BBrilliant STUDY CENTRE PHYSICS (ONLINE) CHAPTER - 00 LAWS OF MOTION Concept of Force Force is pull or push. It is an interaction between two objects or between an object and its environment. Forces in nature 1. Gravitational force : it is the force of mutual attraction between the objects by virtue of their masses 2. Electromagnetic force : force between the objects due to charges on them 3. Strong nuclear force : the strong nuclear force binds protons and neutrons in nucleus 4. Weak nuclear force : it appears only in certain nuclear process such as β decay of a nucleus Based on nature of interaction between two bodies, forces may be broadly classified as under Contact force These forces act between bodies in contact. Tension, normal reaction, friction etc. Noncontact forces [Field forces] These are the forces in which contact between two objects is not necessary Weight, electrostatic forces etc. 1
BBrilliant STUDY CENTRE PHYSICS (ONLINE) Newton’s laws of motion First Law [Law of inertia] If the net force acting on a body is zero, it continues in its state of rest or uniform motion in a straight line ∑If = 0 then a = 0 Fnet When a body is at rest or moving with constant velocity we say that the body is in equilibrium Newton’s first law was called the law of inertia by Galileo. Inertia The tendency of a body to remain at rest or keep moving once it is set in motion results from a property known as inertia. Mass of a body is the measure of its inertia Types of inertia Inertia of rest : It is the inability of a body to change by itself its state of rest Example 1 : A person who is standing freely in bus, thrown backward when bus starts suddenly Inertia of motion : it is the inability of a body to change by itself its state of uniform motion. i.e. a body in uniform motion can neither accelerate nor retard by its own Example 1 : A person jumping out of a moving train may fall forward Example 2 : An athlete runs a certain distance before taking a long jump. Inertia of direction : it is inability of a body to change by itself its direction of motion Example : the rotating wheel of any vehicle throw out mud, if any, tangentially, due to directional inertia. Linear momentum : of a body is the amount of motion possessed by the body. Mathematically it is equal to the product of mass and velocity of the body. Thus Momentum = mass × velocity p = mv Momentum is a vector quantity, its direction is along the direction of velocity. Its SI unit Kg m/s Newton’s Second Law of Motion According to this law, the rate of change of linear momentum of a body is directly proportional to external force applied on the body and this change takes place always in the direction of force applied = dp p = mv Fext dt = d (mv ) = m dv + v dm Fext dt dt dt IF m = constant dm = 0 dt 2
BBrilliant STUDY CENTRE PHYSICS (ONLINE) = ma Fext SI unit of force is newton 1N = 1 kg ms–2 Gravitational unit of force Kilogram weight or kilogram force 1 kg–wt = gN [g = 9.8 m/s2] 1 kg–f = gN Note : If mass of a body is constant the acceleration of a body is inversely proportional to mass and directly proportional to resultant force acting on it. ∑ = ma F = ∑ Fx ma x = ma y ∑ Fy ∑ = ma z Fz Newtons third Law of motion According to this law, to every action, there is always an equal and opposite reaction i.e. force of action and reaction are always equal and opposite. Whenever a body exerts a force on another body, the second body also exert a force on the first. This force is equal in magnitude, it is in the opposite direction, and has same line of action, acting simultaneously Any of the two forces making action - reaction pair can be called action, and other reaction Force always occur inpair, ifa body exerts a force F on body B, then B will exert equal and opposite force on body A. i.e. FBA = −FAB Equilibrium A force can change the state of motion of a body in two ways. It can cause translation or it can cause translation as well as rotation Conditions of equilibrium 1) If vector sum of forces acting on a body is zero, the body is said to be in translational equilibrium for translational equilibrium ∑F = 0 3
