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G481MECHANICS MODULE 2:

FORCES IN ACTIONNOTES AND QUESTIONS

©2010 science-spark.co.uk

Lesson 15 and 16 notes – Newton’s 2nd LawObjectives(a) Solve problems using the relationship: net force = mass × acceleration (F = ma) appreciating that acceleration and the net force are always in the same direction; (b) define the newton; (c) apply the equations for constant acceleration and F = ma to analyse the motion of objects; (d) recall that according to the special theory of relativity, F = ma cannot be used for a particle travelling at very high speeds because its mass increases.

Newton’s LawsNewton helped describe the physical Universe with 3 Laws to describe the motion of objects.His 1st law stated hat a body would stay stationary or carry on with constant velocity unless acted on by another force.

His 2nd law said that the force applied to an object is directly proportional to its acceleration and that as an object grew in mass it would be harder to make accelerate. So mass becomes the property of a body that resists change in motion. This is summed up by the equation:

Force (N) = mass (kg) x acceleration (ms-2)

Since force and acceleration are both vectors they must act in the same direction.This law gives us a definition for the Newton (the unit of force):

1 N = 1 kg m s-2.

His 3rd law states that when an object exerts a force on another object, it will feel an equal (and opposite) force of the same type on it. (this will be covered in A2)


Lesson 16 questions – Newton’s 2nd Law1a) Define a Newton.…………………………………………………………………………………………………….…………………………………………………………………………………………………….……………………………………………………………………………………………… (1)

b) Fig 1.1 shows a car of mass 1200kg, pulling a caravan of mass 400kg along a horizontal road.

Fig 1.2Calculate, for the first 20s of the journey

i) The acceleration of the car

acceleration = ………………..ms-2 (2)


ii) the resultant force acting on the car

resultant force on the car = …………N (2)

iii) the resultant force acting on the caravan.

resultant force acting on the caravan = ………………. N (2)c i) Use the graph to describe how the acceleration changes after the first 20s of the journey.…………………………………………………………………………………………………….…………………………………………………………………………………………………(1)

ii) Suggest the possible reason for this change.…………………………………………………………………………………………………….……………………………………………………………………………………………… (1)

Total [9]2 a) Define acceleration.…………………………………………………………………………………………………….…………………………………………………………………………………………………….……………………………………………………………………………………………… (2)

b) An aircraft of total mass 1.5 x 105 kg accelerates, at maximum thrust from the engines, from rest along a runway for 25s before reaching the required speed for take-off of 65ms-1. Assume that the acceleration of the aircraft is constant.

Calculate

i) the acceleration of the aircraft

acceleration………………. ms-2 (3)ii) the force acting on the aircraft to produce this acceleration

force = ………………….. N (2)


iii) the distance travelled by the aircraft in this time.

distance = ………………….. m (2)

c) The length of runways at some airports is less than the required distance for take-off by this aircraft in (b)(iii).State and explain one method that could be adopted for this aircraft so that it could reach the required take-off speed on shorter runways.…………………………………………………………………………………………………….…………………………………………………………………………………………………….…………………………………………………………………………………………………….…………………………………………………………………………………………………….……………………………………………………………………………………………… (2)

Total [11]


Lesson 17 notes – Non-linear MotionObjectives(a) explain that an object travelling in a fluid experiences a resistive or a frictional force known as drag; (b) state the factors that affect the magnitude of the drag force; (c) determine the acceleration of an object in the presence of drag; (d) state that the weight of an object is the gravitational force acting on the object; (e) select and use the relationship: weight = mass × acceleration of free fall (W = mg); (f) describe the motion of bodies falling in a uniform gravitational field with drag; (g) use and explain the term terminal velocity.

Air resistanceAir resistance affects the flight of a projectile since some energy is lost as heat and sound because of the force of the air acting on it.

The graphs below show the trajectory, vertical speed and horizontal speed of a projectile fired from a cliff respectively.

The red line shows the projection of a projectile without the affects of air resistance taken into account.

The blue line shows the projection of a projectile with the affects of air resistance taken into account.

