Transcript Connected Particles (2)
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
F d Fr
1
Mg
The force due to the car’s engine is called the driving force. F d Suppose the car is towing a trailer using a towrope to connect the two.
N v a R
car
T T F d Fr
2
Fr
1
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.
Fr
2
N mg T v a R T Fr
1
Mg
car
F d
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.
Fr
2
N mg T v a R T Fr
1
Mg
car
F d
You can find the driving force,
F
if you are given the values of the acceleration and the resisting forces using F = ma.
d
You must consider the forces on each mass in turn.
e.g.1. A car of mass mass 200 1000 kg is towing a trailer of 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 resistance: 600 1000 kg newtons
trailer
v R T
600 1000g
car
F d
trailer: mass resistance: 100 200 kg newtons
trailer
100
N T v R T
600 1000g
car
F d
200g We always show all the forces, even though the normal reactions are not used in this type of problem.
acceleration: 0·5 m s -2
trailer
100
N T v
a = 0·5
T R
600 (a) Find
T
200g 1000g N2L: Resultant force
=
Trailer:
T
-
100
=
T
=
mass 200 200
acceleration 0·5
car
F d
acceleration: 0·5 m s -2
trailer
100
N T v
a = 0·5
T R
600 (a) Find
T
200g 1000g N2L: Resultant force
=
mass
acceleration Trailer: (b) Find
F d
Consider Car alone:
F
Subs. for
T d
:
-
T T
-
100
T
600
F d
= = =
200 200
1000
= 1
300 and the driving force is 0·5 0·5 The tension in the towrope is 1300 200 newtons newtons.
car
F d
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 tension.
T
there is a thrust instead of a 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.
1500 kg 800 N 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 1500 kg
car
500 kg EXERCISE 100 N
N a
driving force: 1700 newtons
R
caravan
Solution: 1700
T T
(a) Find Car caravan:
adding
a
.
800 100 1500g N2L: Resultant force
=
1700
-
T
-
800
=
T
-
100
=
500g mass 1500a
500a 1700-800-100 = 2000a
a
=
800 2000
acceleration
a
=
0·4 m s -2
Solution: (b)(i) Find
T
car
F d
1700 EXERCISE
N T a
1 =
-
0·3
T R
caravan
800 100 1500g 500g N2L: Resultant force
=
mass
acceleration caravan:
T
-
100
T
=
500(
= -
0·3) 150 + 100
The negative sign shows that the force is the opposite direction to that shown in the diagram.
( We say there is a
T
= -
50 thrust of 50 newtons
. )
EXERCISE Solution:
car
N a
1 =
-
0·3
R
caravan
(b)(ii) Find
F d F d
1700
T T
800 100
T
= -
50 1500g 500g N2L: Resultant force
=
car:
F d
-
T
-
800
F d F d
= =
mass
= -
450
+
300
acceleration 1500(
-
0·3) 800
-
50 The driving force is 300 newtons
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 tension.
T
there is a thrust instead of a 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.