BMSAPB.M.SHARMA ACADEMY OF PHYSICS 1InertiaInertiaThe inherent property of all bodies, by virtue of which they cannot change their state of rest or of uniform motion along straight line on their own is called Inertia. It depends on mass of the body. Monday, August 31, 2009B.M.sharma Academy of Physics22Inertia of RestIt is the inability of a body to change by itself, its state of rest. This means a body at rest remains at rest and cannot start moving on its own.It is the inability of a body to change by itself its direction of motion i.e. a body continues to move along the same straight line llldb Inertia of Motion It is the inability ofa body to change by itself its state of uniform motion i.e. a body in uniform motion can neither accelerate TYPES OF INERTIA3Monday, August 31, 20093B.M.sharma Academy of PhysicsForce is external form of push or pull which, (a)Produces or tries to produce motion in a body at rest.(b)Stops or tries to stop a moving body.(c)Changes or tries to change the direction of motion of body. (d)Produces a change in shape of body.ForceForceMonday, August 31, 2009B.M.sharma Academy of Physics4Classification of ForcesClassification ForcesBased on the nature of the interaction between two bodies, forces may be broadly classified as under (a)Contact Forces :Tension, Normal Reaction, Friction, etc. Forces that act between bodies in contact.(b)Field Forces (Non-Contact Forces) :Weight, electrostatic forces, etc. Forces that act between bodies separated by a distance without any actual contact.Monday, August 31, 2009B.M.sharma Academy of Physics55First Law: According to this law, a body continues to be in its state of rest or uniform motion along a straight line, unless it is acted upon by some external force to change the state.SecondLaw: According to this law, the rate of change of linear momentum of a body is directly proportional o the external force applied on the body, and this change takes place always in the direction of the force applied.Third LawAccording to this law, to every action, there is always an equal and opposite reaction i.e. the forces of action and reaction are always equal and opposite. NEWTON'S LAWS OF MOTION6Monday, August 31, 20096B.M.sharma Academy of PhysicsNewtonNewton’’s Second Law of Motions MotionWhen an unbalanced force is applied on a body, momentum of the body changes, the rate of change of momentum with respect to time is defined as the net external force acting on the body. i.e., =rr;extdPFdtMonday, August 31, 2009B.M.sharma Academy of Physics77=rr,PmVwhere linear momentumm= mass of the body, = instantaneous velocity rV()extddvFmvmdtdt⇒==rrrextFma= r rMonday, August 31, 2009B.M.sharma Academy of Physics88Σ= r r.Fma=ΣxxFma…(i)=ΣyyFma…(ii)=ΣzzFma…(iii)This is a local law. This means that it applies to a particle at a particular instant without taking into consideration any part history of the particle or its motion. Monday, August 31, 2009B.M.sharma Academy of Physics99θΣF=XYΣFXΣFYXYmaXmaY=Monday, August 31, 2009B.M.sharma Academy of Physics1010NewtonNewton’’s Third Law of Motions MotionTo every action there is an equal and To every action there is an equal and opposite reaction. opposite reaction. Whenever a body exerts a force on another body, the second body also exerts a force on the first that is equal in magnitude, is in the opposite direction and has the same line of action, acting simultaneously. Monday, August 31, 2009B.M.sharma Academy of Physics11Free body diagramIn this diagram the object of interest is isolated from its surroundings and the interactions between the object and the surroundings are represented in terms of forces.Monday, August 31, 2009B.M.sharma Academy of Physics1212The Common Forces Encountered