Your SlideShare is downloading. ×
Physics
Physics
Physics
Physics
Physics
Physics
Physics
Physics
Physics
Physics
Physics
Physics
Physics
Physics
Physics
Physics
Physics
Physics
Physics
Physics
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Text the download link to your phone
Standard text messaging rates apply

Physics

4,420

Published on

0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total Views
4,420
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
Downloads
43
Comments
0
Likes
0
Embeds 0
No embeds

Report content
Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
No notes for slide

Transcript

  • 1. Turning Effects of Forces / MomentsNotesMomentThe moment of a force is the turning effect of a force, or the ability of the force to makingsomething turn.Moment of a force (M) about a point O is the product of the force (F) and theperpendicular distance (D) from the point to the line of action of the force.Moment = Force x DistanceSI Unit: Newton (N)The turning effect of a force depends on- location of applied force- perpendicular distance between the point of application of the force and the pivotType 1,2,3 LeversPrinciple of MomentsWhen a body is in equilibrium, the sum of clockwise moments about the balanced point isequal to the sum of anticlockwise moments about the same point (pivot).
  • 2. Total clockwise moment = Total anticlockwise momentWhen the clockwise moment is not equal to the anticlockwise moment, there is aresultant moment and the object will rotate in the direction of resultant moment.If there is no resultant moment, the object is balanced.Centre of gravityThe centre of gravity (CG) of a body is an imaginery point where the whole weight of thebody seems to act in any orientation.The CG of a regular object is at the centre.The CG of an irregular object is determined using a plumb line.If a body is hanging freely at rest, its CG is always vertically below the pivot, thus theplumb line method works. It can only be used for flat, irregular objects.StabilityStability is a measure of the bodys ability to maintain its original position.3 types of stability:1. Stable equilibriumObject will return to original position after slight disturbance.2. Unstable equilibriumObject will fall after slight disturbance3. Neutral equilibriumObject remains in new position after slight disturbance.To increase the stability of a body, its base area should be increased, and the height of its centreof gravity should be decreased.
  • 3. Quick SummaryExampleA light metre rule is allowed to pivot freely at the zero end. The other end is supported by a springbalance. A weight of 200N is then hung at the 40cm mark. The metre rule stays horizontal. Whatis the reading on the spring balance?SolutionBy the principle of moments, taking moments about the pivot
  • 4. Anticlockwise moment = Clockwise momentF x 1m = 200N x 0.4mF = 80NThe reading on the spring balance is 80N.MCQ Questions1. Which one of the following activities does not apply the turning effect of a force?a. swinging on a swingb. sliding down a slidec. moving up and down on the see-sawd. rowing a boat2. Which one of the following quantities is zero when a uniform rod is supported in the middle?a. massb. weightc. pressured. moment3. When a body is at rest, it obeys thea. principle of momentumb. Archimedes principlec. principle of momentsd. principle of inertia4. A uniform metre ruler of weight 0.2N balances at the 60-cm mark when a weight W is placed atthe 80-cm mark. What is the value of W?a. 0.1Nb. 0.15Nc. 0.2Nd. 0.2667N5. Which one of the following measuring instruments works on the principle of moments?a. spring balanceb. single pan beam balancec. micrometerd. vernier calipers6. A uniform rod of weight 5N and length 1m is pivoted at a point 20cm from one of its ends. Aweight is hung from the other end so that the rod balances horizontally. What is the value of theweight?a. 0N
  • 5. b. 0.05Nc. 5Nd. 7.5N7. An object will not turn if the applied force on ita. does not reach its maximumb. does not produce a momentc. passes through its centre of massd. passes through its centre of gravity8. Levers are classified into different types according to the position of itsa. fulcrum, load and effortb. centre of gravityc. centre of massd. moment and load9. Which one of the following statements does not describe a pair of scissors?a. its fulcrum lies between the load and the effortb. it is a lever of type 1c. it works on the turning effect of a forced. it does not have a centre of mass10. Which of the following levers is of type 2?a. wheelbarrowb. scissorsc. fishing rodd. ice tongs11. The centre of mass of a bodya. has a fixed positionb. depends on the pull of gravityc. is always outside the bodyd. must be in a solid part of the body12. A drinking glass has a low centre of gravity becausea. it is heavyb. it is tallc. it has a broad based. its contents are heavy13. When a body is in neutral equilibrium, any displacement willa. raise its centre of gravityb. lower its centre of gravityc. neither raise nor lower its centre of gravityd. return the body to its original position
