Train and one carriage
Connected particles
Horizontal motion of connected particles
Rn1
Rn2
T
F1
m 1g
T
F2
m 2g
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
The 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:
a ms-2
T
(c) the trailer
(d) R1 and R2
T
F
m1 g
m2 g
Solution
(a)
F = (m1 + m2)a
(b)
F – T = m1a
(c)
T = m2 a
(d)
R1 = m1g and R2 = m2 g
(a) the system as a whole,
(b) the car,
R2
R1
R2
R1
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 2200 N
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.
1 ms-2
T
200 N
1100 g
T
100 N
800 g
Solution
Car: 2200 – T – 200 = 1100a
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.
Rn1
120 N
Rn2
10 kg
T
10g
F1
F1 = R1 = 39.2
System as a whole:
5 kg particle:
T
5 kg
5g
F2
F2 = R2 = 19.6
120 - 39.2 – 19.6 = 15a  a = 4.08ms-2
T – 19.6 = 5a
 T = 40 N
Both particles accelerates at 4.08 ms-2 with a tension 40 N.
Rn1
120 N
Rn2
10 kg
T
10g
F1
T
5 kg
5g
F2
10 kg particle:
5 kg particle:
R1 – 10g = 0
R2 – 5g = 0
R1 = 98
R2 = 49
F1 = R1 = 0.4  98 = 39.2
F2 = R2 = 0.4  49 = 19.6
120 – 39.2 – T = 10a
T – 19.6 = 5a
[1]
[2]
[1] + [2]
120 - 39.2 – 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 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
8000N
T2 T2
10000kg
6000 N
T1 T1
60000kg
45500 N
12000 N
Find the acceleration of the train and tensions T1 and T2.
System as a whole:
45500 – 12000 – 6000 – 8000 = 78000a
 a = 0.25 ms-2.
Train:
45500 – T1 – 12000 = 60000a
 T1 = 18 500 N
Truck B:
T2 – 8000 = 8000a
 T2 = 10 000 N