question 23a (1 mark) A golf ball travels a distance of 250 m through the air as the result of a very powerful hit. If it is travelling to the right, which of the diagrams below gives the best representation of the forces acting on it when it is at the highest point of its flight. (Air resistance can be ignored.)

Example 24 1984 Question 23b (1 mark) A golf ball travels a distance of 250 m through the air as the result of a very powerful hit. If it is travelling to the right, which of the diagrams below gives the best representation of the forces acting on it when it is at the highest point of its flight, if air resistance needs to be considered.


A steel ball is dropped vertically on to a steel bench-top. W is the weight force acting on the ball and FC is the contact force exerted on the ball by the bench at the instant of rebound when the ball is at rest. Example 25 1984 Question 24 (1 mark) Which one or more of the following statements about W and FC are correct? A W and FC form an action-reaction pair and so are equal in magnitude. B The ball is instantaneously at rest, so W and FC cancel exactly to provide the necessary

zero resultant force. C the ball is accelerating upwards, so FC must be greater than W. D FC may be greater or less in magnitude than W depending on whether the collision is

elastic or inelastic. Five identical blocks each of mass 1.0 kg are on a smooth, horizontal table. A constant force of 1 N acts on the first block as shown in the figure below.

Example 26 1985 Question 14 (1 mark) What force does block 4 exert on block 5? Example 27 1985 Question 15 (1 mark) What force does block 3 exert on block 4? Newton’s third law of motion may be stated as follows: ‘To every action there is an equal and opposite reaction, or the mutual reactions of two bodies upon each other are always equal and directed in contrary directions’.

The figure above shows a block being pulled along a rough surface at constant velocity. aF

is the

applied force on the block, rF

the friction force between the block and the surface, W

the weight

force on the block and cF

the normal contact force exerted by the surface on the block. Example 28 1985 Question 17 (1 mark) Which pair of forces (A – D) in this situation are action-reaction pairs in the sense of Newton’s third law? A W

and the gravitational force exerted by the block on the earth.

B aF

and rF

.

C W

and cF

. D none of these.


The graph above gives the velocity-time relationship for a block of mass 4.0 kg which slides across a rough, horizontal floor, coming to rest after 1.0 s. Example 29 1986 Question 1 (1 mark) What is the magnitude of the frictional force of the floor on the block? Example 30 1986 Question 2 (1 mark) What is the magnitude of the frictional force of the block on the floor?


The diagram above shows a velocity-time graph of the motion of a parachutist of mass 100 kg, when he jumps from an aircraft. After jumping, he waits for 2.0 seconds before pulling the ripcord, opening his parachute. (Take g: 10 ms-2) Example 31 1987Question 5 (1 mark) What is his acceleration at time t =1.0 s? Example 32 1987 Question 6 (1 mark)

What is the value of the ratio: Force of gravity on the parachutist at t = 1.0 s

?Force of gravity on the parachutist at t = 6.0 s

Example 33 1987 Question 7 (1 mark) What is the resultant force on the parachutist at t =1.0s? Example 34 1987 Question 8 (1 mark) What is the resultant force on the parachutist at t = 20.0 s?


A road train consists of a large truck towing a trailer, as shown below. The truck and the trailer have a mass of 2.0 x 104 kg each. When moving along a level road, the truck and the trailer experience a constant retarding force of 2.5 x 103 N each.

Example 35 1991 Question 4 (1 mark) If the driving force on the road train when it is accelerating is 3.9 x 104 N, what is the magnitude of the acceleration? Example 36 1991 Question 5 (1 mark) What is the tension force in the coupling between the truck and trailer when the road train is moving at constant speed? When the road train is travelling along a straight, level road at 20 m s-1 the truck is put into neutral gear and the train allowed to roll to a stop. Example 37 1991 Question 6 How far will it travel before coming to rest? A car is tested on a straight level track on a day when there is no wind. When the car reaches 20.0 m s-1 the driver puts the car into neutral gear. The wheels are no longer driving the car, and it gradually slows due to frictional forces. Measurements show that the car's speed decreases uniformly from 20.0 m s-1 to 18.0 m s-1 in 4.0 s. The mass of the car is 1100 kg. Example 38 1996 Question 5 (1 mark) Calculate the magnitude of the net force on the car when it is slowing down. Example 39 1996 Question 6 (1 mark) Calculate the power required at the wheels of the car for it to be driven at a constant speed of 20.0 m s-1 on the straight level track.