  • 6. MCQ Answers1. b2. d3. c4. a5. b6. d7. b8. a9. d10. a11. a12. c13. cStructured Questions1. A uniform metre rule AB is supported at its centre of gravity by a knife edge. A force of5N is applied at a point which is 30cm from end A of the rule. Calculate the force whichmust be applied to point B to restore equilibrium.[2.0N]2. A boy of weight 600N sits on the see-saw as shown at a distance of 1.5m from the pivot.What is the force F required at the other end to balance the see-saw?[450N]3. A very light rod 40cm long is pivoted at the centre. A weight of 50N is placed at one end.Where is the place to put a weight of 200N in order that the rod is in equilibrium?[5cm from the centre]4. A very light rod 20cm long has weights of 60N and 40N at its ends. About which point canthe rod balance horizontally?
  • 7. [8cm from the 60N weight]5. A uniform rod 1m long has masses of 100g and 40g at its ends. If it balances 30cm fromone end, what is the weight of the rod?[0.1N]6. The figure shows a uniform metre rule pivoted at the 50cm mark. 125g and 200g weightshang from the rule as shown.a. Calculate where you would hang a 25g mass in order to balance the rule horizontallyb. State, without calculation, how the rule with the two masses hanging as shown in thefigure could be balanced without using any extra mass.[40cm from the pivot on the side of the 200g mass]https://sites.google.com/site/urbangeekclassroomsg/chemistry-classroom/stoichiometry#TOC-Molar-Volume-and-Molar-Mass
  • 8. Kinematics: Speed, Velocity, and AccelerationNotesScalar vs Vector quantities • Scalar quantities:described by a magnitude only.  eg. distance, mass, length, temperature • Vector quantities: quantities described by a magnitude and direction  eg. displacement, weight, acceleration, force, momentumSome terms • Displacement: The distance measured along a straight line in a stated direction w.r.t. the • original point (vector). • Velocity: Rate of change of displacement • Acceleration: Rate of change of velocity  Note: Negative Acceleration = RetardationAcceleration of free-fall • The acceleration of free-fall near the surface of the Earth is constant and is approximately 10m/s2. It is derived from the gravitational force felt by objects near the Earth surface and independent of the mass of any object. • Speed of a free-falling body (experiencing no other forces other than gravity) increases by 10m/s every second or when the body is thrown up, it decreases by 10m/s every second. • The higher the speed of an object, the greater the air resistance. • Terminal Velocity: When an object is moving at constant velocity, acceleration is 0. • As an object falls, it picks up speed, increasing air resistance. Eventually, air resistance becomes large enough to balance the force of gravity where the acceleration of the object is 0, reaching constant velocity.
  • 9. Displacement-Time Graphs • Used to show displacement over time. • Horizontal line: Body at rest. • Straight line with positive gradient: Uniform velocity. • Straight line with negative gradient: Uniform velocity in the opposite direction. • Curve: Non – uniform velocity. • The gradient of the tangent of this graph gives the instantaneous velocity of the object.Velocity-Time Graphs • Used to show velocity over time. • Such a graph can be used to find:  Velocity  Acceleration: Gradient  Distance travelled: Area under the graphTicker TapeDownload the presentation herechapter2-speed_velocityThe EquationsThey are called the suvat equations because the quantities s, u, v, a and t are used inthe equations, with four of the symbols used in each equation. = displacement (measured in metres) = initial velocity (measured in metres per second, ms -1)
