4) Newton’s Law of Motion 1 Newton’s Law of Motion Newton’s Third Law: When two objects interact, the forces on the bodies from each other are always equal in magnitude and opposite in direction. Example: 10 and 5 Kg masses are attached at the ends of a massless cord that goes over a massless and frictionless pulley as shown, find the acceleration of the system and the tension in the card. If the two masses start from rest 10m above the ground, find the maximum height reached by the 5Kg mass. σ π = ma 2 Newton’s Law of Motion Apply this law on m1 -m1g+T = −π′ a m1g-T = m1a m1 moves upward10m with π acceleration 3.27 π π its velocity 2ay = ππ − πππ , ππ = π π = πππ π π = πππ. πππππ = ππ. π =8.1 π 3 Newton’s Law of Motion Now m2 is projected upward against π gravity with velocity 8.1 π -2gyο’= ππ − πππ π = π , πππ = ππ. π ∴ height reached by m2is: π = ππ + π. ππ = ππ. πππ 4 Newton’s Law of Motion Apply Newton’s second law on m2 π» − ππ π = ππ π Add + ππ π = ππ π = ππ π = ππ π π ππ − ππ = π ππ − ππ π= ππ−π ππ π π. π = From equation 2 π» − ππ π + π π» = π π. π + π. ππ π» = ππ.35 N π.π π = π. ππ π ππ = ππππ. ππ 5 Newton’s Law of Motion Solving problems involving application of Newton’s second law: 1. Draw a clear diagram. 2. Show all forces acting on all objects. 3. Apply Newton’s second law on all objects separately and obtain equations. 4. Solve equations simultaneously to find unknowns. Friction it is the resistance force opposing motion (sliding) of an object on a surface caused by the bonding between the body and the surface. Friction opposed motion. π It does not cause motion. Where π is the coefficient of friction. N is the normal force. = ππ΅ 6 Newton’s Law of Motion There are two types of friction: 1. Static friction ππ = ππ π΅ where ππ is the static frictional force and ππ is the coefficient of static friction and N is the normal force. 2. Kinetic friction ππ = ππ π΅ where ππ is the kinetic friction and ππ is the coefficient of kinetic friction and N is the normal force. ππ > ππ ππ π΅ > ππ π΅ ππ > ππ 7 Newton’s Law of Motion Normal force: N It is the force applied by the surface on the object, perpendicular to the surface. N N=mg f mg Example: A 10Kg box on a rough horizontal surface is pulled by a force F. If the coefficient of static friction is 0.4 and the coefficient of kinetic friction is 0.2. Find the frictional force if F =10N, 20Nm 50N. ππ = ππ π΅ = ππ ππ = π. ππππππ. π = ππ. ππ΅ ππ = ππ π΅ = ππ ππ = π. ππππππ. π = ππ. ππ΅ 8 Newton’s Law of Motion a) π = πππ΅ < ππ ∴ πππ πππ ππππ πππ ππππ ∴π=π πΊπ = π π−π=π π = πππ΅ b) πππ΅ < ππ , π = π πΊπ = π ππ − π = π π = πππ΅ c) ππ > ππ ∴ The box moves and friction is kinetic π = ππ ππ = ππ. ππ΅ πΊπ = π¦π ππ − ππ. π = πππ π¦ π = π. ππ π π 9 Newton’s Law of Motion Example: A 10Kg box is pulled by a force 30N inclined at an angle of 37ο° to the horizontal. If the coefficient of friction is 0.2, find the normal force and find the acceleration of the box. πΊππ = π π΅ + πππππ½ − ππ = π π΅ = ππ − πππππ½ π΅ = ππππ. π − ππππ. π = πππ΅ πΊπ = π¦π πππππ½ − π = ππ πππππ½ − ππ΅ = ππ 30x0.8-0.2x80=10a 24-16=10a π¦ a=0.8 π π 10 Newton’s Law of Motion Example: In the figure m1=5kg m2=10Kg the coefficient of friction between m2 and the surface is 0.2 The pulley is massless and frictionless, find the acceleration of the system and the tension in the string. m1g−T = m1a 1 T−π = ππ π T−π = ππ π T−ππ΅ = ππ π π΅ = ππ π T−πππ π = ππ π Add 1 + 2 m1g−πππ π = ππ + ππ π ππ−π ππ πππ.