A car of mass 1000 kg is being towed on a level road by a van of mass 2000 kg. There is a constant retarding force, due to air resistance and friction, of 500 N on the van, and 300 N on the car. The vehicles are travelling at a constant speed.

Example 40 2004 Pilot Question 9 (2 marks) What is the magnitude of the force driving the van? Example 41 2004 Pilot Question 10 (2 marks) What is the value of the tension, T, in the towbar? When travelling at a speed of 15.0 ms-1 the van driver stops the engine, and the van and car slow down at a constant rate due to the constant retarding forces acting on the vehicles. Example 42 2004 Pilot Question 11 (3 marks) How far will the van and car travel before coming to rest?


The figure shows a car of mass 1600 kg towing a boat and trailer of mass 1200 kg. The driver changes the engine power to maintain a constant speed of 72 km h-1 on a straight road. The total retarding force on the car is 1400 N and on the boat and trailer 1200 N. Example 45 2004 Question 1 (2 marks) Calculate the driving force exerted by the car at this speed. To overtake another car the driver accelerates at a constant rate of 1.20 m s-2 from 72 km h-1 until reaching 108 km h-1 Example 46 2004 Question 2 (3 marks) Calculate the distance covered during this acceleration. Example 47 2004 Question 3 (3 marks) Calculate the tension in the coupling between the car and trailer during the acceleration. (Assume the same retarding forces of 1400 N and 1200 N respectively.)


Two students are discussing the forces on the tyres of a car. Both agree that there must be a friction force acting on the tyres of a car. The first student claims that the friction force acts to oppose the motion of the car and slow it down, for example, when braking. The second student claims that friction acts in the direction of motion as a driving force to speed the car up when accelerating. Example 48 2004 Question 8 (3 marks) On the diagram of the front-wheel drive car in the figure below, clearly show all the forces acting on the tyres of the car when it is accelerating forwards in a straight line. Use arrows for the force vectors to show both the magnitude and point of action of the different forces.

Example 49 2004 Question 9 (2 marks) On the diagram of the same car below clearly show all the forces acting on the tyres of the car when it is braking in a straight line. Use arrows for the force vectors to show both the magnitude and point of action of the different forces.

A recent Transport Accident Commission television advertisement explains the significant difference between car stopping distances when travelling at 30 kmh-1 and 60 kmh-1. Example 50 2000 Question 10 (2 marks) The stopping distance, from when the brakes are applied, for a car travelling at 30 kmh-1 is 10 m. Which one (A – D) is the best estimate of the stopping distance for the same car, under the same braking, but travelling at 60 kmh-1? A. 20 m B. 30 m C. 40 m D. 90 m


A car of mass 1300 kg has a caravan of mass 900 kg attached to it. The car and caravan move off from rest. They have an initial acceleration of 1.25 m s-2.

Example 51 2000 Question 11 (2 marks) What is the net force acting on the total system of car and caravan as it moves off from rest? Example 52 2000 Question 12 (3 marks) What is the tension in the coupling between the car and the caravan as they start to accelerate? After some time the car reaches a speed of 100 kmh-1, and the driver adjusts the engine power to maintain this constant speed. At this speed, the total retarding force on the car is 1300 N, and on the caravan 1100 N.

Example 53 2000 Question 13 (2 marks) What driving force is being exerted by the car at this speed?


Anna is jumping on a trampoline. The figure below shows Anna at successive stages of her downward motion.

a b c d Figure c shows Anna at a time when she is travelling downwards and slowing down. Example 54 1999 Question 6 (2 marks) What is the direction of Anna’s acceleration at the time shown in Figure 4c? Explain your answer. Example 55 1999 Question 7 (3 marks) On Figure c draw arrows that show the two individual forces acting on Anna at this instant. Label each arrow with the name of the force and indicate the relative magnitudes of the forces by the lengths of the arrows you draw. Connected bodies Problems involving the motion of two bodies connected by strings are solved on the following assumptions;

the string is assumed light and inextensible so its weight can be neglected and there is no change in length as the tension varies.

