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FREE BODY DIAGRAM (FBD), , Step 1: Select the body, The first step is to decide the body on which the laws of motion are to be applied. The body may bea, single particle, an extended body like a block, a combination of two blocks-one kept over another or, connected by a string. The only condition is that all the parts of the body or system must have the, same acceleration., , Step 2: Identify the forces, Once the system is decided, list down all the force acting on the system due to all the objects in the, environment such as inclined planes, strings, springs etc. However, any force applied by the system, shouldn’t be included in the list. You should also be clear about the nature and direction of these, forces., , Step 3: Make a Free-body diagram (FBD), Make a separate diagram representing the body by a point and draw vectors representing the forces, acting on the body with this point as the common origin., This is called a free-body diagram of the body., , , , , , , , A, T, T,, R, B, 5 6 R, %, 8, g Ww, w,,, —-100kg platform man, FB.D of Diagram, , , , , , 50 k, Look at the adjoining free-body diagrams for the platform and the man. Note that the force applied by, the man on the rope hasn’t been included in the FBD., Once you get enough practice, you’d be able to identify and draw forces in the main diagram itself, instead of making a separate one, , Step 4 : Select axes and Write equations, When the body is in equillibrium then choose the axis in such a fashion that maximum number of force, lie along the axis., If the body is moving with some acceleration then first find out the direction of real acceleration and, choose the axis one is along the real acceleration direction and other perpendicular to it., Write the equations according to the newton’s second law (F,,, = ma) in the corresponding axis., , 4. APPLICATIONS :, , 4.1 Motion of a Block on a Horizontal Smooth Surface., Case (i) : When subjected toa horizontal pull :, The distribution of forces on the body are shown. As there is no motion along vertical direction, hence,, R=mg, For horizontal motion F = ma ora = F/m R, —a, , Pm |—or, , mg, Case (ii) : When subjected to a pull acting at an angle (0) to the horizontal :, Now F has to be resolved into two components, F cosé@ along the horizontal and F sin 6 along the, , vertical direction. R sino, , é, m Ae Fcos@, TTT, , mg, , , , For no motion along the vertical direction.

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ew:, , we haveR+Fsin8 =mg, or R=mg-Fsine, Hence R#mg. R< mg, , For horizontal motion, , Foos0, Fcos @=ma, a= ——, , Case (iii) : When the block is subjected to a push acting at an, , 4.2, , , , , , , , , , , , , , , , angle @to the horizontal : (down ward) R, The force equation in this case F, R=mg+Fsin6e, Fcos@, R#mg,R>mg, For horizontal motion c, Fcos6 Fsin@, , F cos@ = ma, a=, m, , Motion of bodies in contact., Case (i) : Two body system:, Let a force F be applied on mass m,, , B, , , , A, , —>| fc f, m, m,, , , , , , , , , , , , Free body diagrams :, (vertical force do not cause motion, hence they have not been shown in diagram), , , , , , , , (i) Here f is known as force of contact., (ii) Acceleration of system can be found simply by, , , , , , , , , , , , , , , , , , , , force, , 9 = ‘otal mass, , : If force F be applied on m,, the acceleration will remain the same, but the force of contact will be, , different

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ExX.6, , Sol., , EX.7, , Sol., , Find the contact force between the 3 kg and 2kg block as shown in figure., B, , F, = 100 a 2g [fe 258, 1 = 1OON) 3kg g |<, TITTT TT, , Considering both blocks as a system to find the common acceleration, , vet =F, - Fy = 100- 25=75N, common acceleration Fg, , Fett: 15. 2, , 5kg 5 15m/s', , To find the contact force R, between A & B we draw a, F.B.D of 2 kg block N45} akg —25N, from (SF), = Ma,, => N-25'= (2) (15), => N=55N, , , , , , , , , , , , =75N, 5kg — >a, , , , , , , , , , , , , , , , Case (ii) : Three body system:, Cc, , , , a _8, , , , , , F— ym, | ™]™, TTT TTL., , , , , , , , , , , , , , Free body diagrams:, F ForA ForB ForC, , m,+m, +m,, _(Mp +Mg)F_ Pea eel, (m; +m +mg), , maF F-f,=m,alf,-f,=m,a] f,=m,a, , >a=, , , , , , , , , , , , , , , , , , , , and f, =, , , , , , , , , , , , £ = (mam, +ms), f, = contact force between masses m, and m,, f, = contact force between masses m, and m,, Remember : Contact forces will be different if force F will be applied on mass C, , Find the contact force between the block and acceleration of the blocks as shown in figure., Cc, , , , , , B, A, F, = 30N, F,=SON) akg | 5kg 3kg [te, TTT TITTTTT, , , , , , , , , , , , , , Considering all the three block as a system to find the common acceleration, Fig = 50-30=20N, , , , a= 2 -2m/s* Fra=20N | 10kg |——>a, , , , , , , , To find the contact force, between B & C we draw F.B.D. t a, of 3 kg block., , (Fret), ma J sa, = N,-30=3(2)>N, =36N ng, To find contact force between A & B we draw —a, F.B.D. of 5 kg block, = N,-N,=5a Ne 5kg N,, , N,=5x2+36>N,=46N, , , , Ny 3kg 30N

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4.3 Motion of connected Bodies, , Case (i) For Two Bodies :, F is the pull on body A of mass m,. The pull of A on B is exercised as tension through the string, connecting A and B. The value of tension throughout the string is T only., , B, Waits A, m, re, , TTT TTT, , , , , , , , , , , , , , , , Free body diagrams:, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , For body A For body B, R, —a R, t —>a, e A F B Fert, mg m.g, R,=mg R,=mg, F-T=ma T=ma, F, = m;+my, , Case (ii) : For Three bodies :, , , , , , , , , , , , , , , , , , , , , , a, A B -7C 3, G if E, m, m, ms;, TTT TTT, Free body diagrams:, ForA For B For C, Ry Ry R,, , , , , , , , , , , , ABTIA BRIS CR, T, T, F, , =, , , , , , , , mg m,g mg, , , , R,=m,g, F-T,=m,a, =>F=ma+T,, T, = (m, + m,)a [=Ma+(M,+m,)al, F=(m,+m,+m,)4, , R,=mg, T,=ma, , , , , , , , , , , , , , ~ My +Mz+mM3

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Ex.8 A5kg block has a rope of mass 2 kg attached to its underside and a 3 kg block is suspended, from the other end of the rope. The whole system is accelerated upward is 2 m/s? by an external, force F,,, Fe, , 5kg, 2kg, 3kg, , , , , , , , , , , , , , , , , , (a) What is F,?, (b) What is the force on rope?, (c) What is the tension at middle point of the rope?, (g = 10 m/s?), Sol. For calculating the value of F,, consider two blocks with the rope as a system., F.B.D. of whole system, , F,, (a) amis’, 10g = 100N, F,- 100 = 10x 2, F=120N .(1), (b) According to Newton’s second law, net force on rope., F=ma =(2)(2) =4N (2), (c) For calculating tension at the middle point we draw T, F.B.D. of 3 kg block with half of the rope (mass 1 kg) as, shown., T-4g = 4.(2) =48N 4g, , 4.4 Motion of a body on a smooth inclined plane:, , Natural acceleration down the plane = gsin9, Driving force for acceleration a up the plane, F=m(a+ gsine), and for an acceleration a down the plane, F=m(a - gsin@), , , , Ex.9 Find out the contact force between the 2kg & 4kg block as shown in figure., AB?, , Sol. Onan incline plane acceleration of the block is independent of mass. So both the blocks will move with, the same acceleration (gsin 37°) so the contact force between them is zero.