π π m1g−πππ π = = = = 2.6 π ππ ππ ππ π 11 Newton’s Law of Motion π» = ππ π − π π» = π π. π − π. π π» = ππΏπ. π = πππ΅ Inclined plane An object is moving down in inclined planed of angle of inclination π½ at constant speed find the coefficient of friction, between the object and the plane. The weight of the object is resolved in two directions along the inclined plane and perpendicular to the plane. Acceleration along these two directions is zero. 12 Newton’s Law of Motion πΊππ΅ = π π − π¦π ππ¨π¬π½ = π π = π¦π ππ¨π¬π½ πΊπ = π π¦π π¬π’π§π½ − π = π π¦π π¬π’π§π½ − ππ΅ = π π¦π π¬π’π§π½ = ππ΅ πππππππ½ = π¦π π¬π’π§π½ π = ππππ½ A mass m1 = 8Kg hangs free and it is connected by a cord over a massless and frictionless pulley. The other end is connected to 10Kg mass. The coefficient of friction is 0.2. 13 Newton’s Law of Motion Find the acceleration of the system and the tension in the cord. πΊπ =m1a m1a – T= m1a π» − m2gsinο± −π = m2a π» − m2gsinο± −ππ΅ = m2a π» − m2gsinο± −ππ¦ππ ππ¨π¬ο± = m2a π¨π π 1 + 2 π′g= m2gsinο± −ππ¦ππ π¬π’π§ο± = ππ + ππ π 8g-ππ − π. ππ = πππ π.ππ π.π π π.π π a= = =0.22 π ππ ππ π T= m1(g − a) T= 8(9.8 − 0.22) = 8 x 9.58 = 76.64N 14 Newton’s Law of Motion Example: A 10kg load is weighed in a lift with a spring balance. Find the apparent weight of this load if the lift is: a) Stationary b) Moving up / down with constant speed. π c) Moving up with acceleration of 2 π π d) Moving down with acceleration of 9.8 W π¦ π¬π mg c) σ π = ππ a) a = 0 w - mg = ma σπ = 0 w = m ( g + a ) = 10 (9.8 + 2) =10x 11.8 = 118 N w - mg = 0 w = mg = 98 N d) σ π = −ππ w - mg = - ma b) a = 0 w = m (g – a) σπ = π w=0 w - mg = 0 w = mg = 98 N This weight less- ness, it happens at free fall. Newton’s Law of Motion Example: In a theme park a boy stands with his back in contact with the wall of a cylindrical shape. Elevated room, that rotate about its axis. When its velocity reaches a certain value the base of the cylinder is dropped but the boy does not fall. If the radius of the cylinder is r and the coefficient of friction between the boy and the wall is π , find this velocity. If the boy does not fall frictional force must be equal to the weight of the boy. π = ππ π N = mg N=m π π―= πππ = π ππ π π―π π« ππ Newton’s Law of Motion Example: In a conical pendulum the cord is πΏ π long and the angle the cord makes with the vertical is π. Find the period of pendulum. The mass revolves in a horizontal circle the Centripetal force: ππ π ππ = m = T πππ π½ mg = T πππ π½ ππ πππ π½ = ππ πππ π½ L r = L πππ π½ T = 2π π ππππ½ π T π ππ πππ π½ = π³ π¬π’π§ π½π πππ π½ π π³πππ π½ π ππ = πππ π½ π³π π = πππ π½ πππ π½ ππ π ππ π³ πππ π½ period t = π = π³π πππ π½ πππ π½ mg π Newton’s Law of Motion Newton’s First Law: An object at rest remains at rest and an object moving at constant speed in a straight line remains so, unless acted upon by an external force. Newton’s Second Law: The net force on an object is equal to the product of the object’s mass and its acceleration. = ππππ ππ σ π = ma 18