You can see that a projectile loses vertical and horizontal speed and because of this doesn’t go as high or as far as it would without air resistance.

The bigger the surface area, the bigger the air resistance or drag.

WeightWeight is the force due to gravity pulling objects of mass towards each other. i.e.

Weight = mass x g

Therefore as the mass of an object increases so does its weight. This is the reason that all objects fall with the same acceleration.


Lesson 18 notes – Equilibrium definitionsObjectives(a) draw and use a triangle of forces to represent the equilibrium of three forces acting at a point in an object;

Vector trianglesVector triangles can be used to work out resultant vectors or finding components of vectors. They can be a very useful tool if drawn to scale as angle and length measurements can be taken straight from an accurate diagram.In the example below, a mass of 1kg is hanging by a strong string that is held taut by two poor students who can’t even afford shoes.

A vector diagram could be drawn of the forces, showing the tension from Student 1, T1, and the tension from student 2, T2, and the weight W, of the mass.

We know the Weight, W, is 10N (since Weight = mass x gravitational field strength. So W = 1 kg x 10 Nkg-1)Now we can draw a vector triangle:


T2T1

W

10˚ 15˚

T1

T2

W = 10N80˚

75˚25˚

Since it is a closed triangle the forces must be balanced. The system is in equilibrium. The vertical components acting down (the weight) must equal the vertical components up from the tension in the rope. And similarly, the horizontal components left and right must be equal to each other. How do we know this practically? The system isn’t moving. If the forces were unbalanced in any direction, the system would accelerate in that direction.

Knowing the weight and the angles we can use a scale drawing or calculation to work out the tension in the rope.

If a scale drawing is used then measurements are taken directly.

Calculation uses knowledge of the angles and the sine rule:

Using the Sine Rule for our example:

so, T1 = 23N (22.86N)T2 = 23N (23.30N)

This seems about right because of the shallow angles that they have with the horizontal. A similar force (the horizontal force is equal) and big, so that the vertical component is equal to the weight.

To test that it is correct you should work out the vertical component and see that they do add up to 10N.


sinAa

= =sinB sinC

b cC

b

a

A

B C

sin2510

= =sin75 sin80

T1 T2

T1

T2

W = 10N80˚

75˚25˚

Lesson 18 questions – Forces in equilibrium1) Fig 1.1 shows two strings supporting an object of weight 12N. The tension in one of the strings is 6.0 N. The tension in the other string is T.

i) Calculate the magnitude of the vertical component of the tension T in order for the object to be in equilibrium.

vertical component……………………N (2)ii) Hence calculate the magnitude of T.

magnitude of T……………………N (2)Total [4]

2) A girl travels down a pulley-rope system that is set up in an adventure playground. Fig 2.1 shows the girl at a point on her run where she has come to rest.


a) Explain why the vector sum of the three forces must be zero.………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (1)

b)i) Sketch a labelled vector triangle of the forces acting on the pulley wheel.

(3)ii) Determine by scale diagram or calculation the forces T1 and T2 the rope exerts on the pulley wheel.

T1=……………………..NT2=……………………..N

(3)Total [7]


Lesson 19 notes – moments and couplesObjectives(b) state that the centre of gravity of an object is a point where the entire weight of an object appears to act; (c) describe a simple experiment to determine the centre of gravity of an object;(d) explain that a couple is a pair of forces that tends to produce rotation only; (e) define and apply the torque of a couple; (f) define and apply the moment of force; (g) explain that both the net force and net moment on an extended object in equilibrium is zero;

MomentsA moment is the rotation that is caused by the force multiplied by the perpendicular distance to the pivot point.

Moment (Nm)= Force(N) x distance(m)M = F x s

A couple is the rotation caused by 2 parallel forces where the pivot is in the middle of these two forces.

Couple (Nm) = 2xForce(N) x distance(m)/2 (since there are 2 forces and the pivot is at the middle of that distance)

So,C = F x s

This is how the Torque (T/Nm) of a couple is defined.

If the system is in equilibrium, then it is not rotating. This means that there is no resultant torque.