BBrilliant STUDY CENTRE PHYSICS (ONLINE) Note : If vector sum of moment of force acting on a body about any axis is zero, the body is said to be in rotational equilibrium. Thus for rotational equilibrium ∑τ = 0 For complete equilibrium of a body we have ∑τ = 0 0 and ∑F= Static and dynamic equilibrium 1) ∑ F = 0 and body is at rest. it is known as static equilibrium 2) If ∑ F = 0 and body moves with constant velocity it is known as dynamic equilibrium Free Body Diagram In Free body diagrams (FBD), the object of interest is isolated from its surroundings, and the interactions between the object and surroundings are represented in terms of forces. The common forces encounted in mechanics and representing these forces through free body diagrams are 1) weight 2) normal force 3) tension 4) frictional force 5) elastic spring force Weight Weight is the gravitational force with which earth pulls an object Normal force : whenever two surfaces are in contact, they press [or push] each other by a force called contact force. The component of contact force perpendicular to the surface is called normal reaction and along the surface of contact is called frictional force Note : When contact between two bodies breaks the normal reaction vanishes 2) The weighing equipments measure normal reaction Representing normal reaction in different situations * Normal force will be perpendicular to surface of contact * Normal reaction is a pushing force * The number of normal force acting on a body depends on number of points of surfaces of contact 4
BBrilliant STUDY CENTRE PHYSICS (ONLINE) Example 1 A cylinder of weight w is resting on a v-groove as shown in figure. Draw its free body diagram. FBD N1 and N2 are normal reactions between the cylinder and the two inclined walls Example 2 * internal forces always act in pairs * vector sum of all the internal forces on a system is zero ∑ Fint ernal = 0 system Tension When a rope (string, cord etc.) is connected to a body and pulled out, the rope is said to be under tension. It is quite practical that we can pull objects by a string, but we cannot push objects by the string. This gives us an idea that a string can pull but cannot push. 5
BBrilliant STUDY CENTRE PHYSICS (ONLINE) Tension force is an inter molecular force between the atoms of a string, which acts or reacts when the string is stretched The force of tension acts on a body in the direction away from the point of contact or tied ends of the string pulling force Important Points * Tension force always pulls a body * Tension can never push a body * Rope become slack when tension force becomes zero Ideal string An ideal string is considered to be massless [negligible mass] inextensible [does not stretch when pulled]. String is assumed to be massless unless stated. However, if a string has a mass, the tension at different points will be different. * If a string is inextensible the magnitude of acceleration of any number of masses connected through string is always same 6
BBrilliant STUDY CENTRE PHYSICS (ONLINE) Problem solving by applying Newton’s Laws 1. Identify the object you are considering make a simple sketch of the object 2. Draw arrows on your sketch to show the direction of each force acting on the object. Arrows are drawn to represent direction of forces acting on the body 3. Assign a coordinate system to your free body diagram 4. Resolve all the forces acting on a body into x, y and z components 5. Apply Newton’s law in component form as ∑ Fx = max ∑ Fy = may ∑ Fz = maz 6. Solve the set of equations for any unknowns Motion of Blocks connected by massless string Example Find acceleration and tension in string (m1 ) ⇒ T = m1a ............ (1) (m2 ) ⇒ F − T = m2a .......(2) _________________________ (1) + (2) 7
BBrilliant STUDY CENTRE PHYSICS (ONLINE) F = (m1 + m2 ) a a= F ...........(3) m1 + m2 Substitute equation (3) in equation (1) ∴T = m1F ........(4) m1 + m2 Find acceleration and tension on each string (m1 ) ⇒ F − T1 = m1a ............... (1) (m2 ) ⇒ T1 − T2 = m2a ..............(2) (m3 ) ⇒ T2 = m3a .....................(3) ––––––––––––––––––––––––––––– (1) + (2) + (3) F = (m1 + m2 + m3 ) a a= F ....................(4) m1 + m2 + m3 ∴ T2 = m3a = m1 m3F m3 + m2 + From (2) T1 = m2a + T2 = m2a + m3a T1 = (m2 + m3 ) a T1 = (m2 + m3 ) F m1 + m2 + m3 8
BBrilliant STUDY CENTRE PHYSICS (ONLINE) Example : with what minimum acceleration can a monkey slide down a rope whose breaking strength is two third of his weight mg – T = ma ...........(1) Tmax = 2mg 3 mg − 2mg = ma min 3 mg 1− 2 3 = mamin g = a min 3 Example : A monkey of mass 20kg is holding a vertical rope. The rope will not break when a mass of 25 kg is suspended from it but will break if mass exceeds 25kg. What will be maximum acceleration with which the monkey can climb up along the rope [g = 10 m/s2] 9
