Mechanics 1 - Mathsbox

1 Mechanics 1 Revision notes1. Kinematics in one and two dimensionsEQUATIONS FOR CONSTANT ACCELERATION ARE NOT GIVEN Learn Them! Always list the variables you have - write down the equation you intend to use. Sketch graphs essential for multi-stage journeys Retardation / deceleration don t forget the negative signDistance/ Displacement time graphVelocity Time graphv = u + at s = ut +12at2 s = vt 12at2s =12(u + v)t v2 = u2 + 2ass : displacement (m)u : initial velocity (ms-1)v : final velocity (ms-1)a : acceleration (ms2)t.

2 Time (s)Travels 12m from a point X, turns roundand travels 15 m in the opposite directionfinishing 3m behind = VELOCITYS traight line constant velocity zero accelerationChanged directionafter 18secondsMOST USEFUL GRAPH TO SKETCHGRADIENT = ACCELERATIONS traight line constant accelerationDisplacementm vin f rw rogoadsDisplacement movingbackwardsDISPLACEMENTis representedby theareaunder the FALL UNDER GRAVITYA cceleration dueto ms-2 Unless given in thequestionAssumptions- the body is a point mass- air resistance can be ignored- the motion of a body is in a

3 Vertical line- the acceleration due to gravity is constantEXAMPLE : A ball is thrown vertically upwards from ground level with a velocity of 28 ms-1a) What was its maximum height above the ground ?u = 28 ms-1a = ms-2v2 = u2 + 2asv = 0 (top of balls flight)s = ?0 = 282 + 2 ( 9 8)ss = 40 mb) How long did it take to return to the ground ?u = 28 ms-1s = ut +1at220 = 28t +1( 9 8) t220 = t(28 4 9t)t = 0 or t = 5 71t = 0 : time at which ball thrownClearly identify t = asthe final answera = ms-2s = 0 VECTORS Vectors have both magnitude and directionor A = 4i 2jMagnitude : |A| = 2 SPEED= magnitude of the velocity vectorDirection : q = tan 1 2 DIRECTION OF TRAVEL= direction of the velocity vector Unit Vector.

4 A vector with magnitude = 1ji2 42 22+ to the horizontal (i) in anegative directionMagnitude = LengthDirection = AngleIf working inbearings don tforget the 3 digits 025 A = 4 2 4= 26 6 equations for constant acceleration the 5 equations involving, displacement,velocity etc can be used If asked to write an equation in terms of t for displacement/ velocity etc simplifyyour equation as far as possible bycollecting the i terms and j = 2i + 5ja = 4i 8jDisplacementr =ut + at2R = (2i + 5j)t + (4i 8j) t2= (2t + 2t2)i + (5t 4t2)jExampleTwo particles A and B are moving in a plane with the following propertiesA is at point (0,3), has velocity (2i + j) ms-1 and acceleration (i 2j) ms-2B is at point (2,1), has velocity (3i - j) ms-1 and acceleration (2i) ms-2 Find the vector AB six seconds later, and the distance between the particles at that timeDisplacement : in vector formr is used instead of sUsingr = ut +1at220 For A.

5 R= (2i + j) 6 + (i 2j) 36AB= 30i 30jAs A started at (0,3) six seconds later OA = 30i = 56i 5jThis gives AB = OB OA = 26i + 22jDistance AB = magnitude of AB=262 + 222= m Forces can be represented as vectors If forces are inequilibriumthen theresultant (sum of vectors) = 0 Alli components sum to zero and allj components sum to drawn the forces will form aclosed polygon3 forces in equilibriumThe system is in T1 and T2 Method 2 Lami s TheoremMethod 1 Triangle of forcesSketching the 3 forcesgives a can now use thesine rule to findT1 and = 102 NT2 = 116 NsiT1n 52=T2sin 63= 12gsin questions involving rivercrossings and current you mayneed to use the cosine rule aswell as the sine rule to calculatemissing lengths and anglesCosine Rulea2 = b2 + c2 2bc Cos AMethod 2 Lami s theoremT1 =

6 102 NT2 = 116 NT1sin 128=T2sin 117= 12gsin 115 Psin a=Qsinb=Rsin gLAMI S THEOREMFor any set ofthree forces P,Q and R inequilibriumMore than 3 forces in Equilibrium Resolve the forcesExampleFind the resultant of the following system and state the force needed to (i) = -12sin 40 + 10cos 25 6 cos 65 = - (j) = 12cos 40 + 10sin 25 6sin 65 5= = + needed to maintain equilibrium = of N with direction to the positive x-directionTan 1 2 891 1 186 1 1862 + 2 9812 TYPES OF FORCE ALWAYS DRAW A DIAGRAM SHOWING ALL FORCES (with magnitude if known)Weight: mass x (gravity)Reaction(normal reaction).

7 At right angles to the plane of contactSYSTEMS inEquilibrium resolving in the vertically (or in the j direction)40 TR5 kgR + T sin 40 = 5 cos 30 g30 Verticalor j-directionR = 49 NR5 kgR = 5g cos 30 = NR5 kg30 / Thrust pulling or pushing force on the bodyFriction Always acts in a direction opposite to that in which the object is moving ortending to move Smooth contact friction is small enough to be ignored Maximum Friction (limiting friction) - object ismoving or just on the pointof moving : F =mRwherem is the coefficient of friction F <m Rbody not movingIN EQUILIBRIUM resolving horizontally or in the i direction8g8kgFP30 T8kgF40 FT8kgF = T Cos 40 F = TP Cos 30 = F + 8gSin 30 For questions looking for theminimum and maximum forceneeded to for a blockon a slope to move look at.

8 A)P is too small the block is about to slide down the slope (limiting friction)B)P is too large the block is on the verge of sliding up the slopeChange indirection offrictionResolving in the i directionPcos 30 F mg sin 30 = 0 Resolving in the j directionR Psin 30 mg cos 30 = 0 Resolving in the i directionF + Pcos 30 mg sin 30 = 0 Resolving in the j directionR Psin 30 mg cos 30 = 0 NEWTON S LAWS OF MOTION1st LawEvery object remainsat rest or moves withconstant velocity unless anexternal force is applied Constant velocity - system is inequalibrium- net force (resultant force)

9 = 0- in vector form equate the i and j components to LawF = maNet Force = mass x acceleration Always work out and stateNetforceclearly before equating toma Check - if acceleration is positive net force should also be positiveExample : A taut cable 25m long is fixed at 35 to the horizontal. A light rope ring is placedaround the cable at the upper end. A soldier of mass 8 kg grabs the rope ring and slidesdown the the coefficient of friction between the ring and the cable is , how fast is the soldiermoving when he reaches the bottomjii- direction : 784 cos 35 = Rj direction.

10 784 sin35 F = maMotion friction is limiting so F = sin 35 x 784 cos 35 = 80av2 = u2 + 2asa = ms-2u = 0v2 = 02 + 2 2 41 25v = 11 0 ms 1s = 253rd Law For every action there is an equal and opposite reactionConnected Particles Trains and trailersFinding the acceleration (F=ma)10000 kg3000 kgNet Force =14000 4000 1500= 8500 N8500 = (3000+10000)aa = ms-2 Finding the Tension in the coupling To keep it simple - use the body which has no direct force applied the trailerNet Force = T 1500 T - 1500 = 3000 x = N Stings and PulleysAlways draw a diagram if known show direction of accelerationFinding the g T = (+)T = = = ms-2 Force on the pulley = T + TFinding the tensiona= into either equation(or both just to check)