In these cases, tension is the condition of a body subjected to equal but opposite forces which attempt to increase its length, and tension forces are pulls exerted by a string on the bodies to which it is attached. To solve these problems you need to consider the vertical direction first. m1g – T = m1a The direction of this acceleration must be downwards. This leads to: T = m1g – m1a The tension in the string is the same in both directions, therefore T = m2a. Since both bodies are connected by an inextensible string, both bodies must have the same acceleration.


The vertical forces acting on m2, (not shown) cancel each other out, and do not impact on its motion.

Combing these two equations gives a = 2

1

1

mg

m m+

T =

2

1 2

1

m mg

m m+

Two blocks, each of mass m, are connected by means of a string which passes over a frictionless pulley. One is at rest on a frictionless table; the other is held at rest in the position shown.

Example 56 1981 Question 9 (1 mark) What is the force of the string on Block X? A. zero B. mg/2 C. mg D. 2mg Example 57 1981 Question 10 (1 mark) Block Y is now released and the blocks move. What will then be the net force on Y?

A. zero B. mg/2 C. mg D. 2mg


A block of mass M2 is held at rest on a horizontal frictionless table. A string is attached to a mass M1 over a light frictionless pulley as shown below.

Example 58 1983 Question 6 (1 mark) What is the tension in the string while the block is being held? Example 59 1983 Question 7 (1 mark) If the block is then released, which of the statements below best describes the subsequent state of the block?

A The block starts to move with an acceleration 1 2

2

M +Mg

M

B The block starts to move with an acceleration 1

1

2

Mg

M +M

C The block starts to move with an acceleration g D The direction of the motion depends on whether M2 is greater or less than M1.


A mass M1 is accelerated from rest at X along a horizontal frictionless surface by a tight string passing over a frictionless pulley and attached to mass M2.

M2 falls from its rest position a distance, d, where it strikes the floor. g is the acceleration due to gravity. Example 60 1986 Question 21 (1 mark) What is the acceleration of M1 in the section XY? Example 61 1986 Question 22 (1 mark) What is the speed of M1 at Y?


In a laboratory experiment two blocks, M1 (of mass 4.0 kg).and M2 (of mass 2.0 kg) are connected by a light inextensible string as shown below. There is a constant frictional force of 2.0 N between M1 and the table. All other friction forces should be ignored. The masses are released from rest at the positions shown in the diagram.

Example 62 1988 Question 12 (1 mark) What is the magnitude of the acceleration of the masses? Example 63 1988 Question 13 (1 mark) What is the magnitude of the tension in the string? Example 64 1988 Question 14 (1 mark) What is the speed of M1 at the moment when M2 hits the floor?


Solutions Example 1 1973 Question 8

Using a = Fm

, the maximum acceleration is

caused by the maximum force that the car can exert on the road, (hence the maximum force that the road will exert on the car to accelerate the car). If Fmax is fixed, then tripling the mass will result in 1

3 the original acceleration.

1 ms-2 (ANS) Example 2 1976 Question 1 The distance travelled is the area under the graph. Use the trapezium formula, A = ½(a +b)h to get d = ½ x (3000 + 3500) x 30 d = 97500 m d = 9.8 x 104 m (ANS) (this is a bit difficult to express when you take sig figs into consideration) Example 3 1976 Question 2 Use F = ma to find the net force acting The gradient of the graph over the first 300 seconds will give the acceleration.

a = 30

300 = 0.1

F = 5.0 x 105 x 0.1 F = 5.0 x 104 N (ANS) Example 4 1976 Question 3 The engine is required to provide a net force of 5.0 x 104, so the engine must also provide the force required to overcome the frictional force of 1.5 x 104N. Fengine = 5.0 x 104 + 1.0 x 104 Fengine = 6.5 x 104 N (ANS) Example 5 1976 Question 4 Power is the rate of doing work, it is given by the formula P = Fv. When the train is travelling at a constant speed the engine just needs to overcome the frictional forces. P = 1.5 x 104 x 30 P = 4.5 x 105 P = 450 kW (ANS)