BBrilliant STUDY CENTRE PHYSICS (ONLINE) T – mg = ma ..............(1) T = ma + mg .............(2) ma = T – mg .............(3) 20a = 25g – 20g .......(4) 20a = 5g a max = 5g = 50 = 2.5m / s2 20 20 Example : A block of mass M is suspended through a light string. A horizontal force F = 3mg is applied at the middle point of string. Find the angle of string with the vertical in equilibrium and tension in two points of string FBD of Block FBD of point P of string Block is under equilibrium Consider vertical direction T1 - Mg = 0 T1 = Mg ............(1) 10
BBrilliant STUDY CENTRE PHYSICS (ONLINE) For point P along horizontal (equilibrium) F − T2 sin θ = 0 F = T2 sin θ ...........(2) For vertical (equilibrium) T2 cos θ = T1 ............(3) dividing eq. (2) ÷ (3) tan θ = F = F = 3mg T1 Mg mg ( )θ = tan−1 3 = 60o Acceleration of block on Horizontal smooth surface Along Y Along X N = mg F = ma a= F m When a pull is acting at an angle θ to horizontal Along Y N + Fsin θ = mg ∴ N = mg − Fsin θ Along X Fcos θ = ma a = F cos θ m 11
BBrilliant STUDY CENTRE PHYSICS (ONLINE) When a push is acting at an angle ( θ ) to horizontal Along Y Fsin θ + mg = N Along X F cos θ = ma a = F cos θ m Fsin θ + mg = N Motion of Blocks in contact [Contact force between two blocks without any string] Case 1 : Find the contact force between m1 and m2 if a horizontal pushing force is applied on block m1. (m1) ⇒ F − N = m1a ..........(1) (m2 ) ⇒ N = m2a ...............(2) ––––––––––––––––––––––––––– (1) + (2) F = (m1 + m2 ) a a= F ...............(3) m1 + m2 From eq. (2) N = m2a = m2F m1 + m2 12
BBrilliant STUDY CENTRE PHYSICS (ONLINE) Case 2 : Find the contact force between m1 and m2 if a horizontal pushing force is applied on block m2. (m1 ) ⇒ N = m1a ............. (1) (m2 ) ⇒ F − N = m2a ––––––––––––––––––––––––– (1) + (2) F = (m1 + m2 ) a a= F ...........(3) m1 + m2 From eq. (1) N = m1a = m1F ............. (4) m1 + m2 Example Find acceleration and contact force between blocks (m1 ) ⇒ F − N1 = m1a ............. (1) (m2 ) ⇒ N1 − N2 = m2a ..........(2) (m3 ) ⇒ N2 = m3a ..................(3) ––––––––––––––––––––––––––––– (1) + (2) + (3) F = (m1 + m2 + m3 ) a 13
BBrilliant STUDY CENTRE PHYSICS (ONLINE) a= F ..............(4) m1 + m2 + m3 From (3) N2 = m3a = m3F m1 + m2 + m3 From (1) N1 = m2a + N2 N1 = m2a + m3a N1 = (m2 + m3 ) a N1 = (m2 + m3 ) F m1 + m2 + m3 Example The blocks are attached by an inextensible light string and pulled vertically upward by force 100N as shown. Find common acceleration and tension in string ↑ (2kg) ⇒ 100 − 20 − T = 2a ..........(1) ↑ (3kg) ⇒ T − 30 = 3a ............(2) _____________________________ (1) + (2) 100 - 50 = 5a 14
BBrilliant STUDY CENTRE PHYSICS (ONLINE) 50 = 5a ⇒ a = 10 m/s2 ...........(3) From eq. (2) T = 3a + 30 = 30 + 30 = 60 N Example : The block of masses 2kg, 3kg and 5 kg are connected by light, inextensible strings as shown. The system of blocks is raised vertically upwards by applying a force F0 = 200N. Find common acceleration and tension in the strings (2kg) 200 – T1 – 20 = 2a .............. (1) (3kg) T1 – T2 - 30 = 3a ...................(2) (5 kg) T2 – 50 = 5a ........................(3) –––––––––––––––––––––––––––––– (1) + (2) + (3) 100 = 10a a = 10 m/s2 From (1) T2 = 5a + 50 = 100N 15
BBrilliant STUDY CENTRE PHYSICS (ONLINE) From (2) T1 = 3a + T2 + 30 = 30 + 100 + 30 = 160N Ideal pulley An ideal pulley is assumed to be massless, frictionless. Action of the pulley to change the direction of force. The ideal pulley does not change the magnitude of tension in the rope. Tension is same in the string on both sides of the pulley Example : In the arrangement shown find acceleration of each block, tension in the string and reaction in the pulley. The string and pulley are light (massless) m1 > m2. Neglect the friction in pulley 16
BBrilliant STUDY CENTRE PHYSICS (ONLINE) (m1) ⇒ m1g – T = m1a ...........(1) (m2) ⇒ T – m2g = m2a ...........(2) ––––––––––––––––––––––––––– (1) + (2) m1g − m2g = (m1 + m2 ) a a= m1g − m2g ..............(3) m1 + m2 (1) ÷(2) m1g − T = m1 T − m2g m2 m1m2g − m2T = m1T − m1m2g 2m1m2g = (m1 + m2 ) T T = 2m1m2g ..............(4) m1 + m2 17
BBrilliant STUDY CENTRE PHYSICS (ONLINE) Pulley R – 2T = 0 [Since the pulley is massless]; R = 2T R = 4m1m2g m1 + m2 Example The pulley is light and smooth the strings are inextensible and light. The system is released from rest, find the acceleration of each block, tensions in the strings and reaction in pulley. (5kg) ⇒ 5g − T1 = 5a ..................... (1) (3kg) ⇒ T1 + 3g − T2 = 3a ..............(2) (2kg) ⇒ T2 − 2g = 2a ...................(3) ––––––––––––––––––––––––––––––– (1) + (2) + (3) 18