in MechanicsThe Mechanics1.Weight2.Normal Force3.Tension4.Frictional Force5.Elastic Spring ForceMonday, August 31, 2009B.M.sharma Academy of Physics1313Another system: the EarthOne system : youGravitational force of the Earth on youGravitational force of you on the Earth(b)Monday, August 31, 2009B.M.sharma Academy of Physics1414F.B.D of a particle in gravitational field. The earth pulls the particle of mass mby a force mg.EarthmgmMonday, August 31, 2009B.M.sharma Academy of Physics1515Now consider a ball projected upon the surface of earth. BallmgmgBallmgSystemSystemEarthEarthMonday, August 31, 2009B.M.sharma Academy of Physics1616Whenever two surfaces are in contact they exert force on each other. Such forces are called contact forces. Contact ForceNFfric.Horizontal componentNormal componentMonday, August 31, 2009B.M.sharma Academy of Physics17We resolve these contact forces into components, one parallel to the contact surface, the other perpendicular to that surface. Contact ForceNFfric.Horizontal componentNormal componentMonday, August 31, 2009B.M.sharma Academy of Physics18NOne system : youForce of the surface on youMonday, August 31, 2009B.M.sharma Academy of Physics1919Force of you on the surfaceAnother system: the surfaceN(a)Monday, August 31, 2009B.M.sharma Academy of Physics2020Force of you on the surfaceAnother system: the surfaceNNOne system : youForce of the surface on you(a)Monday, August 31, 2009B.M.sharma Academy of Physics2121BANBNABANBNABANBNABANBNAMonday, August 31, 200922B.M.sharma Academy of Physics22•If direction of contact force cannot be determined, it should be shown as two components (as shown in figure).NYNXMonday, August 31, 2009B.M.sharma Academy of Physics2323• When contact between two bodies breaks, When contact between two bodies breaks, the normal reaction vanishes. the normal reaction vanishes. • The weighing equipments measure the The weighing equipments measure the normal reaction. normal reaction. Monday, August 31, 2009B.M.sharma Academy of Physics2424mMSystem IMSystem IIThe system that have been identifiedFMonday, August 31, 2009B.M.sharma Academy of Physics2525mMSystem INF(M+ m)gMonday, August 31, 2009B.M.sharma Academy of Physics2626MgSystem IINFN1fMonday, August 31, 2009B.M.sharma Academy of Physics2727mgSystem IIIN1fMonday, August 31, 2009B.M.sharma Academy of Physics2828XYPABANAmAgPBmBgNANBNBGroundMonday, August 31, 2009B.M.sharma Academy of Physics2929XYNmg cos θmgmg sin θθfMonday, August 31, 2009B.M.sharma Academy of Physics3030MmFBody is moving togetherMonday, August 31, 2009B.M.sharma Academy of Physics3131mMFree body diagrammSystem ISystem IIFN1(M+m)gmgRfMonday, August 31, 2009B.M.sharma Academy of Physics3232MAlternate IImSystem ISystem IIMonday, August 31, 2009B.M.sharma Academy of Physics3333MmSystem ISystem IIFN1MgmgRRffMonday, August 31, 2009B.M.sharma Academy of Physics3434Problem 7.Two blocks of masses m1and m2are placed side by side on a smooth horizontal surface as shown in the figure. A horizontal force Fis applied on the block m1.(i) Find the acceleration of each block.m2m1F(ii) Find the normal reaction between the two blocks.Monday, August 31, 2009B.M.sharma Academy of Physics3535Solution.Since the two blocks always remain in contact with each other so they must move with same acceleration.m2m1FMonday, August 31, 2009B.M.sharma Academy of Physics3636m2m1FFN1F.B.D. of the block m1mg1 a= ax a= 0y Monday, August 31, 2009B.M.sharma Academy of Physics3737m2m1FNN2F.B.D. of the block m2mg2 a= ax a= 0y Monday, August 31, 2009B.M.sharma Academy of Physics3838Using Newton’s second law,For block m1F –N = m1a…(i)For block m2N = m2a…(ii)FN1F.B.D. of the block m1mg1 a= ax a= 0y NN2F.B.D. of the block m2mg2 a= ax a= 0y Monday, August 