Example 6 1976 Question 9 The force of the road on the bicycle is always perpendicular to the surface of the road. C (ANS) The road cannot push sideways. Example 7 1976 Question 10 Since the speed is constant the acceleration must be zero. D (ANS) Example 8 1976 Question 11 The sum of the forces must be zero, therefore the force of the road on the bicycle must be the sum of the normal reaction (C) and the frictional force on the tyres (A). B (ANS) Example 9 1977 Question 1 Use v2 = u2 + 2ax 202 = 102 + 2x a x 200 400 = 100 + 400a 300 = 400a a = 0.75 ms-2 (ANS) Example 10 1977 Question 2 The net force is found from F = ma F = 800 x 0.75 F = 600 N (ANS) Example 11 1977 Question 3 The towing car also needs to overcome the frictional forces. Ftowing car = 600 + 500 Ftowing car = 1100 N (ANS) Example 12 1977 Question 4 Since the speed is constant, the net force must equal zero. The magnitude of the net force is the same as the magnitude of the frictional force F = 500 N (ANS) Example 13 1979 Question 8 Using F = m x a We get 48 = (4 + B) x 4 12 = 4 + B B = 8 kg (ANS)


Example 14 1979 Question 9 The force that A exerts on B is the force that accelerates B. Using F = ma F = 8 x 4 FA on B = 32 N (ANS) Example 15 1979 Question 10 This is the classic example of Newton’s third law. FB on A = - FA on B

We are asked for the magnitude, FB on A = 32 N (ANS) Example 16 1979 Question 11 Since the force is the same as before and the total mass hasn’t changed, the acceleration will also be 4.0 ms-2. Therefore to accelerate 4.0 kg at 4.0 ms-2 requires a force of 16 N FB on A = 16 N (ANS) Example 17 1980 Question 14 The net force to accelerate a 1.0 kg mass at 2.0 ms-2 is 2N. To overcome the weight, the spring balance needs to supply a 10 N force. Therefore the total force required to accelerate the mass upwards at 2.0 ms-2 will be 10 + 2 =12N

12 N (ANS) Example 18 1980 Question 15 If everything is moving up at a constant velocity, then the spring balance only needs to overcome the weight (otherwise the mass would fall to the floor of the lift).

10 N (ANS) Example 19 1981 Question 5 The man and the parachute are travelling at a constant speed, so the net force acting on the man is zero

D (ANS) Example 20 1981 Question 6 The direction of the girl’s acceleration is north. Therefore the net force acting on her must be northerly.

A (ANS)

Example 21 1981 Question 7 In this case the unbalanced acceleration on the block is given by g sinθ.

a = h

gL

k = g C (ANS) Example 22 1983 Question 4 With a constant thrust the rocket will accelerate in the direction of the thrust. This means that the answer needs to be D, E or F. Once the thrust stops, there will not be any forces acting on the rocket, so it will continue to travel in the same direction.

E (ANS) Example 23 1984 Question 23a Without air resistance the only force acting on the golf ball is its weight. This acts vertically downwards.

A (ANS) Example 24 1984 Question 23b With air resistance, there are now two forces acting on the golf ball its weight, and air resistance in the opposite direction to the motion. (to the left).

D (ANS) Example 25 1984 Question 24 The ball is momentarily stationary, but it is about to rebound. This means that to get it to rebound the net force acting on it must be upwards.

C (ANS) Example 26 1985 Question 14 The force the block 4 exerts on block 5, causes block 5 to accelerate. The system is accelerating at 0.2 ms-2. Therefore the net force on block 5 is

F = ma F = 1 x 0.2 0.2N (ANS)

g g sinθ

g cosθ

N


Example 27 1985 Question 15 The force the block 3 exerts on block 4, causes blocks 4 and 5 to accelerate. The system is accelerating at 0.2 ms-2. Therefore the net force on block 4 is

F = ma F = (1 + 1) x 0.2 0.4N (ANS)

Example 28 1985 Question 17 Newton’s third law states that FA on B = -FB on A The weight of an object can be written as FEarth on Mass. From an action reaction pair perspective, this means that the opposite to the weight is: FMass on Earth.