BBrilliant STUDY CENTRE PHYSICS (ONLINE) 6g = 10a; a = 60 = 6 m / s2 From eqn. (3) 10 From eqn (1) T2 = 2a + 2g FBD of Pulley = 2× 6 + 2×10 = 12 + 20 = 32N 5g – T1 = 5a = 5g – 5a = T1 = 50 – 30 = 20 T1 = 20 R − 2T2 = 0 (Pulley is massless) R = 2T2 = 64N Example Two blocks are connected by an inextensible light string is passing over smooth light pulley as shown. Find the acceleration of blocks and tension in the string reaction in pulley 19
BBrilliant STUDY CENTRE PHYSICS (ONLINE) (m1 ) ↓ 2mg − T = m1a .........(1) (m2 ) T = m2a ...........................(2) ––––––––––––––––––––––––––––– (1) + (2) m1g = (m1 + m2)a a = m1g ....................(3) m1 + m2 Substitute (3) in (2) T = m1m2g m1 + m2 Pulley R − 2T = 0 R = 2T Example The strings are inextensible and light. The pulleys are smooth and light. Find the acceleration of each block and tensions in the strings 20
BBrilliant STUDY CENTRE PHYSICS (ONLINE) FBD ↓ (5kg) ⇒ 5g − T1 = 5a .............. (1) ← (2kg) ⇒ T1 − T2 = 2a .............(2) ↑ (3kg) ⇒ T2 − 3g = 3a ..............(3) ––––––––––––––––––––––––––––– (1) + (2) + (3) 2g = 10a a = 2g = 2m / s2 10 From eq. (2) T2 = 3a + 3g = 6 + 30 = 36N From (2) T1 = 2a + T2 = 4 + 36 = 40N Tension at different points of string, when the string is not light Apply a cut at that point where tension is required, make free body diagram and solve the equations to determine the tension Example : A uniform rope of length L is pulled by a force F on a smooth surface. Find tension in the rope at a distance x from the end where force is applied 21
BBrilliant STUDY CENTRE PHYSICS (ONLINE) Let M be the mass of rope acceleration of rope a = F M Now we divide the rope into two parts by applying a cut at P m1 = M x L m2 = M (L − x) L m1 + m2 = M a= F M T = M (L−x)a F − T = M xa L L =Mx F = M (L−x) F LM LM T = F − Fx = F 1− x L L = F 1− x L We can find tension at point P by taking either part. Example : A block of mass M is attached to a uniform rope of mass m. This arrangement is pulled by applying a force F at the rope as shown. Find the tension in the rope a) at the junction of rope and the block b) at the midpoint of rope 22
BBrilliant STUDY CENTRE PHYSICS (ONLINE) F = (M + m)a a = F ...............(1) M+m at junction of the block and the rope T = ma = MF M+m Midpoint of rope ( )T1 = M + m 2 a ( )= F M+m2 M+m = 2M + m F 2 + M m T1 = 2M + m F + 2(M m) IF m = 0 T = T1 = F If the rope is light, tension in the rope at each point is same. 23
BBrilliant STUDY CENTRE PHYSICS (ONLINE) Spring Force Force & stretch [or compression] i.e. F = kx i.e. restoring force is linear. K is called the force constant of spring Example : Two blocks are connected by a spring. The combination is suspended at rest from a string attached to the ceiling as shown in fig. The string breaks suddenly immediately after the string breaks, what is the initial downward acceleration of the upper block of mass 2m ? 24
BBrilliant STUDY CENTRE PHYSICS (ONLINE) Step 1 : Discuss the problem before cutting the string From the free body diagram of lower block Kx0 = mg From the free body diagram of upper block T = 2mg + Kx0 “After cutting the string tension on string = zero elongation in spring do not change just after cutting the string” 2mg + Kx0 = 2ma 3mg = 2ma a = 3g 2 25
BBrilliant STUDY CENTRE PHYSICS (ONLINE) Motion on an inclined plane If a block os moving on an inclined plane we choose two directions along the plane and perpendicular to the plane. Resolve the forces along these directions and solve. Fixed Apply Newtons second law perpendicular to the plane N − mgcos θ = 0 N = mgcos θ Apply Newtons second law along the plane mg sin θ = ma a = gsin θ acceleration of the block along the smooth inclined plane a = g sin θ Example A block is released on an smooth inclined plane of inclination θ . After how much time it reaches to the bottom of the plane. 26
BBrilliant STUDY CENTRE PHYSICS (ONLINE) Along the plane sin θ = h ⇒ = h sin θ a = g sin θ = ut + 1 g sin θt2 2 h = 0 + 1 gsin θ t2 sin θ 2 h = 1 g sin2 θ t2 2 2h = gsin2 θ t2 t2 = g 2h θ t = 1 2h sin2 sin θ g Example Consider the situation shown in the figure. The surface is smooth and string and the pulley are light. Find the acceleration of each block and tension in the string. [m1g > m2g sin θ] Sol 27
BBrilliant STUDY CENTRE PHYSICS (ONLINE) (m1) ⇒ m1 g − T = m1 a − − − (1) (m2 ) ⇒ T − m2 gsin θ = m2 a − − − (2) (1) + (2) m1g − m2g sin θ = (m1 + m2 ) a a = m1 g −m2 gsin θ − − − (3) m1 + m 1÷ 2 m1 g − T = m1 T − m2 g sin θ m2 m1 m2 g − m2T = m1T − m1m2 g sin θ m1 m2 g + m1m2 g sin θ = (m1 + m2 ) T m1 m2 g[1+ sin θ] = (m1 + m2 ) T T = m1 m2 g[1+ sin θ] m1 + m2 Example: In the arrangement all the surfaces are smooth, strings and pulleys are light. Find acceleration of each block and tension in the string. [m1g sin α > m2g sin β] 28