31, 2009B.M.sharma Academy of Physics3939On adding the two equations, we getF = (m1+ m2) a12Famm⇒=+Substituting the value of ain equation (ii), we getFN1F.B.D. of the block m1mg1 a= ax a= 0y NN2F.B.D. of the block m2mg2 a= ax a= 0y 2212FmNmamm⇒==+Monday, August 31, 2009B.M.sharma Academy of Physics4040Alternatively,the situation may be considered as follows: Instead of drawing the free body diagrams of each block we can draw the free body diagram of both blocks together as shown in the figure.m2m1FaF.B.D. of both the blocksconsidered as one.Monday, August 31, 2009B.M.sharma Academy of Physics4141The net force acting on the system is Fand the total mass of the system is (m1+ m2), thusFamm=+12m2m1FaF.B.D. of both the blocksconsidered as one.Monday, August 31, 2009B.M.sharma Academy of Physics4242To find out the normal reaction N between the two blocks we can imagine like this :The block m2is moving with an acceleration am2m1FaF.B.D. of both the blocksconsidered as one.Monday, August 31, 2009B.M.sharma Academy of Physics4343therefore, the net force acting on it must be (m2a) which is nothing but the normal reaction applied by the block m1.m2m1FaF.B.D. of both the blocksconsidered as one.Monday, August 31, 2009B.M.sharma Academy of Physics4444mFNmamm==+2212Thus, m2m1FaF.B.D. of both the blocksconsidered as one.Monday, August 31, 2009B.M.sharma Academy of Physics4545Problem 11.The figure shows an arrangement of two blocks.10 kg5 kg30 N(i) The acceleration of each block is ……(ii) The normal reaction between the two blocks is ……(iii) If the 30 N force acts on the 10 kg block instead of the 5 kg block, then the magnitude of normal reaction is ………..Monday, August 31, 2009B.M.sharma Academy of Physics4646Solution.netFamsmm−===++212302510(i)mNFNmm⎛⎞⎛⎞===⎜⎟⎜⎟++⎝⎠⎝⎠212103020510(ii)mNFNmm⎛⎞⎛⎞===⎜⎟⎜⎟++⎝⎠⎝⎠1125'3010510(iii)10 kg5 kg30 NMonday, August 31, 2009B.M.sharma Academy of Physics47473.Tension :We introduce a physical quantity, tension, which is a property of rope or a point of the rope, and which equals the magnitude of the force with which the rope pulls the body connected to it. Monday, August 31, 2009B.M.sharma Academy of Physics4848F.B.D. of a ball suspended by a string. Two vertical forces act on the ball: the earth pulls the ball downward by mg and the string pulls the ball upwards by T.MgTmMonday, August 31, 2009B.M.sharma Academy of Physics4949Force of the string on manThe system is man(a)Monday, August 31, 2009B.M.sharma Academy of Physics50Force of the string on the wallThe system is the wall(a)Monday, August 31, 2009B.M.sharma Academy of Physics51Force of the wall on the stringThe systemForce of your friend on string(b)Monday, August 31, 2009B.M.sharma Academy of Physics52Problem 10.A spring balance is equally pulled at the two ends as shown in the figure. The reading on the balance is ……10 N10 NSolution.A single force cannot stretch a spring it can simply accelerate it. A spring gets stretched by a force of 10 N if it is pulled equally and oppositely with a force of 10 N at its ends.Monday, August 31, 2009B.M.sharma Academy of Physics5353TTABSystemPullTTABSystemPullSystemMonday, August 31, 2009B.M.sharma Academy of Physics5454m1m2m2> m1`F.B.D.of m1m1gTF.B.D.of m2m2gTMonday, August 31, 200955B.M.sharma Academy of PhysicsBAPulley 2String 1String 3Pulley 1String 2mB(III) F.B.D of Pulley 2mA(I) F.B.D of Mass A(II) F.B.D of Mass B(III) F.B.D of Pulley 1Monday, August 31, 2009B.M.sharma Academy of Physics56BAPulley 2String 1String 3Pulley 1String 2mBmAT1T1T21T2mBgT1T3AT1T1mAgMonday, August 31, 2009B.M.sharma Academy of Physics57(I) F.B.D. of AString 1BAString 2String 3String 1(II) F.B.D. of B(III) F.B.D. of C(III) F.B.D. of pulleyCMonday, August 31, 2009B.M.sharma Academy