A (ANS) Example 29 1986 Question 1 Using F = ma, we need to find ‘a’. The acceleration is the gradient of the velocity-time graph. The question asks for the magnitude. The

gradient is: ΔvΔt

= 51

a = -5 Therefore F = 4 x 5 F = 20N (ANS) Example 30 1986 Question 2 This is equal to (in magnitude), but in the opposite direction to the force of the floor on the block. F = 20N (ANS) Example 31 1987 Question 5 (72%) The acceleration is the gradient of the velocity time graph.

gradient is: ΔvΔt

= 14.4

2

a = 7.2 ms-2 (ANS) Example 32 1987 Question 6 (76%) This ratio must always be 1. Examiners comment This question simply asked candidates to realise that the force of gravity would be the same irrespective of speed. Example 33 1987 Question 7 (76%) Using F = ma F = 100 x 7.2 F = 720 N (ANS)

Example 34 1987 Question 8 (77%) At 20 secs, the velocity is constant, this means that the net force must be zero. Therefore the acceleration is zero.

a = 0 (ANS) Examiners comment Questions 7 and 8 tested the relationship between resultant force and acceleration. These two questions, in conjunction with others, seem to reflect many candidates' lack of appreciation of Newton's laws. “lf the velocity is constant there is no acceleration and hence no resultant force. Example 35 1991 Question 4 Driving force – friction forces = net force FNet = ma 3.9 x 104 – (2 x 2.5 x 103) = 4.0 x 104 x a

a = 4 4

4

0.5 103.9 104.0 x 10

a = 0.85 ms-2 (ANS) Example 36 1991 Question 5 The road train is moving at constant speed, therefore, the net force acting on the trailer must be zero. The tension needs to overcome the frictional force. T = 2.5 x 103 N (ANS) Example 37 1991 Question 6 It is simplest to treat the truck and trailer as the one system. The net retarding force is 5 x 103 N.

The deceleration is Fm

= 3

4

5 104.0 10

= 0.125 ms-2 Now use v2 – u2 = 2ax -202 = 2 x 0.125 x ‘x’ x = 400 ÷ 0.25 x = 1600 m x = 1.6 x 103 m (ANS) Example 38 1996 Question 5

The acceleration is ΔvΔt

= 20 18

4

= 0.5 ms-2 F = 1100 x 0.5 F = 5.5 x 102 N (ANS)


Examiners comments Students needed to calculate the acceleration (change in velocity divided by time) and then apply Newton's 2nd Law (F= ma) to determine the net force, resulting in an answer of 550 N. Eighty percent of students were able to do this correctly, a pleasing result indeed. Example 39 1996 Question 6 Power is the rate at which work is being

done, P = Fdt

Power is found from P = Fv P = 5.5 x 102 x 20 P = 1.1 x 104 W (ANS) Examiners comments In order to maintain a constant speed the driving force needed to be equal and opposite to the resistance force, as calculated in the previous question. The power at the wheels is then calculated using the formula, P = Fv. where F is the driving force at the wheels of the car. Assuming the correct answer to the previous question, this calculation resulted in a power of 11 000 W. Consequentially correct answers, arising from an incorrect answer to question 5, were also scored as correct. Forty per cent of students were able to correctly answer this question. The most common incorrect answer (220 000 W) arose from students applying the formula, P = Fv, but substituting in the weight force for F. Example 40 2004 Pilot Question 9 Since they are travelling at a constant speed, the net force must be zero. The only driving force is the due to the van’s wheels pushing the van forward, so this must equal 500 + 300

= 800N (ANS) Example 41 2004 Pilot Question 10 The tension in the towbar is the only force acting forwards on the car being towed. T = 300N, so that the net force acting on the towed vehicle is zero. T = 300N (ANS) Example 42 2004 Pilot Question 11 With the driving force from the van = zero, then the only forces acting are the frictional forces that total 800N. This is acting on a 3000 kg mass. We treat both as one,

because the deceleration of both must be the same as they are joined by the towbar. Use v2 = u2 – 2ax