BBrilliant STUDY CENTRE PHYSICS (ONLINE) (m1) ⇒ m1g sin α − T = m1 a − − − (1) (m2 ) ⇒ T − m2 g sinβ = m2 a − − − (2) (1) + (2) m1g sin α − m2 g sinβ = (m1 + m2 ) a a = m1g sin α − m2 g sinβ − − − (3) m1 + m2 1÷ 2 m1gsin α − T = m1 T − m2 gsinβ m2 m1 m2 g sin α − m2 T = m1 T − m1 m2 g sin β m1 m2 g[sin α + sin β] = (m1 + m2 ) T T = m1 m2 g[sin α + sinβ] m1 + m2 LAWS OF CONSERVATION OF LINEAR MOMENTUM = d F P dt In case the external force applied to a particle or a (body) be zero. We have = =0 or = constant F dP P dt In the absence of an external force linear momentum of particle remains constant. For a system of particle F net force [vector sum of all the forces] on a system of particle os zero. The vector sum of linear momentum of all the particles remains conserved. P1 + P2 + P3 ......Pn = a constant Conservation of linear momentum is equivalent to Newton’s third law of motion. For a system of two particles in absence of external by law of conservation of linear momentum. P1 + P2 = constant m1 v1 + m1 v2 = constant Differentiating above with respect to time 29
BBrilliant STUDY CENTRE PHYSICS (ONLINE) d v1 dv2 m1 dt + m2 dt =0 m1 a1 + m1 a2 = 0 F1 + F2 = 0 F2 = − F1 i.e., for every action there is equal and opposite reaction which is Newton’s third law of motion Recoiling of a gun For bullet and gun system the force exerted by trigger will be internal so the momentum of system remains conserved. Let mG= mass of gun mB= mass of bullet VG= velocity of gun VB= velocity of bullet Initial momentum of system = 0 Final momentum of system = mG VG + mB VB mG VG + mB VB = 0 mG v G = −mBv B v G = − mBv B mG Negative sign indicates that the velocity of recoil VG is opposite to velocity of bullet. Example An explosion blows a rock in to three parts. Two parts go off at right angles to each other. These two are, 1 kg first part moving with a velocity of 12 ms–1 and 2 kg second part moving with a velocity of 8 ms–1. if the third part flies off with a velocity of 4 ms–1, its mass would be 30
BBrilliant STUDY CENTRE PHYSICS (ONLINE) Since explosion is internal total linear momentum of the system is conserved. P1 + P2 + P3= 0 P1 + P2 = −P3 P1 + P2 = −P3 P12 + P22 = P3 m v 2 + m v2 2 = m3v3 (1×12)2 + (2×8)2 = m3 × 4 144 + 256 = 400 20 = 4m3 m3 = 20 = 5kg 4 Variable mass In our discussion of conservation of linear momentum, we have so far deal with systems whose mass remains constant. We now consider those systems whose mass is variable. i.e., those in which mass enters or leaves the system. Problems related to variable mass can be solved in following three steps. 1) Make a list of all the forces acting on the main mass and apply them on it. ± dm and direction 2) Apply an additional thrust force Ft , on the mass, the magnitude of which is Vr dt is given by direction of Vr in case the mass is increasing and otherwise the direction of −Vr , it is decreasing. 3) Find net force on the mass and apply = m (m = mass at that particular instant) Fnet dv dt Rocket Propulsion Let m0 be the mass of rocket at time t = 0. m it mass at any time t and v its velocity at that moment. Initially let us suppose that the velocity of the rocket is u. 31
BBrilliant STUDY CENTRE PHYSICS (ONLINE) −dm Let dt be the mass of the gas ejected per unit time and Vr the exhaust velocity of the gases. −dm Usually dt and Vr are kept constant through the journey of the rocket. At time t = t 1) Thrust force on the rocket F = Vr dm (upwards) dt 2) Weight of rocket W = mg (downwards) 3) Net force on the rocket Fnet = Ft − W Fnet = Vr dm − mg (upwards) dt 4) Net acceleration of rocket a = F m dv = vr −dm − g dt m dt dv = vr −dm − g dt m ∫ ∫vdv = vr m −dm − g dt m0 m u v − u= vr n m0 − gt m v = u − gt + vr n m0 m Note: 1) Ft = vr −dm is upwards, as vr is downwards and dm dt dt is negative 32
BBrilliant STUDY CENTRE PHYSICS (ONLINE) 2) If gravity is ignored and initial velocity of rocket u = 0 v = vrn m0 m Mass of rocket at any instant m = m0 − dm t dt * Here vr is the velocity of the mass gained or mass ejected relative to main mass. In case of rocket it is sometimes called exhaust velocity of gas. Force exerted by jet on wall Mass of element dm = volume × density dm = Adx ρ Change in momentum of liquid jet = 0 − ∆mv = Adxρv dx = v dt ( )dP = Adxρv wall ( )F = ( )dP = Adxρv = Av2ρ wall wall dt dt F = Av2ρ here we consider a jet of area A strikes the stationary wall with velocity v and stop. Impulse Impulse of a force F acting on a body is defined as J = ∫ Fdt = ∫m dv dt = ∫ m dv dt 33