The clockwise moments = The anticlockwise moments

Σ Fs = Σ Fs

Centre of GravityAn objects weight can be calculated by multiplying its mass by the gravitational field strength. This weight can then be used as the force in the equations above if for example we were looking at two people on a see-saw. But what distance do we apply that force from. Each part of an object will have mass and therefore some weight. To simplify this problem we can work out the objects centre of gravity. For uniform objects like a plank of wood, this is easy, it’s the centre point, for non-uniform objects it is more difficult and practical methods may have to be used.


moment = Fs

s

F

couple = Fs

s

F

F

Worked exampleA uniform horizontal shelf of width 0.38 m is attached to a wall a shown in the diagram. The total weight of the shelf and books is 70 N. This weight acts from the middle of the shelf (0.19 m from the wall).

Calculate the turning effect of the weight about point P.

Moment = 70 N 0.19 m = 13.3 N m

Calculate the tension in the support wire.

The component of T at 90 to the shelf must provide the moment to balance the moment of the weight.

T cos 50° 0.38 m = 13.3 N m

T cos 50° = 35 N

T = 54.5 N

Note that we take moments about point P. This is because there is a third force which acts on the shelf; this is the contact force (or ‘reaction’) of the wall on the shelf. We do not know its magnitude or direction but, since it acts through point P, it has no turning effect about P.


T

40o

70 N

P

Lesson 20 notes – moments and couplesObjectives(h) apply the principle of moments to solve problems, including the human forearm;

MomentsA moment is the rotation that is caused by the force multiplied by the perpendicular distance to the pivot point.

Moment (Nm)= Force(N) x distance(m)M = F x s

A couple is the rotation caused by 2 parallel, equal, forces where the pivot is in the middle of these two forces.

Couple (Nm) = 2xForce(N) x distance(m)/2 (since there are 2 forces and the pivot is at the middle of that distance)

So,C = F x s

This is how the Torque (T/Nm) of a couple is defined.

If the system is in equilibrium, then it is not rotating. This means that there is no resultant torque.

The clockwise moments = The anticlockwise moments

Σ Fs = Σ Fs

Centre of GravityAn objects weight can be calculated by multiplying its mass by the gravitational field strength. This weight can then be used as the force in the equations above if for example we were looking at two people on a see-saw. But what distance do we apply that force from. Each part of an object will have mass and therefore some weight. To simplify this problem we can work out the objects centre of gravity. For uniform objects like a plank of wood, this is easy, it’s the centre point, for non-uniform objects it is more difficult and practical methods may have to be used.

Worked exampleA uniform horizontal shelf of width 0.38 m is attached to a wall a shown in the diagram. The total weight of the shelf and books is 70 N. This weight acts from the middle of the shelf (0.19 m from the wall).


moment = Fs

s

F

couple = Fs

s

F

F

Calculate the turning effect of the weight about point P.

Moment = 70 N 0.19 m = 13.3 N m

Calculate the tension in the support wire.

The component of T at 90 to the shelf must provide the moment to balance the moment of the weight.

T cos 50° 0.38 m = 13.3 N m

T cos 50° = 35 N

T = 54.5 N

Note that we take moments about point P. This is because there is a third force which acts on the shelf; this is the contact force (or ‘reaction’) of the wall on the shelf. We do not know its magnitude or direction but, since it acts through point P, it has no turning effect about P.


T

40o

70 N

P

Lesson 20questions – moments and torque1a) Define centre of gravity………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (2)

b) Define moment of a force………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (2)

c) Fig 1.1 shows a student being weighed. The student, of mass 50kg, stands 0.30 m from end A of a uniform plank of mass 8.0 kg and length 2.0 m. The plank is pivoted 0.50 m from end A, and an object of mass 5.0 kg is moved from and B until the plank balances.

fig 1.1

i) On fig 1.1, draw and label the forces acting on the plank.

ii) Show by using the principle of moments that the 5.0 kg mass must be placed 0.30m from end B in order for the plank to balance.

(4)iii) The arrangement shown in Fig 1.1 has a maximum mass that it can determine. Suggest two changes that can be made to the arrangement to increase this maximum possible mass. Explain your answers.1. ………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………2. ………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (3)

Total [13]


2a) Using a sketch define the moment of a force.