Let v = 0, u = 15 and a = 8003000

= 0.2666m/s2

0 = 225 – 0.53 x x = 421.9m = 422m (ANS) Make sure that you don’t round off to early in this solution as it can lead to an answer of 417m Example 43 2004 Sample Pilot Question 9 Since the car and van are moving at a constant speed, the resultant force acting on either is zero.. From Newton's first law of motion, F = ma, where a = 0, because of the constant speed. The frictional forces acting on the car must be 300 N (to the right), to oppose the effect of the towing force. Example 44 2004 Sample Pilot Question 10 Consider the car (ignoring friction). Using F = ma, gives T = 1000 x 0.5 = 500N. We need to assume that the frictional forces remain constant, so the 300 N resistive force needs to be overcome. Therefore the tension in the rope is 500 + 300 = 800 N

800 N (ANS) Example 45 2004 Question 1 Since the speed is constant, the net force is zero. Hence the driving force exerted by the car must equal the sum of the resistive forces. 1400 + 1200 = 2600N (ANS) Example 46 2004 Question 2 The tension in the coupling (towbar) is the only force acting forwards on the trailer. The net force on the trailer must give rise to its acceleration. Using F = ma gives F = 1200 1.20 = 1440N. This gives the net force acting on the trailer to be 1440 (to provide the acceleration) and 1200 (to overcome friction)

Tension = 1200 + 1440 = 2640N (ANS)


Example 47 2004 Question 3 You need to convert both the 72 km h-1 and 108 km hr-1 to m s-1. This is done by dividing by 3.6. (This should be on your cheat sheet)

72 km hr-1 = 20m s-1 and 108km hr-1 = 30m s-1. Use v2 = u2 + 2ax 230 = 220 + 2 1.20 x 900 – 400 = 2.4 x

5002.4

= 208 m. (ANS)

Example 48 2004 Question 8 (37%)

The net force acting on the car MUST be in the direction of the acceleration. This net force can only come from the frictional contact between the tyres and the road. You need to remember that the frictional forces oppose the motion, but in this case, to get the car to move forwards the tyre actually wants to move backwards at ground level. Examiners comment The average score for Question 8 (1.1/3) indicates that the concept of friction as a driving force was poorly understood. While many students were aware that there had to be a net force acting forwards to accelerate the car, the origin and point of application of this force was rarely correctly sketched. In fact, many students still sketched the road–tyre friction force acting backwards so as to oppose the motion of the car. It was also disappointing to note the number of sketches that did not show the correct point of application of the weight, normal or frictional forces. It is quite apparent that VCE Physics students need more practice in drawing free-body force diagrams.

Example 49 2004 Question 9 (45%)

The net force acting on the car MUST be in the direction of the acceleration. This net force can only come from the frictional contact between the tyres and the road. Since the car is decelerating the net force must be in the opposite direction to the motion. Examiners comment The average score for this question (0.9/2) again indicates that students found this nearly as difficult as the previous question and that friction as a braking force was not thoroughly understood. Example 50 2000 Question 10 (57%) C This question was testing you understanding of how speed and stopping distance is related. Let’s assume the mass of the car, and the force stopping it, are constant. That is, if the car has a kinetic energy of

221 mv and it requires a force by a distance to

stop it then: 21

2 mv = F d

if the speed of the car is doubled then the stopping distance in 4 times larger. This is known as a square relationship. For this question the speed of the car is 30kmh-1 and the stopping distance is 10m. The speed of the car is then doubled to 60kmh-1 then because the speed has doubled the stopping distance is now 4 times the 10m, which is 40m so C is the correct answer. Examiners comment Students needed to understand that work done equals change in kinetic energy. If a car is travelling at twice the speed then it will have four times the kinetic energy and the brakes will need to do four times as much work in bringing it to rest. Hence, for the


same braking force the stopping distance will be about four times as far. Distance C (40 m) was the best estimate of the stopping distance. Example 51 2000 Question 11 (58%) Just Fnet = ma Using the mass of the system. Fnet = (1300 + 900)1.25

= 2.75103N (ANS) Example 52 2000 Question 12 (58%) For this question think of the caravan as an object that is accelerating that is pulling it. So F = ma