  • 10. = final velocity (also measured in ms-1) = acceleration (measured in metres per second per second, ms -2) = time (measured in seconds, s)Below are the equations:Note • It is important to bear in mind that these equations can only be used for CONSTANT ACCELERATION ONLY. When acceleration is not constant, these equations do not work. For variable acceleration, either graphical methods or calculus would be needed. • Furthermore, these equations can only be used for motion in a straight line or one- dimensional motion. • Thus these equations are known as the equations of rectilinear motion. • Rectilinear motion is one-dimensional motion with uniform acceleration.MCQ Questions1. Which of the following is a vector?a. areab. volumec. densityd. force2. The displacement of an object from a fixed point is the distance moved by the objecta. in a particular interval of timeb. in a particular directionc. at a constant speedd. at a constant velocity3. A car accelerates from rest at 5ms-2 for 0.5 minute. The final velocity of the car isa. 150ms-1b. 5.5ms-1c. 10ms-1d. 2.5ms-1
  • 11. 4. When the brakes of a bicycle were applied, the bicycle was brought to rest from 4ms -1 in 2 minutes. What is the acceleration of the bicycle?a. -1/30 ms-2b. 1/30 ms-2c. -2ms-2d. 2ms-25. A free falling object is said to be in linear motion. This is because the object is fallinga. due to its weightb. at constant velocityc. at constant accelerationd. in one direction6. An object has been falling freely from rest for 3 s. The maximum velocity of the objectisa. 30ms-1b. 3.3ms-1c. 10ms-1d. 13ms-17. A body is moving in a circle at a constant speed. Which of the following statementsabout the body is true?a. There is no accelerationb. There is no force acting on itc. There is a force acting at a tangent to the circled. There is a force acting towards the centre of the circlee. There is a force acting away from the centre of the circle8. What must change when a body is accelerating?a. the force acting on the bodyb. mass of the bodyc. speed of the bodyd. velocity of the bodyMCQ Answers1. d2. b3. a4. a5. c6. a7. d8. d
  • 12. Structured Questions Worked Solutions1. A car accelerates uniformly from a speed of 20ms-1 to a speed of 25ms-1 in 2s.Calculatea. the average speed for this period of 2sb. the distance travelled during this periodc. the accelerationSolution1a. 22.5 m/s1b. 45 m1c. 2.5 m/s22.Describe the motion of the lorry over the following sections of graph along...a. PQb. QRc. RSSolution2a. moving with decreasing speed2b. stationary/zero speed2c. moving with increasing speed
  • 13. 3.a. How far has the object travelled during the first 5 seconds?b. What is the acceleration of the objectc. For how long does the object move at uniform velocity?d. What is the average speed of the object during the first 15 seconds?Solution3a. 25 m3b. 2 m/s23c. 03d. 11.7 m/s24. In an experiment, a student measured the accelerations of a steel ball and afeather falling freely. Will the two accelerations be the same or different? Give areason for your answer.SolutionThe two accelerations are different. The air resistance slowed the downward motion ofthe falling feather much more than that of the falling steel ball.5. The figure below shows the velocity of a bus moving along a straight road overa period of time.
  • 14. a. What does the portion of the graph between O and A indicate?b. What can you say about the motion of the bus between B and C?c. What is the deceleration of the bus between C and D?d. What is the total distance travelled by the bus in 100 s?e. What is the average velocity of the bus?Solution5c. 1 m/s25d. 2000 m5e. 20 m/s6. Two cyclists, A and B, start a race. A accelerates for the first 5 s, until hisvelocity reaches 12ms-1, after which he travels with constant velocity. Baccelerates for the first 10 s, until his velocity reaches 15ms-1, after which hetravels with constant velocity.a. Sketch the velocity-time graphs for the two cyclistsb. Calculate the distance travelled by both cyclists in the first 10 sc. Who is in the lead after 10 s?Solution6b. 90 m; 75 m7. A car travels along a straight road. The speedometer reading after every 5 s istabulated below Time/s 0 5 10 15 20 25 30 35 40 Velocity 0 10 20 30 30 30 30 15 0
  • 15. m/sa. Draw a velocity-time graph to show the variation of velocity with timeb. Describe the motion of the carc. How far from the starting point is the car after 20 sd. What is the total distance travelled by the care. What is the average velocity of the car for the whole journeySolution7c. 373 m7d. 825 m7e. 20.6 m/s8. Find the average velocity of a car which travels 360km in 6 hours ina. km/hb. m/sSolution8a. 60 km/h8b. 16.7 m/s9. Find the average velocity of an athlete who runs 1500m in 4 minutes ina. m/sb. km/hSolution9a. 6.25 m/s9b. 22.5 km/h10. The graph below shows the speed-time graph for a child on a swing