………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (2)

b) Fig 2.1 shows a computer resting on a tabletop that is hinged at A.

fig 2.1The tabletop has a mass of 5.0 kg and its centre of gravity is 0.40m from the axis of the hinge at A. The computer has a weight of 200N acting through a point 0.25 m from the hinge at A. The tabletop is supported to maintain it in a horizontal position by a force F acting vertically at B. The distance AB is 0.80m.

i) Calculate the weight of the tabletop.

weight = …………..N (1)

ii) On fig 2.1 draw and label an arrow to represent the weight W of the tabletop. (1)

iii) Apply the principle of moments about the hinge at A to determine the vertical force F applied at B that is required to maintain the tabletop in equilibrium.

force F = …………..N (3)


iv) The tabletop must experience a resultant force of zero in order to be in equilibrium. Explain how the forces acting on the tabletop fulfil this condition.………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (2)

Total [9]


Lesson 21 notes – Density and pressureObjectives(i)select and use the equation for density: ρ = m / V(j) select and use the equation for pressure P P = F / Awhere F is the force normal to the area A.

DensityDensity is a measure of a things mass per size. If something is very heavy and only takes up a small volume then it is dense. Water has a density of 1000kgm-3, objects that have a greater average density will sink, whilst those that have a smaller average density will float. Oil, for example has a density of 800kgm-3, and so floats on top. Saturn is another example of something that is less dense than water at 120kgm-3, it would float in a swimming pool of water large enough to hold it!

Density is the mass per unit volume of an object.

Density(kgm-3)=mass(kg) / volume(m3)

PressurePressure is the amount of force per unit area on an object.

Pressure (Pascals) = Force (N) / area (m2)

Pressure units can be quite confusing since there are so many. The S.I. (systeme internationale) Units for pressure are the Pascal (Pa) with 1 Pa = 1Nm-2. In an exam Pa or Nm-2 are interchangeable and both are equally acceptable. 1Pa is actually quite a small amount of pressure though so it is not uncommon to use kPa (kilo Pascals - 1000 Pascals) Units or MPa (1000 000 Pascals).

Two other pressure units that you should be familiar with are Atmospheres and the Bar. 1 Atmosphere (atm) is literally the pressure of the weight of the air in the atmosphere pressing down on a square metre on the ground. 101.325kPa = 1 atmosphere. The Bar is used in industry and is roughly equal to 100kPa (actually 1 Bar = 0.9869 atm.


http://en.wikipedia.org/wiki/Blaise_Pascal

http://ptcl.chem.ox.ac.uk/MSDS/glossary/units_of_pressure.html

Lesson 21 questions – density and pressure1 a) Define density.………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (1)

b) Fig 1.1 shows a swimming pool with a base of surface area 80m2. The pool contains water of density 1000kgm-3 to a uniform depth of 1.2m.

fig 1.1i) Show that the weight of water in the pool is 9.4 x 105N.

(2)ii) Calculate the pressure exerted on the base of the pool by the water.

pressure = …………………….. unit ………… (3)

c) A second pool, of base area 40m2, has water of the same density and depth as that of the pool in part (b). Determine the pressure of the water on the base of this pool.

pressure = ……………………..comment on your answer.…………………………………………………………………………………………………….…………………………………………………………………………………………………(2)

Total [8]


Lesson 22 notes – Forces on vehiclesObjectives(a) define thinking distance, braking distance and stopping distance; (b) analyse and solve problems using the terms thinking distance, braking distance and stopping distance; (c) describe the factors that affect thinking distance and braking distance;

Motive ForceMotive comes from the word “motion”. So this is a force which causes motion.The engine provides a torque to the driving wheels however, this is not really the motive force. The motive force occurs at the wheels. The wheels turning push backwards on the road. By Newton’s third law, the road must be pushing forwards on the wheel, like this:

So the engine provides torque to turn the wheels so the wheels push back on the road. Motive Forces make objects accelerate.