of Physics58T1mBgT2BT1T3AmA gT1T1T3T2ABCmCgCMonday, August 31, 2009B.M.sharma Academy of Physics59m1m2m2> m1`F.B.D.of m1m1gTF.B.D.of m2m2gTMonday, August 31, 200960B.M.sharma Academy of Physicsm1m2m2> m1`F.B.D.of m1m1gTF.B.D.of m2m2gTEquation of motion of m222mgTma−=…(i)for m111Tmgma−=…(ii)2112()()mmgamm−=+Monday, August 31, 200961B.M.sharma Academy of PhysicsF.B.D.of m1211112mmTmgmgmm⎛⎞−−=⎜⎟+⎝⎠T12122()mmgTmm=+m1gF.B.D.of m2Tm2gMonday, August 31, 2009B.M.sharma Academy of Physics62m1m2gm1gm2Monday, August 31, 2009B.M.sharma Academy of Physics6363m1m2gm1gm2m2m1m1gm2gSystemMonday, August 31, 2009B.M.sharma Academy of Physics6464Net external force Fext= m2g –m1gFext= (m2–m1)gFext= (m2–m1)g = (m1+ m2)a2121()()mmagmm−=+m2m1m1gm2gSystemMonday, August 31, 2009B.M.sharma Academy of Physics6565calculate the acceleration of the system.m1m2Monday, August 31, 2009B.M.sharma Academy of Physics6666Free body diagrams:-NaTm1m1gTam2m2gm1m2Monday, August 31, 2009B.M.sharma Academy of Physics6767Equation of motion 2Tma=For ‘m1’…(i)For ‘m2’22mgTma−=…(ii)Free body diagrams:-NaTm1m1gTam2m2gm1m2Monday, August 31, 2009B.M.sharma Academy of Physics6868NaTm1m1gTam2m2gm1m2From (i) and (ii)212()mgmma=+212()mgamm⇒=+Monday, August 31, 2009B.M.sharma Academy of Physics6969Method 2 :-Taking m1and m2together as a systemm2m1m2gMonday, August 31, 2009B.M.sharma Academy of Physics7070m2m1m2gFor external force Fext= m2gFext= (m1+ m2)a212()mgmma=+212()mgamm⇒=+Monday, August 31, 2009B.M.sharma Academy of Physics7171Find the acceleration of the system.m2m(2)(1)θMonday, August 31, 2009B.M.sharma Academy of Physics7272Free body diagrams :-NTmg sin θmg cos θTa2m2mgF.B.D of (2)F.B.D of (1)Monday, August 31, 2009B.M.sharma Academy of Physics7373Equation of motion sinTmgma−θ=…(i)22mgTma−=…(ii)from equation (i) and (ii)12sin(2)mgmgmma−θ=+(2sin)3ag−θ⇒=NTmg sin θmg cos θTa2m2mgMonday, August 31, 2009B.M.sharma Academy of Physics7474θmg sin θ2mg22m1mMethod ‘2’Monday, August 31, 2009B.M.sharma Academy of Physics7575Taking (1) and (2) as a system togetherm2m1Mg sin θ2mgm2m(2)(1)θmg sin θ2mgMonday, August 31, 2009B.M.sharma Academy of Physics7676m2m1Mg sin θ2mgm2m(2)(1)θmg sin θ2mg2sin(2)mgmgmma−θ=+(2sin)3ga−θ⇒=Net force Fext= 2mg –mg sin θMonday, August 31, 2009B.M.sharma Academy of Physics7777Illustration 1.Three boys, each of mass 45 kg, pull simultaneously a block on a smooth surface. Mass of block is 20.0 kg.(a)Find the acceleration of the block.(b)Find the instantaneous acceleration of the boy exerting the force F1.45°37°F3F2F1Given F1=90N, F2=114N and F3=1282NMonday, August 31, 2009B.M.sharma Academy of Physics787845°37°F3F2F1−+45°37°F1F2+−F3Given F1=90N, F2=114N and F3=1282NMonday, August 31, 2009B.M.sharma Academy of Physics797945°37°F1+−F3+SolutionSolutionFirst we will resolve all the forces acting on the block into xand ycomponents.xFFFΣ=°+°13cos37cos45F2Monday, August 31, 2009B.M.sharma Academy of Physics808045°37°F1+−F3+SolutionSolutionFirst we will resolve all the forces acting on the block into xand ycomponents.yFFFFΣ=°+−°123sin37sin45xFFFΣ=°+°13cos37cos45F2Monday, August 31, 2009B.M.sharma Academy of Physics818145°37°F1+−F3+SolutionSolutionFirst we will resolve all the forces acting on the block into xand ycomponents.yFFFFΣ=°+−°123sin37sin45Now,xFFFΣ=°+°13cos37cos45and yxxyFFaammΣΣ==2(90.0)cos371282cos45.010m/s20.0xa°+°==2(90.0)(sin37)114(1282sin45.0)2m/s20.0ya°+−°==F2Monday, August 31, 2009B.M.sharma Academy of Physics8282F1F’1According to Newton’s third law the force exerted on the block by the boy must be equal to the force exerted on the boy by the block.Therefore,211'90.02m/s45Fam===Monday, August 31, 2009B.M.sharma Academy of Physics8383Illustration 2.A bucket is suspended by two light ropes aand bas