Newton's Law of Motion - Mind Map - Arjuna Neet 2024 PDF

INERTIA NEWTON‛S THIRD LAW Inclined Forces A body cannot change its state of rest or -To every action, there is always an equal...

INERTIA NEWTON‛S THIRD LAW Inclined Forces A body cannot change its state of rest or -To every action, there is always an equal Fsin If, Fsinθ M2 T unit and no dimension. - A single isolated force is not possible. T reaction M Fcos M1- M2 a= F = If, Fsinθ =Mg a a Reading of - Counter force experienced by a body- reaction = force exerted b INERTIA block just leaves g weighing machine contact with ground M M1 + M2 T weighing machine - Action and reaction never act on the same body Mg 2M1M2 M2 T Apparent weight ,(Wapparent) = Reaction If, Fsinθ >Mg T = g M2g force (R) Inertia Inertia Inertia of * Force exerted on body A by body B (action ) M1 + M2 M1 Case 1: Lift is at rest the block leaves contact with ground and M1g of Rest of Motion Direction * force exerted on body B by body A (reaction) R = mg , Wapparent = Wactual = mg it begins to accelerate obliquily. - Inertia of Rest Reaction (on A) Inability to change state of rest by itself. FAction=-FReaction Case 2: Lift moving up or down with A MOTION OF CONNECTED BODIES N - Inertia of Motion a a constant velocity B F1 > F2 R = mg , W Inability of a body to change its state of Some common forces Action (on B) T apparent = Wactual = mg 1.Normal Reaction M2 uniform motion by itself. F1 - F2 = Ma Case 3: Accelerated upward at - Occurs when two surfaces are in contact with F2 M F1 M1g - Inertia of Direction each other F1 -F2 a a= a rate of ‘a‛ a= M1 + M2 Inability of a body to change direction of Always perpendicular to the surface. M T R-mg=ma = R=m(g+a)=Wapp M2g motion by itself. NBW M1 a M1M2 Wapparent > Wactual → Feels over weight Newton‛s Second Law NBg=NgB NBg NBW=NWB T = g Accelerated upward at a rate of ‘g‛ Fnet= Rate of change of linear momentum. (normal on block by NWB NWg=NgW M1 + M2 ground) NWg R -mg = mg , R = 2mg , W = 2 x W M1g app act Instantaneous Case 4: Accelerated downward at (normal on Average NgB ground by Tension block) a rate of ‘a‛ p pf-pi NgW dp F= Fav= = mg - R = ma , R = m(g-a) = Wapp , W < W dt t t app act INCLINED PLANE + PULLEY Restoring force developed Accelerated downward at a rate of ‘g‛ [ Freefall] MOMENTUM T T F when a longitudnal force is mg - R = mg , R = mg - mg =0 , W = 0 app P=mv applied on a body If a > g : -It is a vector quantity having direction body leaves contact from ground and begins free-fall same as that of velocity -Unit is kg m/s. Ideal Rope Lift moves from ground floor to first floor T v=0, a=0 N=mg (true weight) here,f, f1 and f2 are C L LIQUID JETS normal reactions v decrease N=m(g-a2) first floor *Massless T between blocks When jet is stopped at wall *Tension is same everywhere a2 (retardation) u wall *Opposes only elongation B v Increase Area=A m *On compression it becomes slack. F N=m(g+a1) x *Tension always acts away from the object. MOTION OF BLOCKS CONNECTED BY MASSLESS STRING T=m2a T=m1m2 sinθg a1 v=0 Ground floor m1+m2 A L Pf-Pi mx0-mu Rod v=0, a=0 Fjet= = N=mg (true weight) Δt Δt Graph T F T F -mu = =-ρAu (m=ρAΔx,2 Δx = u) Δt B Δt a=0 ) 1 +a Elongation Compressive ) } = ρAu2 (g 1 Fwall +a N2 N2 =m lift moves =m =m (g can support both elongation and compression =m with constant 1 N (g (g V V THICK ROPE -a 2 -a 2 speed 1 N When liquid jet bounce back ) ) C Tension will be different at different points. 1 2 1 2 +ve A Pf-Pi -2mu t t Fjet= = x Δt Δt u F Acceleration,a1=tanθ1 Fjet = -2ρAu 2 u L Retardation,a2=tanθ2 -ve M Fwall = 2ρAu 2 Mass per unit length= L M When liquid jet strikes obliquely Mass of x length of rope = L x FRAME OF REFERENCE & PSEUDO FORCE SINGLE BLOCK usin Note : Frame of Reference Horizontal Force total mass given Mass of given length= x u total length length A frame in which observer is situated Acceleration is along x-axis only ucos mass and makes his observation Along y-axis ay=0, = constant length x N Inertial frame of Non-Inertial frame arope = F F usin N - Mg = May a M L reference of reference u ay=0 M M (L-x) = m2 x= m1 