Braking ForceThe car is moving – so the wheels are turning – usually at high speed. The braking force is provided by the brake pads gripping the wheel. The brakes provide a torque opposite to the torque driving the wheels round. This reduces the force of the wheels on the road, and hence the push of the road on the car.


Wheel pushing backRoad pushing forwards

Both types of brakes use friction to slow the wheels, decreasing the push from them on the road and therefore the push on the road to the wheels (by Newton’s 3rd Law) so the vehicle slows down.

Stopping DistancesThis is the distance moved by the vehicle between the driver seeing the obstacle and the vehicle stopping. The vehicle keeps moving at a steady speed whilst the driver reacts (thinking distance). The vehicle slows down whilst the brakes are applied (braking distance).

Stopping = Thinking + Braking

Speed(m/s)

Time (s)

Thinking distance

Braking distance

Reaction time

This speed-time graph shows the two sections of thinking – at a constant speed and braking – decelerating to zero. The total braking distance can be found from the area under this trapezium or by adding the two separate sections.

Thinking DistanceFactors that increase thinking distance:Speed;Alcohol/drug consumption will increase reaction time and therefore increase thinking distance;Distractions like mobile phones, radio, kids in the back could also increase reaction time and therefore increase thinking distance.


Braking DistanceFactors that increase braking distance:Initial speed;Friction between tyres and road: icy or wet roads decrease friction and therefore increase braking distance; tyres with very little tread will decrease friction in poor conditions and increase stopping distance.


Lesson 22 questions – Forces on vehicles1. Use the internet to help you answer the following questions. A good start would be: http://www.howstuffworks.com

a) Label the following diagram:

(7)b) What happens when you hit the brake pedal?

………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (2)

c) Label the following diagram:

d) How are disc brakes different to normal bike brakes?………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (2)

e) What force between the pads and the disc slows the disc down?…………………………………………………………………………………… (1)

Total [12]


http://www.howstuffworks.com/

2 a)i) Define pressure…………………………………………………………………………………………..…………………………………………………………………………………………..…………………………………………………………………………………………..

ii) State the SI unit of pressure.……………………………………………………………………………………… (3)

b) Fig 2.1 shows the arrangement of the master cylinder and the slave cylinder for part of a braking system. The master cylinder piston, of surface area 8.0cm2, exerts a force of 28N on the brake fluid when the driver presses the brake pedal.

Fig 2.1

i) Calculate the pressure on the brake fluid produced by the master piston.

pressure = ………….. unit ……… (2)This pressure is transmitted to the slave cylinder piston that has a surface area of 40cm2.ii) Calculate the force exerted on this piston.

force = ……………….. N (2)iii) State two methods of increasing the force produced by the slave cylinder piston.…………………………………………………………………………………..…………………………………………………………………………………………..…………………………………………………………………………………… (2)

Total [9]3) a) Explain the following terms when associated with driving a car.

i) thinking distance…………………………………………………………………………………………..…………………………………………………………………………………………..

ii) braking distance.…………………………………………………………………………………………..…………………………………………………………………………………… (2)


b) Fig 3.1 shows the variation with speed v of the thinking distance d for the driver of a car.

fig 3.1i) Explain why the graph is a straight line through the origin.…………………………………………………………………………………………..…………………………………………………………………………………… (1)ii) Use the graph to determine the time taken for the driver to react when he is travelling at 16ms-1.

time taken = ……………………. s (2)iii) The driver is travelling at 30ms-1 in a car which brakes with an acceleration of -6.5ms-2. Calculate

1. the thinking distance

thinking distance = …………………… m


2. the overall stopping distance.

stopping distance =…………………….. m(5)

c) Explain the effect of the road conditions and tyre tread on the stopping distance.…………………………………………………………………………………………..…………………………………………………………………………………………..…………………………………………………………………………………………..…………………………………………………………………………………………..…………………………………………………………………………………………..…………………………………………………………………………………………..…………………………………………………………………………………………..…………………………………………………………………………………………..…………………………………………………………………………………… (3)

Total [13]


Lesson 23 notes – Car safetyObjectives(d) describe and explain how air bags, seat belts and crumple zones in cars reduce impact forces in accidents; (e) describe how air bags work, including the triggering mechanism; (f) describe how the trilateration technique is used in GPS (global positioning system) for cars.