  • 16. a. Write downi. the maximum speedii. the time at which the maximum speed occursbi. On the graph, mark with Z the point where the magnitude of acceleration of thechild is maximum.bii. Mark with M one point at which the acceleration is zeroc. Estimate the distance travelled by the child in 1.2sd. Describe briefly the changes in acceleration during the period shown on thegraphSolution10ai. 6.0 m/s10aii. 0.60 s10b.
  • 17. 10c. 3.6 m10d. The acceleration increases from zero initially to a maximum and decreases to zero.The body then decelerates with deceleration increasing to a maximum and decreases tozero.11. A body is accelerated uniformly from rest and in the first 8.0s of its motion ittravels 20m. Calculatea. the average speed of this period of 8 sb. the speed at the end of this periodc. the accelerationSolution11a. 2.5 m/s11b. 5 m/s11c. 5/8 m/s212. A car of length 6.0m accelerates from rest along a straight level road as shown.The car takes 2.0s to pass the point P.10.0s later the car has just passed point Q.The car takes 0.40s to pass point Q.Calculate
  • 18. ai. the average speed of the car as it passes Paii. the average speed of the car as it passes Qaiii. the average acceleration of the car between P and Qbi. Estimate the distance between P and Qbii. What assumption did you make when you estimated the distance between Pand Q?Solution12ai. 3 m/s12aii. 15 m/s12aiii. 1.2 m/s212bi. 90 m12bii. The car accelerates uniformly between P and Q. The acceleration remainsconstant.13a. What is meant by the period of a simple pendulum?13b. The period of a simple pendulum 1 m long is approximately 2 s. Stateaccurately how you would determine the period of such a pendulum as accuratelyas possible, using a stopclock accurate to within 0.1 s.Solution13a. The period of a simple pendulum is the time taken for one complete oscillationmade by the pendulum.13b.The pendulum as shown above is slowly brought to point X and then released.Simultaneously, the stopclock is started. For every time the bob returns to X, it isconsidered one oscillation. Count to 30 oscillations and read the time taken. The wholeexperiment is repeated three times. Then average the three readings. The period isfound by taking average time over 30.14. Students, investigating motion down an inclined plane, measure the speed of a
  • 19. steel ball at one second intervals after the ball starts to roll from rest down onesuch plane: time in s 0.00 1.00 2.00 3.00 speed in m/s 0.00 0.60 1.20 1.80a. calculate the average acceleration over the first 3.00 sb. calculate the average speed over the first 3.00 sc. what was the distance travelled by the ball in the first 3.00 s?d. how do the numbers in the table show that the acceleration was constant?Solution14a. average acceleration = (final speed - initial speed) / time = (1.80 - 0.00) / 3.00 =0.60 m/s214b. average speed = 1/2 x (u + v) = 1/2 x (0.00 + 1.80) = 0.90 m/s14c. distance travelled = average speed x time = 0.90 x 3 = 2.7 m14d. The speed increases by a constant value of 0.60 m/s every one second, hence theacceleration can be seen to be constant.15. A metal box, attached to a small parachute, is dropped from a helicopter.a. Explain in terms of the forces acting, whyi. its velocity increased immediately after being droppedii. it reached a uniform velocity after a short timeb. The total force opposing the motion of the box and parachute at a particularinstant during its fall is 30N. The combined mass of the box and parachute is5.0kg. Calculate the resultant downward force on the box and parachute. (g = 10N/Kg)Briefly describe the motion of the box and parachute at this timec. At the end of this fall the parachute is caught on a tall tree. The box is then cutloose and falls from rest to the ground. The time of fall is 2.4 s. Calculatei. the velocity with which the box strikes the groundii. the average velocity during its falliii. the distance fallen (g = 10 m/s2)Solutions
  • 20. 15ai. This is because at the moment the box left the helicopter, the force of gravity pulledit downwards, causing an increase in velocity15aii. The presence of the parachute increased the air resistance which acted as aconstant retarding force on the box. Once this force was equal to the weight of the box,velocity became constant.15b. resultant downward force = (5 x 10) - 30 = 20 NFrom Newtons Second Law of Motion, the box and parachute at this time will accelerateat a rate of 4 m/s215ci. given t = 2.4 s, u = 0, g = 10 m/s2,v = u + atv = 10(2.4) = 24 m/s15cii. average velocity = (u + v)/2 = (0 + 24)/2 = 12 m/s15ciii. distance = 12 x 2.4 = 28.8m

×