Presentation on theme: "Train and one carriage. Connected particles Horizontal motion of connected particles T F1F1 F2F2 m1gm1g m2gm2g R n1 R n2."— Presentation transcript:
Train and one carriage
Connected particles Horizontal motion of connected particles T F1F1 F2F2 m1gm1g m2gm2g R n1 R n2
A car and a trailer A car pulls a trailer along a level road at a constant velocity v. The trailer is pulled forward by the tension T in the tow bar. The trailer will exert an equal and opposite force T on the car. TT If the car is accelerating, then there must be a force acting on the trailer to produce this acceleration, and this will be provided by the tension T in the tow bar. On the other hand, when the car is slowing down, so is the trailer. In the absence of brakes on the trailer, some force must act in the opposite direction to the motion of the car and trailer. In this case the tow bar will exert thrust on both the car and the trailer.
Example The diagram shows a car of mass m 1 pulling a trailer of mass m 2 along a level road. The engine of the car exerts a forward force F, the tension in the tow bar is T and the reactions at the ground for the car and the trailer are R 1 and R 2 respectively. If the acceleration of the car is a, write down the equation of motion for: (a) the system as a whole, (b) the car, (c) the trailer (d) R 1 and R 2 Solution (a) F = (m 1 + m 2 )a TT R1R1 R2R2 m 2 gm 1 g a ms -2 F (b) F – T = m 1 a (c) T = m 2 a (d) R 1 = m 1 g and R 2 = m 2 g
Example A car of mass 1100 kg tows a caravan of mass 800 kg along a horizontal road. The engine of the car exerts a forward force 2.2 k N. The resistance to the motion of the car and caravan are 200 N and 100 N respectively. Given that the car accelerates at 1 ms -2 find the tension in the tow bar. Solution Car: 2200 – T – 200 = 1100a TT R1 R g 1 ms N200 N 800 g 100 N 2200 – 200 – 1100 = T So T = 900 N Caravan: T – 100 = So T = 900 N
Two particles of mass 5 kg and 10 kg are connected by an inextensible string. The particle of mass 10 kg is being pulled by a horizontal force of 120 N along a rough, horizontal surface. Given that the coefficient between each particle and the surface is 0.4, find the acceleration of the system and the tension in the string. T 120 N F2F2 10g 5g R n1 R n2 Example 10 kg 5 kg F1F1 F 1 = R 1 = 39.2F 2 = R 2 = 19.6 System as a whole: – 19.6 = 15a a = 4.08ms -2 5 kg particle:T – 19.6 = 5a T = 40 N Both particles accelerates at 4.08 ms -2 with a tension 40 N.
T 120 N F2F2 10g 5g R n1 R n2 10 kg 5 kg F1F1 Both particles accelerates at 4.08 ms -2 with a tension 40 N. 10 kg particle:5 kg particle: R 1 – 10g = 0R 2 – 5g = 0 R 1 = 98R 2 = 49 F 1 = R 1 = = 39.2 F 2 = R 2 = = – 39.2 – T = 10a  T – 19.6 = 5a   +  – 19.6 = 15a a = 4.08ms -2 5 kg particle:T – 19.6 = 5a T = 40 N Both particles accelerates at 4.08 ms -2 with a tension 40 N.
Train and two carriages
A train consists of an engine of mass kg coupled to two trucks A and B of masses kg and kg respectively. The couples are light, rigid and horizontal. The train moves along a horizontal track with a constant acceleration. The resistance to motion of the engine, truck A and truck B are N, N and N respectively. The engine exerts a driving force of N. T 2 T kg10000kg60000kg N 8000N 6000 N N Find the acceleration of the train and tensions T 1 and T 2.
System as a whole: – – 6000 – 8000 = 78000a a = 0.25 ms -2. Train: – T 1 – = 60000a T 1 = N Truck B: T 2 – 8000 = 8000a T 2 = N