CrashesWhen a car crashes it goes from having a speed to having zero speed very quickly. That’s what a crash is! Car safety devices have to find a way to increase the time it takes to come to a complete stop so that impact force on the vehicle and more importantly, the driver and passengers is reduced.Devices such as airbags, seatbelts and crumple zones do this by deforming and slowing down the impact, increasing the crash time. This is very important as we can see from the 3 equations that help us analyse crashes.

Analysing crashesUltimately to reduce injuries we must reduce the energy transferred from the impact on the car to the energy transferred to the driver and passengers.

Kinetic energyThe best way of doing this is to reduce speeds.

The equation for kinetic energy explains why:

KE=1/2 m v2,

This equation has two variables: mass and velocity

So why not get car manufacturers to reduce mass?This may help, a little, but look at the equation and you see that velocity is squared. Reducing velocity of the vehicle reduces the transfer of energy proportionally by the square of the reduction in velocity.


Example.

A car of mass 700kg is travelling at 70mph (32ms-1) It crashes. How much energy is transferred to the passengers assuming it all gets transferred from the impact on the car to the passengers?

It’s KE = ½ m v2, KE = ½ 700 x 322,KE = 358400 Joules

Let’s try reducing the mass of the car to 600kg travelling at the same speed.


Now let’s reduce the car’s speed by 10 mph, so it is now travelling at 60mph (27ms-1) but keep its mass at 700kg.


So, the results are clear. If we reduce the mass or the speed by the same factor, the reduction in speed has a much greater affect on the energy that can be transferred in an impact.

Apart from educating drivers to drive slower, how else can cars be made safer?

Newton 2nd LawNewton 2nd Law, the equation for kinetic energy and the equation for work done can all be used to analyse crashes.

Newton 2 states that the acceleration of an object is directly proportional to the force that it is pushed depending on its inertial mass.

or, Force = mass x acceleration (F=ma)

Acceleration is the rate of change of velocity. In crashes we are looking at negative acceleration (deceleration or retardation). (a=Δv/Δt)If we have a large change in velocity and a small time we will have a large acceleration. This is what happens in crashes. The speed goes to zero over a very short amount of time.As acceleration a gets bigger, the force F gets bigger:

F=ma©2010 science-spark.co.uk

That force is the impact force on the car which is ultimately the force on the people inside the car. If we can make the acceleration smaller, the force felt by the people in the car will be reduced:

F=ma

We cannot alter the value of the change in velocity so car manufacturers have to try to change the time it takes to reach zero. since a=Δv/Δt if we make Δt larger a will decrease and therefore decrease the force on the vehicle and people:

a=Δv/ΔtF=ma

So crumple zones crumple to increase the time of the crash, seatbelts extend to increase the time of the crash and airbags increase the distance of the impact, increasing the time of the crash.

The energy would then be reduced if we look at the work done in the collision.

Work Done = Force x distance.

The distance over which the collision takes place stays constant but the force is smaller (but constant), so the work done, or energy transferred in the collision, is reduced:

W = Fs


GPSGlobal positioning systems use 3 (possibly 4) satellites to pinpoint a receivers position. It’s easy to see how it works in 2D:

You know you’re in Devon but you don’t know where.

Somebody tells you that you’re 20 km from Buckfastleigh.

So you could be anywhere on the red circle with a radius of 20km from Buckfastleigh.


You meet somebody else that tells you that you are 26km from Exeter

So you could be near Princetown on Dartmoor or in Torquay. The two places that the red and yellow circles intersect.

And a third person who tells you that you are 45km from Plymouth

So because the three circles all meet at Torquay, unfortunately you must be there.

It works very similarly in 3D apart from the circles become spheres and where they meet is where you are.


Check out this youtube video that explains it: http://uk.youtube.com/watch?v=3zRlbboMvb0


http://uk.youtube.com/watch?v=3zRlbboMvb0

http://uk.youtube.com/watch?v=3zRlbboMvb0

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