Resultant Forces and Newton’s Second Law
1. Resultant Forces and Newton’s Second Law
An object is in equilibrium if the forces acting on it balance meaning that there is no
resultant
This means that if an object is not in equilibrium it will be accelerating at a rate that is
proportional to the size of the resultant force. Conversely, if an object is accelerating
it cannot be in equilibrium and their must be a resultant force in the direction of
acceleration.
That is
a
F
m
Newton’s second law states that:
When there is a resultant force there will be an acceleration and
Resultant Force = mass × acceleration
Examples Involving Newton’s Second Law
Example 1 : A particle of mass 3kg is accelerating at 5ms-2. Find the magnitude of the
resultant force acting on the particle.
F  ma
 3 5
 15 N
Example 2 : A cart of mass 200kg is pulled by a horse with a force of 300N. Find the
acceleration of the cart.
F  ma
300  200a
a  1.5 ms -2
Example 3 : Suppose you hold a book that weighs 0.8kg upwards in the palm of your
hand and accelerate it upwards at a rate of 0.5ms-2. What force does your
hand exert on the book if g = 9.8ms-2 ?
R
0.8g
↑ R – 0.8g = 0.8 × 0.5
R = 0.4 + 0.8 × 9.8
R = 8.24 N
Example 4 : A person of mass 50 kg stands on the floor of a lift which accelerates
a) upwards at 5 ms-2
b) downwards at 5 ms-2
Find the reaction, in each case, between the floor and the person’s feet.
R
50g
a) By Newton’s second law, and taking the positive direction to be upwards,
F  ma
R  50 g  50  5
R  50 g  250
 740 N
So in a lift accelerating upwards, the person feels a larger reaction than normal, as
we would expect.
b) For the lift accelerating downwards, we take the positive direction to be
downwards,
F  ma
50 g  R  50  5
R  50 g  250
 240 N
So in a lift accelerating downwards, the person feels lighter. This causes the ‘heart
in the mouth’ feeling.
Example 5 : A car travels a distance of 32m along a straight road while uniformly
accelerating from rest to 16ms-1. Find the acceleration of the car. Given
that the mass of the car is 640 kg, find the magnitude of the accelerating
force.
We must first find the acceleration, using an equation of uniform acceleration.
v 2  u 2  2as
162  02  2  a  32
256  64a
a4
Therefore,
F  ma
 640  4
 2560 N
Example 6 : A bullet of mass 0.02 kg is fired into a wall with velocity 400ms-1. If the
bullet penetrates the wall to a depth of 0.1m, find the resistance of the
wall, assuming it to be uniform.
v 2  u 2  2as
02  400 2  2  a  0.1
160000  0.2a
a  800000
F  ma
 0.02  800000
 16000 N
The resistance of the wall is 16000 N.
Exercis 4B page 63 questions 1 to 9
2. Newton’s Second Law for Connected objects
Activity 1
A car is using a tow bar to pull a trailer along a straight, level road. There are
resisting forces R acting on the car and S acting on the trailer. The driving force of
the car is D and its braking force is B.
Draw diagrams to show the horizontal forces acting on the car and the trailer:
i) When the car is moving at a constant speed
ii) When the speed of the car is increasing
iii) When the speed of the car is is decreasing
In each case write down the resultant force acting on the car and the resultant force
acting on the trailer.
R
i)
D
S
T
T
No resultant force as no acceleration on either the trailer or the car
R
ii)
S
T
Resultant on trailer = T – S
Resultant on car = D – R – T
T
D
R
iii)
B
S
T
T
Resultant on trailer = -T – S
Resultant on car = -B - R + T
When two moving particles are connected by a string which is light and inextensible,
there will be a tension in the string. By Newton’s third law, the forces acting on the
particles will have the same magnitude but will act in equal and opposite directions,
as shown.
T
T
In problems involving connected particles, we can consider the forces acting on each
particle separately, or on the system as a whole.
Example 1 : Two particles A and B of masses 3 kg and 4 kg respectively, connected
by a light inextensible string, are at rest on a smooth horizontal surface.
A force of magnitude 7 N is applied to particle B in the direction AB.
Find the acceleration produced and the tension in the string.
A
T
T
3kg
B
7N
4kg
We can use Newton’s second law for the whole system, or for each particle
separately, as they all have the same acceleration a.
For the system the resultant force is 7 N, as the two tensions cancel each other out,
F  ma
7  7a
a  1 ms-2
For A,
F  ma
T  3 1
3N
Example 2 : A car of mass 1000 kg tows a caravan of mass 750 kg along a horizontal
road. The engine of the car exerts a forward force of 2500 N. The
resistances to the motion of the car and caravan are each k × their masses,
where k is a constant. Given that the car accelerates at 1ms-2, find the
tension in the tow-bar.
1000 k
750 k
caravan
T
T
car
2500
For the system,
F  ma
2500  1750k  1750  1
750  1750k
k
3
7
For the caravan,
F  ma
T  750  73  750 1
T  1071 N (n.w.n.)