BBrilliant STUDY CENTRE PHYSICS (ONLINE) It is also defined as change in momentum J = ∆p (impulse momentum theorem) Instantaneous impulse There are many occasions when a force acts for such a short time that the effect is instantaneous. eg, a bat striking a ball. ∫J = Fdt = ∆P =Pf − Pi It is important to note that impulse applied to an object in a given time interval can also be calculated from the area under Force - time graph in the same time interval. Impulse is a vector quantity SI unit Ns or kgms-1 ∫ magnitude is equal to area under F-t graph J = Fdt = Fav∆t It is not a property of any particle, but it is a measure of the degree to which an external force change the momentum of the particle. Impulsive force A force of relatively higher magnitude and acting for relatively shorter time is called Impulsive force. An impulsive force can change the momentum of a body in a finite magnitude in very short time interval. “Colliding forces are impulsive in nature”. Example: Two identical balls strike a rigid wall with same speed but at different angles, and get reflected without any loss of speed as shown in figure what is 1) the direction of force on the wall due to each ball. 2) ratio of the magnitudes of impulse imparted on two balls by the wall. 34
BBrilliant STUDY CENTRE PHYSICS (ONLINE) Ans. Let the ball strikes the wall with speed u. Its mass is m, then momentum = mu ˆi Pi = − mu ˆi Pf impulse in × direction = −muˆi − muˆi Jx = − 2muˆi No motion along Y axis, impulsive along Y direction is zero. resulting impulse = = −2muˆi J1 Jx + Jy The direction of force along direction of impulse. So force experienced by wall on ball is along negative x axis. By Newton’s third law the direction of force exerted by ball on wall is along positive x axis. ( ) Jx = Pf − Pi x = −mu cos 30 ˆi − mu cos 30 ˆi = −2mu cos 30 ˆi ( ) Jy = Pt −Pi y = mu sin 30ˆj −− mu sin 30 ˆj =0 3muˆi resulting impulse J1 = Jx + Jy = − The force exerted by wall on ball is along negative x axis. By Newton’s third law the direction of force on wall by ball is along positive x axis. J1 = 2mu = 2 J2 3 mu 3 35
BBrilliant STUDY CENTRE PHYSICS (ONLINE) A machine gun fires a bullet of mass m with a speed v m/s. The person holding the gun exert a maximum force F on it. What is the number of bullets that can be fired from the gun per second? The change in momentum of each bullet ∆P = m[v − u] As u = 0 ∆P = mv If n is the number of bullets fired per second, then rate of change of momentum of gun ∆P = nmv ∆t thus by Newton’s second law F = ∆P = nmv ∆t n= F mv FRAME OF REFERENCE A frame in which an observer is situated and makes his observation is known as frame of reference. Frame of reference are two types 1) inertial frame of reference 2) non inertial frame of reference Inertial frame of reference A frame of reference which is at rest or which is moving with a uniform velocity along a straight line is called an inertial frame of reference. Non Inertial frame of reference Accelerated frame of reference are called non inertial frame of reference. Pseudo or Fictious force In non inertial frame of reference a pseudo force acts on the body present in accelerated frame in the direction opposite to the acceleration of frame. Magnitude of Pseudo force = mass of object × acceleration of frame of reference. Direction of pseudo force is opposite to acceleration of frame. Important points regarding pseudo force 1) Pseudo force is an observer dependent force 2) Apply a pseudo force on an object if and only if its is observed by a nonintertial frame 3) The direction of pseudo force must be opposite to the direction of acceleration of non-inertial frame 4) The magnitude of pseudo force is the product of mass of the body and acceleration of noninertial frame 36
BBrilliant STUDY CENTRE PHYSICS (ONLINE) In noninertial frame Newtons second law take the form + = ma Freal Fpseudo a is acceleration of object in noninertial frame Fpseudo = mass of object × acceleration of frame Example: A pendulum is hanging from the ceiling of a car having an acceleration a0 with respect to road. Find the angle made by string with vertical. solve this question with respect to two observers A and B with respect to observer A [inertial frame of reference] FBD of pendulum T cos θ = mg .............(1) T sin θ = ma0 ................(2) (2) ÷(1) tan θ = a0 g θ = tan −1 a0 g 37
