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A set of slides created to teach Forces & Motion to learners at Bishops Diocesan College in Cape Town.

A set of slides created to teach Forces & Motion to learners at Bishops Diocesan College in Cape Town.

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    Forces & Motion Forces & Motion Presentation Transcript

    • For FULL presentation click HERE >> www.warnescience.net Forces & Newton’s Laws K WARNE
    • For FULL presentation click HERE >> www.warnescience.net Newton’s First Law ”A body continues in its state of rest or of uniform motion in a straight line, unless acted on by a resultant force."  This law describes - Inertia. IMPORTANT POINTS: (A moving object…) • Continues in its state • Rest or uniform motion in a straight line • Unless acted upon by (external) forces
    • SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Inertia 1. A body has a natural tendency to resist any changes to its state of motion. • This resistance is known as Inertia. • If the card in the picture is flicked the peg should fall into the glass. Inertia keeps the peg stationary when the card is moved quickly. • The peg’s Inertia is overcomes the friction forces which try to keep it’s position on the card. 2. The moon was moving past the earth in a straight line but became trapped by the earth’s gravity. • Gravity does not act against the direction of motion (90o) so the motion continues because there is no unbalanced force to stop it. • (The question is who threw it in the first place!) Every object in a state of uniform motion tends to remain in that state of motion unless an external (unbalanced)force is applied to it. Gravity Motion Earth 1. 2. Motion Inertia Friction
    • SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net The Normal or Reaction Force • This is the Reaction Force (Newton 3) exerted by a surface on any object that rests (exerts a force on) that surface. • Always EQUAL in magnitude and to the initial force and at 900 to the SURFACE. Weight (Fg) F┴ N = - F┴ Normal Reaction Force (N) Weight (Fg) Normal Reaction Force (N) N = - Fg Calculate the normal force in each example here if the ball has a 2 kg mass. 30o F┴ = Fg cos  = 19.6 cos 30 = 16.97 N FN = 16.97 N
    • SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net FRICTION • Friction forces exists between two surfaces in contact. • There are two types …………… FRICTION and ………………(moving) FRICTION. • Static friction is ………………. than dynamic friction. • The maximum force exerted on a stationary object …………..it begins to move (accelerate) is equal to the maximum static frictional force. • The force required to ………. an object moving with ……………… velocity on a surface is equal to the dynamic friction. Max force Static friction V = 0 Force dynamic friction constant velocity
    • SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net  = = Coefficient of FRICTION STATIC FRICTION and • Can vary from zero to a MAX IMUM of Fs = μs........ μs = coefficient of .......... Friction DYNAMIC (moving) FRICTION. • Always constant value given by Fd = μdN μd = coefficient of .......... friction ....... force Static friction V = .... Constant Force Dynamic friction constant velocity N = -Fg N = -Fg Ff N FA Fg
    • SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net FN (N) Ff N μs = = gradient Ff (N) Ff N μs = = gradient DRY surface WET surface static kinetic Graphs of Frictional force against normal force. 0 1 2 3 0 1 2 3 Ff (N) static kinetic The force of friction Ff is directly proportional to the Normal force (weight) The gradients of these graphs would be EQUAL to the coefficients of friction. FN (N)
    • SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Friction Examples 5kgFA A block is held against a vertical surface by a horizontal force. What is the coefficient of friction if the minimum applied force is 20 N?
    • SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Friction Examples Object on an inclined plane.  What is the maximum angle that the slope can have before the block begins to slide if the coefficient of friction is 0.8?
    • SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net A B Exam Question The diagram shows a truck with a crate immediately behind the cab in position A.The truck suddenly accelerates forward and the crate slides towards the back of the truck and comes to rest in position (B) a) Draw a force diagram, indication and labeling the horizontal force(s) acting on the crate while the truck accelerates. (2) Friction between the crate and truck
    • SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Newton’s Laws Second Law A ..............................on an object causes it to ...................in the direction of that force. The magnitude of the acceleration will be directly proportional to the ...................and inversely proportional to the ..............of the object. Fres = F + f and Fres = m x a F m a = ? F m=1kg T = ? Fg Fres = Fg + T Fres = m x a And Fg = mg These are the two most basic scenarios - any of the variables can be calculated if all the others are known. a=2m.s-2
    • SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Newton’s - Connected objects In both these cases the tension gets ………………… as more blocks are connected I.e. T1 ….. T2 …. T3 The systems can be treated as a …….. …………: F… = ………………………….. Where F … = ………. or F … = ……….. So …………………………………….. T1 T2 T3 m1 m2 m3f m3 m2 m1 T3 T2 T1 Fg To find the tension between blocks you must consider them ……………. I.e. ……………………………………………….
    • SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Newton’s Laws - Pulley systems The accelerations of these systems can be easily found by considering them as a …………….: Fres = ………… where mtotal = ………… And Fres = ………………….. (always opposed) To find tension; ……………………………….. Individual masses or tensions can be found by considering masses individually: ……………………………………... Tension in any one string is the same throughout! m2 m1 T T Fg1 Fg2 m1 m2 m3 T1 T2 Fg1 Fg2
    • SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Examples Newton 1 & 2 &3 5. The frictional force on the system is 10 N (6 N & 4 N). Calculate the force that the 4kg block exerts on the 6kg block. 4kg6kg a = 2m.s-2 (Ff=10N) (6N of 6kg block 4N on 4kg block) Consider 4 kg block: Fnett = ma = (4)*(2) = 8 N Fnett = Fa + Ff .: 8 = Fa + (-4) Fa = 12 N .: F exerted by 4kg on 6kg = 12 N Consider whole system: Fnett = ma = (10)*(2) = 20 N Fnett = F + Ff .: 20 = F + (-10) F = 30 N F
    • SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Examples Newton 1 & 2 2. A ball with a mass of 18kg is suspended by a string. a) Calculate the tension in the string if the ball is i) stationery, ii) accelerates upwards at 3m.s-2; iii) accelerates downwards at 3m.s-2. T = 176.4N upwards T = 230.4 upwards T = 122.4N upwards ii) Fres = ma = (18)*(3) = 54 N Fres = T + fg .: 54 = T + fg but fg = mg = (18)*(-9.8) = -176.4 N .: 54 = T + (-176.4) .: T = 54 + 176.4 = 230.4 N upwards
    • SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Examples Newton 1 & 2 3. Two blocks rest on a frictionless horizontal surface connected by a string. A force of 80N is exerted on the 7kg block. Calculate: a) the acceleration of the blocks b) the force exerted by the string on the 13 kg block. a) Fres = ma .: 80 = (13+7)*a .: a = 80/20 = 4 m.s-2 right 13 kg 7kg 80 N 13 kg T b) Fres = ma .: T = (13)*4 .: T = 52 N right
    • SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Examples Newton 1 & 2 4. A constant force of 120N is used to lift two masses of 4 kg and 6kg which are attached to each other with string. Ignore any mass of the string and calculate the acceleration of the system and the tension in the string. 4kg 6kg 120 N
    • For FULL presentation click HERE >> www.warnescience.net Newton’s 3rd Law "For every action there is an ………… and ………………… reaction.” The gasses experience a force ………………… out of the rocket, this has an equal but opposite …………………. which drives the rocket forward! The force exerted by the hand on the head is the same …………………. as the force exerted by the head on the hand! The head hits the hand just as hard as the hand hits the head!
    • SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net An apple on a table. Non contact forces Gravity • Exerted by the ……… on the apple • Reaction by the ……….. on the earth. Contact forces: Table Apple exerts a force ……… on the table. Table exerts a …………. or …………. force on the apple. • All these forces are …………. so there is no …………… force. • Forces on apple: gravity (down) & reaction (up)
    • SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Newton’s 3rd ANS: Since every action has an equal and opposite reaction, when BOB pushes the van, the van pushes BOB with the same size force but opposite direction. BOB weighs 50 kg, and the van weighs 2,000 kg. Taking right as positive calculate the acceleration experianced by both Bob and the van as well as the distance moved by both. abob= F/m =(-100/50)= -2m/s2 sbob = ut + 1 /2 at2 =0(1)+1/2(-2)(1)= -1m avan= F/m =(100/2000)=0.05m/s2 svan = ut+1/2at2 =0(1)+1/2(0.05)(1)= +0.025m
    • SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Person Walking • The person pushes ……………. on the ground. • The ground pushes …………….. on the person. • The person has an …………. force so moves in the direction of that force. • (The earth………… movement Forces on person
    • SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net 2. Pairs of forces Action-reaction pairs 1. Man against ground - ground against man – greater for …. than for ………. 2. Man on rope – rope on man – all ………….. but less than ……... ground forces and greater than ….. ground forces. 3. A and B both experience a …………………………. 1. A B
    • SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Hi - This is a SAMPLE presentation only. My FULL presentations, which contain a lot more more slides and other resources, are freely available on my resource sharing website: www.warnescience.net (click on link or logo) Have a look and enjoy! WarneScience