Example 3 : A stationary helicopter is raising two people of
masses 90kg and 70kg as shown.
a) Draw a diagram to show all
the forces acting on the people
(ignore the forces acting on the
helicopter)
b) When the force applied to the
90kg is 1800 N find the
acceleration of both people
and the tension in the ropes.
Take g = 10ms-2
90
70
T1
90 kg
↑ T1 – T2 – 900 = 90 × a
70 kg
↑ T2– 700 = 70 × a
But T1 = 1800N
So 1800 – (70a + 700) – 900 = 90a
200 = 160a so a = 1.25ms-2
and therefore T2 = 700 + 70 × 1.25 = 787.5 N
90
T2
T2
90g
70
70g
Example 4 : A block A of mass 100kg is suspended by a vertical cable. A block B of
mass 150 kg is suspended from block A by another vertical cable. The
blocks are raised 10 metres in 10 seconds with a constant acceleration.
Find the tension in each cable.
Using an equation of motion,
T1
s  ut  12 at 2
10  0  12  a  102
A
a  0.2 ms -2
Using Newton’s second law for the system,
F  ma
T1  250 g  250  0.2
T1  2500 N
Using Newton’s second law for B,
100g + T2
T2
B
150g
F  ma
T2  150 g  150  0.2
T2  1470  30
T2  1500 N
2. Problems Involving Pulleys
If two particles are connected by a string passing over a smooth pulley or peg, the
heavier particle will move downwards and the lighter one upwards. Three modelling
assumptions are important here (and often appear in examination questions).
• As before, if they are connected by an inextensible string, they will have the
same acceleration.
• If the pulley or peg is smooth, the tension in the portions of string either
side of the pulley or peg will be the same. Anyone who has made a guitar
from an empty shoe box and an elastic band is relying on the friction
between box and band to keep one part of the band more taut than the other.
• If the string is light, it has no mass, and therefore the tension will be
constant along its length. If the string had mass, the portion of the string
higher up would be more tense as it would have more mass to support
beneath it.
Example 1 : Two particles of mass 5 kg and 8 kg are connected by a light inextensible
string which passes over a smooth fixed pulley. Find the acceleration of
the system and the tension in the string.
Using N2 upwards for the 5 kg mass,
F  ma
T  5 g  5a
T
Using N2 downwards for the 8 kg mass,
F  ma
8 g  T  8a
T
5g
Adding the equations,
8g
3g  13a
a  133 g
 2.26 ms-2 (2 d.p.)
Substituting in the first equation,
T  5 g  5a
 6 132 g
 60.31 N (2 d.p.)
Example 2 : Two particles of masses 2 kg and 7 kg are connected by a light inelastic
string passing over a fixed pulley. The system is released from rest with
both portions of the string vertical, and both particles at a height of 3m
above the ground. In the subsequent motion, the 7 kg mass hits the
ground. Find the greatest height reached by the 2 kg mass if the pulley is
of such height that the mass never reaches the pulley.
For the 7kg mass,
7 g  T  7a
For the 2kg mass,
T  2 g  2a
T
Adding these equations,
5 g  9a
a  95 g
2g
 5.44 ms -2 (2 d.p.)
We can now find the velocity of the 2 kg mass when the 7 kg
mass hits the ground.
v 2  u 2  2as
v 2  0  2  95 g  3
v
10
3
g
T
7g
3m
The 2kg mass is now a projectile, with initial upward velocity
10
3
g , and decelerating
under gravity.
Considering when it comes to instantaneous rest,
v 2  u 2  2as
02  103 g  2 gs
s  53 g
So the total height of the 2kg mass is 3  3  53  7 23 m.
Example 3 : Two particles P and Q of masses 6 kg and 3 kg respectively are
connected by a light inextensible string. Particle P rests on a rough
horizontal table. The string passes over a smooth pulley fixed at the edge
of the table, and Q hangs vertically. The coefficient of friction between P
and the table is 13 . The system is released from rest. Find in terms of g
a) the acceleration of Q,
b) the tension in the string,
c) the force exerted on the pulley by the string.
a) For Q,
3g  T  3a
R
T
F
For P,
T   R  6a
P
6g
T
T  13  6 g  6a
Q
T  2 g  6a
3g
Adding the equations,
g  9a
a  19 g
The acceleration of Q is
1
9
g vertically downwards.
b) Substituting this value of a in the second equation,
T  2 g  6  19 g
T  83 g
The tension in the string is
8
3
g N.
c) By Newton’s third law, the forces on the pulley are T horizontally, and T vertically,
giving the resultant shown.
T
So we have
T
R
R  T2 T 2
 2T 2
 2T
 83 2 g N
The force exerted on the pulley by the string is 83 2 g N , directed at 45° below the
horizontal. Of course, this force is opposed by the reaction of the table.
Exercise 4C page 71 questions 1 to 8