shown in figure. Determine the tension in the ropes aand b.mφθTbTaφθTaTbYmgTasin θTbsin φTacos θTbcos φXMonday, August 31, 2009B.M.sharma Academy of Physics8484SolutionSolutionLight rope implies that weight of rope is negligible as compared to the force it exerts. Since the bucket is at rest, its acceleration is zero. Thus Newton’s second law gives0and 0xyFFΣ=Σ=φθTaTbYmgTasin θTbsin φTacos θTbcos φXMonday, August 31, 2009B.M.sharma Academy of Physics8585φθTaTbYmgTasin θTbsin φTacos θTbcos φXTherefore,coscos0xabFTTΣ=−θ+φ=sinsin0yabFTTmgΣ=θ+φ−=From equation (i),coscosabTTθ=φ…(i)Monday, August 31, 2009B.M.sharma Academy of Physics8686φθTaTbYmgTasin θTbsin φTacos θTbcos φXOn substituting Tbin equation (ii), we get, cossinsin0cosaaTTmgθφθ+−=φOr,sincostanbmgT=θ+θφMonday, August 31, 2009B.M.sharma Academy of Physics8787Problem 3.State the following statements as TRUE or FALSE(iii) The force of tension on a body always act away from the body.(iii) TRUE : Tension is always a pulling force.(iv) The normal force always act towards the body(iv) TRUE : Normal reaction is always a pushing force.Monday, August 31, 2009B.M.sharma Academy of Physics8888(v) Friction force always opposes the motion of a body.(v) FALSE : Friction force opposes the relative motion between the two bodies.(vi) Spring force is a constant force.(vi) FALSE : Spring force is proportional to deformation of the spring.i.e., F = kxWhere kis a constant.Monday, August 31, 2009B.M.sharma Academy of Physics8989(vii) In equilibrium, a body must be stationary.(vii) FALSE : Equilibrium means no acceleration (i.e., a = 0). The body may be stationary or moving with constant velocity.(viii) A body always moves in the direction of force.(viii) FALSE : A body accelerates in the direction of force.Monday, August 31, 2009B.M.sharma Academy of Physics9090(ix) In a light string, tension is constant throughout its length.(ix) TRUE : Mass of the string changes the tension in the string.(x) If a body is at rest it means there is no force acting on the body.(x) FALSE : If a body is at rest it means there is not net force acting on the body.Monday, August 31, 2009B.M.sharma Academy of Physics9191Problem 4.Determine the tensions T1and T2in the strings..T1= ………..T2= ………..10 KgT1T2Monday, August 31, 2009B.M.sharma Academy of Physics9292Solution.A horizontal string cannot balance a vertical weight.ΣFy=0 T1= 100 NΣFx= 0 T2= 0T2T1100 NF.B.D. of the block10 KgT1T2Monday, August 31, 2009B.M.sharma Academy of Physics9393Problem 2.A ball of mass m = 10 kg is suspended with the help of three strings as shown in the figure. Find the tensions T1, T2and T3.T2T3600300T1m = 10 kgA ball suspendedfrom an arrangementof strings is in equilibrium.Monday, August 31, 2009B.M.sharma Academy of Physics9494T2T3600300T1m = 10 kgA ball suspendedfrom an arrangementof strings is in equilibrium.T1100 NF.B.D. of the ball(b)Solution.T2T3O600300100 N(c)F.B.D. of the point ‘O’Monday, August 31, 2009B.M.sharma Academy of Physics9595Considering the equilibrium of the ball, we getT1= 100 NT1100 NF.B.D. of the ball(b)T2T3600300T1m = 10 kgA ball suspendedfrom an arrangementof strings is in equilibrium.Monday, August 31, 2009B.M.sharma Academy of Physics9696T2T3O600300100 N(c)F.B.D. of the point ‘O’At point ‘O’23sin30sin600xFTTΣ=−+=23cos30cos601000yFTTΣ=+−=23sin30sin60TT=231322TT=233TT=Monday, August 31, 2009B.M.sharma Academy of Physics9797T2T3O600300100 N(c)F.B.D. of the point ‘O’At point ‘O’23cos30cos601000yFTTΣ=+−=233TT=33313100022TT⎛⎞⎛⎞+−=⎜⎟⎜⎟⎝⎠⎝⎠33210050TTN=⇒=2503TN=Monday, August 31, 2009B.M.sharma Academy of Physics9898Problem 5.Determine the tension T4in one step.T4= ………..10 KgT4T2T1T3300600Monday, August 31, 2009B.M.sharma Academy of Physics9999Solution.The free body diagram is drawn so that only the string T4is cut, as shown in the figure.