Train and one carriage

Connected particles

Horizontal motion of connected particles

F 1 R n1 m 1 g R n2 T T F 2 m 2 g

A car and a trailer

A car pulls a trailer along a level road at a constant velocity v.

T T 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.

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

R 1 R 2 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 a ms -2 T T

F

m 1 g m 2 g

Solution (a) F = (m 1 + m 2 )a (b) F – T = m 1 a (c) T = m 2 a (d) R 1 = m 1 g and R 2 = m 2 g

R1 R2

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.

2200 N

Solution Car: 2200 – T – 200 = 1100a

1100 g 1 ms-2 T 200 N T 100 N 800 g

Caravan: T – 100 = 800



1 2200 – 200 – 1100 = T So T = 900 N So T = 900 N

Example

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.

R n1 R n2 10 kg T T 5 kg F 1 =  R 120 N 1 5 kg particle: 10g = 39.2

System as a whole: T – F 1 F 2 5g 120 - 39.2 – F 2 =  R 2 = 19.6

19.6 = 15a  a = 4.08ms

-2 19.6 = 5a  T = 40 N Both particles accelerates at 4.08 ms -2 with a tension 40 N.

R n1 10 kg 120 N 10g T T F 1 R n2 5 kg F 2 5g 10 kg particle: R 1 – 10g = 0 R 1 = 98 F 1 =  R 1 = 0.4  98 = 39.2

120 – 39.2 – T = 10a [1] 5 kg particle: R 2 – 5g = 0 R 2 = 49 F 2 =  R 2 = 0.4  49 = 19.6

T – 19.6 = 5a [2] 120 - 39.2 – 5 kg particle: [1] + [2] 19.6 = 15a T – 19.6 = 5a  a = 4.08ms

-2  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 60 000 kg coupled to two trucks A and B of masses 10 000 kg and 8 000 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 12 000 N, 6 000 N and 8 000 N respectively. The engine exerts a driving force of 45500 N.

8000kg T 2 T 2 10000kg T 1 T 1 60000kg 45500 N 8000N 6000 N 12000 N Find the acceleration of the train and tensions T 1 and T 2 .

System as a whole: 45500 – 12000 – 6000 – 8000 = 78000a  a = 0.25 ms -2 .

Train: 45500 – T 1 – 12000 = 60000a  T 1 = 18 500 N Truck B: T 2 – 8000 = 8000a  T 2 = 10 000 N