Teach A Level Maths
Connected Particles (2)
Volume 4: Mechanics 1
Connected Particles (2)
Suppose a car is moving along a straight horizontal
road. Modelling the car as a particle we have the
following force diagram.
v
a
R
car
Fd
Fr1
Mg
The force due to the car’s engine is called the
We may also have a force resisting the motion.
driving force. Fd
Suppose the car is towing a trailer using a towrope
to connect the two.
v
a
N
R
T
Fr2
T
car
Fd
Fr1
mg
Mg
The tension in the towrope is the force that
accelerates the trailer.
Tip: Exaggerate the length of the tow rope so
that there is room to show the forces.
v
a
N
R
T
Fr2
T
car
Fd
Fr1
mg
Mg
To find an equation for T, we need to use F=ma on
either the car or the trailer.
We draw exactly the same force diagram for a
towbar or coupling but the force in the bar may
be a thrust instead of a tension. If the force is
a thrust, T will be negative.
v
a
N
R
T
Fr2
T
car
Fd
Fr1
mg
Mg
You can find the driving force, Fd
if you are given the values of the acceleration and
the resisting forces using F = ma.
You must consider the forces on each
mass in turn.
e.g.1. A car of mass 1000 kg is towing a trailer of
mass 200 kg in a straight horizontal line. The car is
travelling with constant acceleration of magnitude
0·5 m s -2.
The resistance to motion for the car is of magnitude
600 newtons and the trailer is 100 newtons.
The towrope connecting the car and trailer is
horizontal.
Find the magnitudes of
(a) the tension in the towrope and
(b) the driving force.
car: mass 1000 kg
resistance:
600 newtons
trailer
v
R
T
600
1000g
car
Fd
trailer: mass 200 kg
resistance:
100 newtons
trailer
100
v
N
R
T
200g
T
car
Fd
600
1000g
We always show all the forces, even though the
normal reactions are not used in this type of problem.
trailer
acceleration:
0·5 m s -2
(a) Find T
100
v
a = 0·5
N
R
T
200g
T
600
1000g
N2L: Resultant force = mass  acceleration
Trailer:
T - 100 = 200  0·5
T = 200

car
Fd
trailer
acceleration:
0·5 m s -2
(a) Find T
v
a = 0·5
N
R
T
100
200g
T
600
1000g
N2L: Resultant force = mass  acceleration
Trailer:
(b) Find Fd
T - 100 = 200  0·5
T = 200

Consider Car alone:
Fd - T - 600 = 1000  0·5
Subs. for T:
Fd = 1300
The tension in the towrope is 200 newtons
and the driving force is 1300 newtons.
car
Fd
SUMMARY
 When a vehicle tows another with a rope, there
is a tension in the rope.
 If a towbar or coupling is used, we draw the
same diagram. If the calculations give a
negative value of T there is a thrust instead of a
tension.
 The tension ( or thrust ) provides the force that
gives the towed vehicle an acceleration equal to
that of the towing vehicle.
 We solve problems by using N2L on the bodies
separately and/or together.
EXERCISE
1. The sketch shows a car towing a caravan. The
masses and resistances to motion are shown.
800 N
1500 kg
500 kg
100 N
(a) The driving force of the car is 1700 newtons
Draw a complete force diagram and show that the
acceleration of the car and caravan is 0·4 m s -2.
(b) If the car then decelerates at a constant rate of
0·3 m s -2, find
(i) the force in the tow bar, explaining the sign in
your answer, and
(ii) the new driving force.
800 N
500 kg
1500 kg
car
Solution:
(a) Find a.
EXERCISE
100 N
a
N
T
1700
driving force:
1700 newtons
R
caravan
T
100
500g
N2L: Resultant force = mass  acceleration
Car
1700 - T - 800 = 1500a
caravan:
T - 100 = 500a
adding
800
1500g
1700-800-100 = 2000a

a = 800  a = 0·4 m s -2
2000
EXERCISE
Solution:
(b)(i) Find T
car
Fd
1700
a1 = -0·3
N
T
T
800
1500g
N2L: Resultant force
caravan:
T - 100


R
caravan
T
T
100
500g
= mass  acceleration
= 500(-0·3)
= -150 + 100
= -50
The negative sign shows that the force is the
opposite direction to that shown in the diagram.
( We say there is a thrust of 50 newtons. )
EXERCISE
Solution:
(b)(ii) Find Fd
car
Fd
1700
a1 = -0·3
N
T
R
caravan
T
100
T = -50
500g
N2L: Resultant force = mass  acceleration
car:
Fd - T - 800 = 1500(-0·3)
Fd = -450 + 800 - 50

Fd = 300

The driving force is 300 newtons
800
1500g
The following page contains the summary in a form
suitable for photocopying.
TEACH A LEVEL MATHS – MECHANICS 1
CONNECTED PARTICLES (2)
Summary

When a vehicle tows another with a rope, there is a tension in the rope.

If a towbar or coupling is used, we draw the same diagram. If the
calculations give a negative value of T there is a thrust instead of a
tension.

The tension ( or thrust ) provides the force that gives the towed vehicle
an acceleration equal to that of the towing vehicle.

We solve problems by using N2L on the bodies separately and/or
together.