F For(L-x) rope length, L L At rest or moving with N = Mg LAWS OF MOTION ucos M F uniform velocity along F=ma = T= m2 Accelerated frame Along x-axis F M straight line. of reference. Fjet = -2ρAu cosθ 2 Fwall = 2ρAu cosθ 2 a= i.e unaccelerated F - 0 = Ma M Mg Change in momentum=-2mu cos θ Newton‛s law of motion hold‛s Newton‛s law of motion not applicable. Fnet + Fpseudo = ma Fnet=ma Fpseudo = -mao  a RELATIVE SLIPPING ROD SLIDING ON A WALL HORIZONTAL TRUCK BOX ANGLE OF FRICTION Minimum force required to push the inclined plane Angle(θ) made by resultant of normal iv) When slipping has started, Case-1 m B F such that “m” does not slip with respect to “M” vcos vsin (N) & frictional force(fs) with normal f=fk Box does not slip. M F = (m+M) gtan , a = gtan v fs < µs N Rmax = N 1+µk2 F = ma < µ mg fs T s u usin R N = aT < µsg Minimum acceleration of Minimum mass Mo such that A u x R = N2 + fs2 R is resultant “M” such that there is no there is no relative slipping a velocity of B towards A = vsin ucos θ = Angle of kinetic friction m (fs)max aT relative slipping. fs a velocity of A away from B = ucos m2 these velocities should be equal, tan θ = m F m N M = vsin = ucos f k= N m1 = v = ucot When fs =(fs)max R = k a = g Mo = M+m mg Case-2 m2 M m1 cot - 1 tan θ = µs , µs is Coefficient of friction Box slips F M0 FRICTION aT > µkg M0g i) f = 0 ANGLE OF REPOSE Static friction tanθ=0 R Angle made by inclined - It is a self adjusting force. θ=0 N = mg plane such that a block kept N (fs)max m fk aT EQUILLIBRIUM & LAMI‛S PARALLELOGRAM LAW - The opposing force that comes into play, on it just begins to slide THEOREM when object tends to slip over the surface Depends only on µs and m F1 F2 F3 of other object, but slipping has not m = = fs = 0 is independent of mass. sin sin sin yet started. F1 F - As applied force increases static friction tanθ0 = µs o mgcos o VERTICAL TRUCK BOX F3 F2 mgsin also increases. o o aT - The body doesn‛t move until a maximum mg Variation of R mg value of static friction is attained fs , ii) 0 < fs < s k - This value is called limiting friction or (fs)max (fs)max As angle of inclined plane increases, R remains F1 M N constant (=mg) and when sliding starts R starts F2 R N decreasing. mg F2 F = F1 + F2 , F = F12 + F22 , tan = F (fs)max = µsN F1 N Variation of angle of friction. m MAN-CAGE PROBLEM IMPULSE fs F As angle of inclined plane increases, angle of Case-1 Man holds the cage stationary If a large force acting for short period fs friction will also increase and as sliding Box does not slip. of time, there will be a sudden change mg starts its value becomes constant and tanθ=µk fs < µs N in momentum = mg < µ ma Kinetic friction Minimum & Maximum force (applied parallel to s T µs > g dp 0 < tanθ < µs m+M Fimp = inclined plane) = T = g dt If the applied force is increased further and N < R < N 1+µs2 aT in T 2 m F t slipping between surfaces start, the friction a=0 Fmin = mg ( sinθ - µscosθ) Case-2 T m-M Pf-Pi = Fimp dt = area under F-t graph N N = 2 g o t opposing the slipping is called kinetic friction. iii) fs= (fs)max (fs)max Box slips m µk < g mM o F-t graph (fs)max angle of inclined plane is aT in mg Rmax = N 1+µs gs man cannot hold =µ N greater than angle of repose m the cage stationary s F fk = µN LAWS OF = k Mg N Rmax fs N tan θmax =µs ax m F a=0 u=0 Case-1 f kinetic region Fmax is the maximum force required to MOTION Static region max m F keep the block stationary on the inclined plane if the angle of inclined plane v=0 Impulse Average impulsive (fs)max= N s is greater than angle of repose h1 I = Pf - Pi force in gs h2 = mV2 -(-mV1) impulse P (fs)max m Favg= = t v2 to I = m (V1+V2) o PULLING FORCE & PUSHING FORCE Fmax = mg ( sinθ + µscosθ) v2 = 2gh2 m(v1+v2) = µsmg v1 = 2gh1 v1 to Fsin F Fcosθ ≥ µs(mg-Fsinθ) Fpulling = mg(sinθ - µscosθ) < F < mg (sinθ + µ cosθ) s cosθ + µssinθ µsmg u=0 v=0 Case-2 F≥ Fcos cosθ + µssinθ Average impulsive M Impulse fs force h h m(v+v) = J P 2mv s k Fcos Fcosθ ≥ µs(mg+Fsinθ) Favg= = µsmg µsmg = m(2v) t to m Fpushing = Fsin F≥ v J = 2mv cosθ - µssinθ cosθ - µssinθ fs v Fpushing > Fpulling