ΣFy= 0T4cos 600= 100 NT4= 200 N100 NT430060010 KgT4T2T1T3300600Monday, August 31, 2009B.M.sharma Academy of Physics10010Problem 12.A uniform rod AB of mass m and lengthl is pulled with a constant force Fon a smooth horizontal surface. The tension in the rod at a distance xfrom the end Ais………xLFMPMonday, August 31, 2009B.M.sharma Academy of Physics10110Acceleration of rope a= F/MFind the tension of point PFMP(L-x)Part IPart IIxFTxTxMonday, August 31, 2009B.M.sharma Academy of Physics102Writing equation of motion for part I1()XTMa=1()MMLxL=−XMFFLxTLxLML−⇒=−×=()()FMP(L-x)Part IPart IIxFTxTxMonday, August 31, 2009B.M.sharma Academy of Physics103And rope of mass M, length L, and a block is attached at lower end of the rope and at the top end a force F= 33 Mg is applied vertically. Calculate(i)Tension of P(ii)Tension at A(iii)Tension at C2MPABLCMxF=33 MgMonday, August 31, 2009B.M.sharma Academy of Physics104F.B.D. of system (rope + block)333333FMgMgMgaMM−−==10ag=FSystem (rope + block)3 Mg2MPABLCMxF=33 MgMonday, August 31, 2009B.M.sharma Academy of Physics105M2gFM1gTXPxFSystem 1System 2Monday, August 31, 2009B.M.sharma Academy of Physics106Equation of motion :System I : 11XFMgTMa−−=1()MMLxL⎛⎞=−⎜⎟⎝⎠()()10gxMMFLxgTLxLL−−−=−FM1gTXPxFSystem 1System 211()xMTFLxgL=−−Monday, August 31, 2009B.M.sharma Academy of Physics107At x= 0 (at A)11(0).AMTFLgL=−−1122ATFMgMg=−=At x= L(at C)033CTFFMg=−==2MPABLCMxF=33 Mg11()xMTFLxgL=−−Monday, August 31, 2009B.M.sharma Academy of Physics108Find the tension (i) at A(ii) at A(iii) at Pwhich is at a distance xfrom end B.MxFLAABBPPMonday, August 31, 2009B.M.sharma Academy of Physics109MxFLAABBPPMTAAAxFLBBPPMonday, August 31, 2009B.M.sharma Academy of Physics110Taking ‘rope + block’as a system()FaMm=+Tension at B; TB= F()AMFTMaMm==+MxTAFLMonday, August 31, 2009B.M.sharma Academy of Physics111'XFTMa−='.mMxL=()XmFFTxLMm⎛⎞−=×⎜⎟+⎝⎠()XmxFTFMmL=−+FTxMxFLAABBPPMonday, August 31, 2009B.M.sharma Academy of Physics112If person acc. up with acceleration a, find the tension in rope.(Assume rope to be massless)aMMonday, August 31, 2009B.M.sharma Academy of Physics113Solution :F.B.D. of manT MgaEquation of motion of manTMgMa−=()TMga=+aMMonday, August 31, 2009B.M.sharma Academy of Physics114If man is sliding down with acceleration aMgTMa−=()TMga=−T MgaaMMonday, August 31, 2009B.M.sharma Academy of Physics115Question : If breaking strength of rope is 10 Mg, then calculate maximum possible acceleration of man.aMMonday, August 31, 2009B.M.sharma Academy of Physics116TMgMa−=10MgMgMa−=9ag=T MgaaMMonday, August 31, 2009B.M.sharma Academy of Physics117Calculate maximum possible acceleration of man of mass mso that Mdo not loose contact with ground.Question : MmMonday, August 31, 2009B.M.sharma Academy of Physics118F.B.D. of MSolution : MgTNIf Mloose contact with ground T= MgMmMonday, August 31, 2009B.M.sharma Academy of Physics119F.B.D. of manTmgma−=Mgmgma−=()Mmagm−=T MgaMmMonday, August 31, 2009B.M.sharma Academy of Physics120If man starts moving up with acceleration g/3 with respect to rope then calculate :Question : (i)Acceleration of M(ii)Tension in stringMmaGiven : M= 10 mMonday, August 31, 2009B.M.sharma Academy of Physics121Solution : Acceleration of man,mmroperopeaaa=+rrrGiven : M= 10 m(up)3ga⎛⎞=+↑⎜⎟⎝⎠T MgamMgTaMmaMonday, August 31, 2009B.M.sharma Academy of Physics122Equation of motionFor m3gTmgma⎛⎞−=+⎜⎟⎝⎠…(i)For MMgTMa−=…(ii)T MgamMgTaMmaMonday, August 31, 2009B.M.sharma Academy of Physics123F.B.D. of a block supported by a spring of stiffness k. Two vertical forces act on the block: the gravitational pulling, and the spring force kx.kx MgkMxkMonday, August 31, 2009B.M.sharma Academy of Physics12412