= 9001.25 = 1.13103N (ANS)

Examiners comment Application of Newton’s Second Law for the caravan alone resulted in an answer of 1.125 ·x 103 N for the tension in the coupling. Students needed to be aware that the driving force for the caravan was provided solely by the tension in the coupling. Example 53 2000 Question 13 (63%) If the car is travelling at a constant velocity then there is no net force acting on the car. Therefore the magnitude of the driving force from the motor equals the retarding force of the car and the caravan. Fd = 1300 + 1100 = 2400N (ANS) Examiners comment Constant velocity implies a net force of zero. Hence, the driving force must be equal and opposite to the retarding forces, i.e. 2400 N. Any errors were due to students not realising that constant speed in a straight line implies a net force of zero. Example 54 1999 Question 6 (55%) Up At this time Anna is travelling downwards and slowing down. This means that her acceleration is up, because it is opposing the motion (she is slowing down). Examiners comment Anna was travelling downwards and slowing, hence the direction of her acceleration was upwards. The main errors were to reason that the net force was her weight and thus her acceleration was

downwards, or that since she was moving downwards then her acceleration must also be downwards. It was certainly clear that many students experience great difficulty in distinguishing between velocity and acceleration vectors, (a number of students based their answer on the state of Anna's hair at the instant). Example 55 1999 Question 7 (50%)

The reaction force > weight, because she is slowing down, ie. a net upward force. Examiner’s comment Students were expected to draw two force arrows on Figure C. One arrow was the weight force, acting through Anna’s centre of mass and the other arrow being the normal contact force, acting at her feet. The arrow for the normal contact force should have been longer than the weight force arrow. The average mark for this question was 1.5/3, with the most common errors being in either choosing the incorrect point of application for each force or in not indicating the relative magnitudes as specifically asked in the question. Example 56 1981 Question 9 Block X is initially stationary and remains stationary until the hand supporting Block Y is removed. Therefore the net force on Block X is zero. A (ANS)


Example 57 1981 Question 10 When Block Y is released, its weight accelerates it downward.

As the two blocks are connected by a string, they are going to both have the same acceleration when released. The net force acting on Block Y is given by mg – T = ma The net force acting on Block X is given by T = ma Substituting for T gives mg – ma = ma mg = 2ma

ma = mg

2

B (ANS) Example 58 1983 Question 6

Since the block is being held, M1 is stationary, and going to remain stationary, therefore the net force acting on it is zero. T = M1g

Example 59 1983 Question 7

As the two blocks are connected by a string, they are going to both have the same acceleration when released. The net force acting on Block M1 is given by M1g – T = M1a The net force acting on Block M2 is given by T = M2a Substituting for T gives M1g – M2a = M1a M1g = M1a + M2a M1g = (M1 + M2)a

a = 1

1 2

M g

M +M

B (ANS) Example 60 1986 Question 21 As the two blocks are connected by a string, they are going to both have the same acceleration when released. The net force acting on Block M2 is given by 4g – T = 4a The net force acting on Block M1 is given by T = 1a Substituting for T gives 4g – a = 4a 4g = 4a + a 4g = 5a

a = 4×10

5

8ms-2 (ANS) Example 61 1986 Question 22 Use v2 – u2 = 2ax v2 – 0 = 2 x 8 x 1 v2 = 16 v = 4 ms-1 (ANS)


Example 62 1988 Question 12 As the two blocks are connected by a string, they are going to both have the same acceleration when released. The net force acting on Block M2 is given by 2g – T = 2a The net force acting on Block M1 is given by T – 2 = 4a T = 4a + 2 Substituting for T gives 2g – (4a + 2) = 2a 2g = 2a + 4a + 2 20 = 6a = 2 18 = 6a a = 3 ms-2 (ANS) Example 63 1988 Question 13 Using T = 4a + 2, give T = 4 x 3 + 2 = 14 N (ANS) Example 64 1988 Question 14 Use v2 – u2 = 2ax v2 – 0 = 2 x 3 x 1.5 v2 = 9 v = 3 ms-1 (ANS)

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2. Forces - TSFX