Newton s 3 rd Law Book page 48-49 14/9/2016 cgrahamphysics.com 2016
Newton s 2 nd Law problem Newton s second law does not always work: - does not work when applied to atoms and molecules - does not work for particles moving close to the speed of light
Newton s 3 rd Law Valid for all of Physics Forces appear in equal and opposite pairs: When a force acts on a body, an equal and opposite force acts on another body somewhere in the Universe In order to be a 3 rd law pair, the following requirements must be met: Act for the same time Same line of action Same type Same magnitude Acts on / exerted by different bodies Opposite direction 14/9/2016
Confusion: what is a pair? Consider a book on a table Write down the reaction action pairs The reaction force to the weight is not the normal force - not the same type of force - both act on the same body - both can be drawn on the same FBD If N and W are not equal, they cannot be an action reaction pair C tb C Et Action reaction pairs are W never drawn on the same FBD Eb - using subscripts will help to avoid this trap W Et C bt W be W te C te
Two forces are equal and opposite if x + y = 0 and A + B = 0 A y x B X is the weight of the book W b Y is the force the book exerts on Earth Without table, the book would fall to Earth A is the force the table exerts on book C tb B is the force the book exerts on the table C bt This means for a 3 rd law pair W bt + W te + W Et + W tb where W te = W Et If a system of two isolated particles that exert equal and opposite forces on each other, then the ratio of their acceleration will be in the ratio of their masses cgrahamphysics.com 2016
Example A truck is pulling two trailers starts from rest and accelerates to a speed of 16.2kmh 1 in 15s on a straight level section of the highway. The mass of the truck itself, T, is 5450kg, the mass of the first trailer, A, is 31500kg and the mass of the second trailer, B, is 19600kg. Assume friction is negligible. a) What magnitude of force must the truck generate in order to accelerate the entire vehicle? b) What magnitude of force must each of the trailer hitches withstand while the vehicle is accelerating? truck Trailer A Trailer B Hitch between truck and trailer A Hitch between trailer A and trailer B
5450kg F AT 31500kg 19600kg Draw FBD for each part: F pull on T F AT F TA F BA Given: F AB u = 0 m T a m T a m A a v= 16.2kmh 1 m F A a AT = 4.5ms 1 F BA t = 15s F pull on T F TA m T a + F AT =F pull on T m A a+ F BA = F TA m B a = F AB Truck: F PT = m Total a = 31500 + 5450 + 19600 0.3 = 16965N Solve for F AT = F PT m T a = 16965 5450 0.3 = 15330N m B a F AB a is the same for all m B a a = v t = 4.5 15 = 0.3ms 2
A- trailer: m A a+ F BA = F TA F BA = F TA m A a=15330-31500 x 0.3= 5880N 2 nd hitch F BA : 5.9 10 3 N Check: F BA should be equal to F AB m B a = F AB F AB = 19600 0.3 = 5880 = 5.9 10 3 N
Example with friction A train is pulling two carriages. It is moving along a horizontal track with an acceleration a. The engine has mass M and each carriage has mass m. The combined drag and frictional force on each carriage and on the engine is f. What is the thrust of the engine and the tension in each coupling? Identify any action reaction pairs f T 32 T 23 m m M F f f T 12 T 21 ma + f = T 32 ma + f + T 23 = T 12 Ma + f + T 21 = F
All together as one: (M + 2m)a + 3f = F (M + 2m)a 3f M + 2m The thrust in the engine is (M + 2m)a + 3f F Last carriage: ma + f = T 32 2 nd carriage: ma + f + T 23 = T 12 ma + f +ma + f = T 12 T 12 = 2(f + ma) Check answer using first carriage Ma + f + T 21 = F F = Ma + f + 2(f + ma) = Ma + 2ma + 3f = (M + 2m)a + 3f
Example Two boxes, one of mass 5kg and the other of unknown mass M are pushed along a rough floor with a force of 40N. If the frictional force is 10N and the boxes are accelerated at 2ms 2, what is the value of M? 5kg M F F f Consider the system: (5+M)a + F f = F M + 5 = F F f a = 40 10 2 = 15 M = 15 5 = 10kg
Example A block of mass m 1, lying on an inclined plane, is connected to mass m 2 by a massless cord passing over a pulley. a) Determine the general formula for the acceleration of the system in terms of m 1, m 2, g and θ. If m 1 sin θ > m 2 the block will move down the plane m 2 a T T F f m 2 m 1 g cos θ N m 1 m 1 a m 1 g sin θ Incline: m 1 a + T + F f = m 1 g sin θ T = m 1 g sin θ m 1 a F f m 2 g m 1 g Hanging: T = m 2 g + m 2 a m 2 g + m 2 a = m 1 g sin θ m 1 a F f m 2 a + m 1 a = m 1 g sin θ m 2 g F f a = m 1sin θ m 2 g F f m 1 + m 2
continued b) Suppose the coefficient of static friction between m 1 and the plane is μ S = 0.15 and that each block has an identical mass of m 2.0kg. Determine the acceleration 2 g of the blocks if θ = 30 0 Remember: If m 1 sin θ > m 2 the block will move down the plane: 2sin 30 = 1, not >2 block will move up the plane F s = μn = μm 1 gcosθ m 1 gsin θ + F f + m 1 a = T and T + m 2 a = m 2 g m 1 gsin θ + F f + m 1 a = m 2 g m 2 a m 1 a + m 2 a = m 2 g - m 1 gsin θ F f a m 1 + m 2 = m 2 g - m 1 gsin θ μm 1 gcosθ a = mg 1 sinθ μcosθ 2m = 1 2 m 2 a T = m 2 g m 2 a 10 1 sin 30 0.15 cos 30 = 1.85ms 2 T T m 1 a F f m 1 = m 2 Up the ramp