2 Connected particles Horizontal motion of connected particles Rn1 Rn2 T TF2F1m1gm2g
3 A car and a trailerA car pulls a trailer along a level road at a constant velocity v.TThe 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.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.
4 ExampleTR1R2m2 gm1 ga ms-2FThe diagram shows a car of mass m1 pulling a trailer of mass m2 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 R1 and R2 respectively. If the acceleration of the car is a, write down the equation of motion for:Solution(a) F = (m1 + m2)a(b) F – T = m1a(c) T = m2a(a) the system as a whole,(d) R1 = m1g and R2 = m2 g(b) the car,(c) the trailer(d) R1 and R2
5 Example Solution Car: 2200 – T – 200 = 1100a Caravan: T – 100 = 800 1 1100 g1 ms-22200 N200 N800 g100 NExampleA 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.SolutionCar: – T – 200 = 1100aCaravan: T – 100 = 800 12200 – 200 – 1100 = TSo T = 900 NSo T = 900 N
6 ExampleTwo 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.Rn1Rn2T T10 kg5 kg120 NF1F210g5gF1 =R1 =39.2F2 =R2 =19.6System as a whole:– 19.6 = 15a a = 4.08ms-25 kg particle:T – 19.6 = 5a T = 40 NBoth particles accelerates at 4.08 ms-2 with a tension 40 N.
7 Rn1Rn210 kgT T5 kg120 NF1F210g5g10 kg particle:5 kg particle:R1 – 10g = 0R2 – 5g = 0R1 = 98R2 = 49F1 = R1 = 0.4 98 = 39.2F2 = R2 = 0.4 49 = 19.6120 – 39.2 – T = 10a T – 19.6 = 5a  + – 19.6 = 15a a = 4.08ms-25 kg particle:T – 19.6 = 5a T = 40 NBoth particles accelerates at 4.08 ms-2 with a tension 40 N.Both particles accelerates at 4.08 ms-2 with a tension 40 N.
9 Find the acceleration of the train and tensions T1 and T2. 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.T2 T2T1 T18000kg10000kg60000kg45500 N8000N N NFind the acceleration of the train and tensions T1 and T2.
10 System as a whole:45500 – – 6000 – 8000 = 78000a a = 0.25 ms-2.Train:45500 – T1 – = 60000a T1 = NTruck B:T2 – 8000 = 8000a T2 = N