BBrilliant STUDY CENTRE PHYSICS (ONLINE) with respect to observer B [noninertial frame of reference) FBD of pendulum T cos θ = mg ................(1) T sin θ − ma0 = m × 0 T sin θ = ma0 .................(2) (2) ÷(1) tan θ = a0 g θ = tan −1 a0 g MOTION IN A LIFT/ ELEVATOR Consider a man of mass m standing on a weighing machine placed in the lift. The actual weight of man is mg. The reading of weighing machine indicates the force experienced by it which is equal to reaction on the man standing on it. When a man is in an accelerated lift, his weight appears to change. This changed weight is known as apparent weight. Case I Lift is at rest [a = 0] N - mg = 0 38
BBrilliant STUDY CENTRE PHYSICS (ONLINE) N = mg Apparent weight = actual weight Case II When lift is moving with uniform velocity (a = 0) N - mg = 0 N = mg Apparent weight = Actual weight Case III Lift is moving upwards with uniform acceleration a. N=mg+ ma mgapparent = mg + ma gapparent = g + a Apparent weight > Actual weight 39
BBrilliant STUDY CENTRE PHYSICS (ONLINE) N - mg = ma N=ma + mg Case IV Lift is moving downward with uniform acceleration a (a < g) N + ma = mg N = mg − ma mgapparent = mg – ma gapparent = g – a 40
BBrilliant STUDY CENTRE PHYSICS (ONLINE) mg - N = ma N = mg − ma Case V If the supporting cable of lift breaks, then the lift falls freely with an acceleration a = g N + mg = mg N = mg - mg = 0 mgapparent = 0 gapparent = 0 Apparent weight of man in freely falling lift becomes zero. mg – N = mg mg – mg = N N=0 41
BBrilliant STUDY CENTRE PHYSICS (ONLINE) Example A block of mass m is placed on a smooth wedge of inclination θ . The whole system is accelerated horizontally so that block does not slip on wedge. Determine the force exerted by the wedge on block. Sol. FBD of block in non inertial frame Let the wedge is accelerated towards left with an acceleration ‘a’. The FBD of block in the frame from is shown in figure. The block remains at rest with respect to wedge, so along the inclined plane, we have mg sin θ − ma cos θ = 0 mgsin θ = ma cos θ a = g tan θ Perpendicular to the inclined plane the block is also at rest. ∴N = mgcos θ + ma sin θ N = mgcos θ + mg tan θ sin θ N = mgcos θ + mg sin θ sin θ cos θ N = mgcos2 θ + mg sin2 θ cos θ N = mg cos2 θ + sin2 θ cos θ ∴N = mg cos θ Thus force exerted by wedge on block 42
BBrilliant STUDY CENTRE PHYSICS (ONLINE) N = mg cos θ FRICTION Friction can be defined as a force which opposes the relative motion between surfaces in contact. Friction is the parallel component of contact force between two bodies in contact. These forces are basically electromagnetic in nature. Static Friction The opposing force that comes into play when one body tends to move over the surface of another. But the actual motion has yet not started is called static friction. For example, consider a bed inside a room, when we gently push the bed with a finger, the bed does not moves. This means that the bed has a tendency to move in the direction of applied force but does not move as there exists static friction force acting in the opposite direction of applied force. The static friction force on an object is opposite to its impending motion relative to the surface. The direction and magnitude both are self adjusting such that relative motion is opposed. Note: Here once again the static friction is involved when there is no relative motion between two surfaces. Important points about static friction force 1) Static friction always opposes the primary cause of its motion. 2) It is variable and self adjusting force 3) Magnitude of static friction (fs) lies between 0 and µsN 0 ≤ fs ≤ µsN Minimum value of static friction = 0 Maximum value of static friction = µsN Limiting friction The maximum value of static friction upto which body does not move is called limiting friction. The magnitude of limiting friction between any two bodies in contact is directly proportional to the normal reaction between them fms = µsN Kinetic Friction Force Kinetic friction exists between two contact surfaces only when there is relative motion between the two contact surfaces. It stops acting when relative motion between two surfaces ceases. Direction of Kinetic Friction on an Object It is opposite to the velocity of the object with respect to the other object in contact considered. Note: That its direction is not opposite to the force applied, it is opposite to the motion of the body considered which is in contact with the other surface. 43
BBrilliant STUDY CENTRE PHYSICS (ONLINE) How to decide the direction of kinetic friction force. 1) It is opposite to the relative velocity of the object with respect to other in contact considered Examples 1) Direction of kinetic friction on the block Direction of kinetic friction on the ground 2) For the direction of kinetic friction on the block ‘A’ decide the direction of relative velocity of A w.r to B. v AB = v A − v B = 5 − 3 = 2ˆi Direction of kinetic friction on block B Magnitude of kinetic friction is proportional to the normal force acting between the two bodies. fK = µ K N Value of µS and µK is independent of surface area it depends only on surface properties of contact surface. µS and µK has no unit and no dimension. µS and µK are properties of given pair of surfaces. Graph between applied force and force of friction 44
BBrilliant STUDY CENTRE PHYSICS (ONLINE) TYPES OF KINETIC FRICTION Sliding Friction : The opposing force that come in to play when one body is actually sliding over the surface of the other body is called Sliding Friction. Rolling Friction : When objects such as a wheel (disc or ring) sphere or a cylinder rolls over a surface, the force of friction that comes into play. Rolling friction is quite small as compared to sliding friction. Example Suppose a block of mass 1Kg is places over a rough surface and a horizontal force F is applied on the block as shown in fig. Now let us see what are the values of force of friction f and acceleration of the block a if the force F is gradually increased. Given that µS = 0.5 , µK = 0.4 and g = 10 m/s2 Net force along y ∑Fy = 0 N - mg = 0 N = mg = 1×10 = 10 N fL = µs N= 0.5 ×10 = 5N fK = µs N = 0.4 ×10 = 4N 45
BBrilliant STUDY CENTRE PHYSICS (ONLINE) Angle of Friction Angle of friction may be defined as the angle which the resultant of limitting friction and normal reaction make with normal reaction. tan θ = fms − − − − (1) N tan θ = µS − − − − (2) 46
BBrilliant STUDY CENTRE PHYSICS (ONLINE) θ = tan−1 (µS ) − − − − (3) Coefficient of static friction is equal to tangent of angle of friction Resultant force exerted by surface on block R = ( )N2 + fms 2 R = (mg)2 + (µSmg)2 R = mg µS2 +1 When there is no friction µS = 0 and R will be minimum R = mg Hence the range of R can be given by mg ≤ R ≤ mg µS2 +1 Angle of response ( α ) Suppose a block of mass m is placed on an inclined plane whose inclination θ can be increased or decreased. Let µ be coefficient of friction between the block and the plane. At a general angle θ . Normal reaction N mgcos θ Limitting friction fL µN µ mgcos θ and driving force or pulling force F mg sin θ From these three equations we see that when θ is increased from 0 to 900, normal reaction N and hence the limitting friction fL is decreased while the driving force mg sin θ is increased. There is a critical angle called angle of response α at which these two forces are equal. Now, if θ is further increased, then driving force mg sin θ becomes more than limitting friction fL and the block starts sliding. Thus fL = F [at θ α ] µSmgcos α mgsin α tan α µS 47
BBrilliant STUDY CENTRE PHYSICS (ONLINE) α tan1 µS Angle of friction is numerically equal to angle of repose From above discussion we can conclude that If θ α F fL the block is stationary. If θ α F fL the block is on verge of sliding and if θ α F fL the block slide down with acceleration. Acceleration of block sliding down over a rough inclined plane When angle of inclined plane is more than angle of repose the body placed on the inclined plane slides down with an acceleration a. mg sin θ − µKN = ma mg sin θ − µKmgcos θ = ma a = g sin θ − µkgcos θ Retardation of a block sliding up over a rough inclined plane Net retarding force Fnet = m × –a mg sin θ + fk = m × −a mg sin θ + µkmg cos θ = m × −a Retardation –a = g sin θ + µkg cos θ 48
BBrilliant STUDY CENTRE PHYSICS (ONLINE) Pull is easier than push Consider a block of mass m placed on a rough horizontal surface. The coefficient of static friction between the block and surface is µ . Let a push force F is applied at an angle θ with the horizontal. r Along y F sinθ + mg = N Along x Fcosθ = µN = µ[Fsinθ + mg] Fcos θ = µF sin θ + µmg F[cosθ − µ sinθ] = mg F = mg − − − (1) cos θ − µ sinθ Along Y F sinθ + N = mg N = mg − F sin θ To just move the block along x axis Fcosθ = µN = µ[mg − Fsinθ] Fcos θ = µmg − µF sin θ Fcos θ + µF sinθ = µmg 49
BBrilliant STUDY CENTRE PHYSICS (ONLINE) F[cosθ + µ sinθ] = µmg F = µmg cos θ + µ sin θ It is clear that from above discussion that pull force is smaller than push force. Example 1) What is the maximum possible value of F so that the system move together 2) If there is relative sliding between M and m then calculate the acceleration of M and m Solution Let the system move together then a = F M+m Normal reaction between block m and M N = mg For (M) f = Ma = MF M+m As there is no sliding between M and m friction is static f ≤ µsN 50
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