Forces
Forces - LOs
• (a) Solve problems using the relationship:
• net force = mass x acceleration (F = ma)
appreciating that acceleration and the net
• force are always in the same direction;
• (b) define the newton;
• (c) apply the equations for constant acceleration and F =
ma to analyse the motion of objects;
• (d) recall that according to the special theory of relativity,
F = ma cannot be used for a particle travelling at very
high speeds because its mass increases.
Newton’s Laws of motion
Newton’s First Law
An object stays still or keeps moving in a
straight line at the same velocity unless a
resultant force makes it change.
For speeding up, slowing down
or changing a direction there
is always a force involved.
Newton’s Second Law
Acceleration of an object is directly
proportional to the resultant force
exerted on it and inversely proportional
to its mass.
F
a= m
The harder you push something
the faster it will speed up, and
the smaller the object is, the
faster it will speed up.
Newton’s Third Law
For every action there is an equal and
opposite reaction.
If you jump in the air. As you push your feet against
the ground, the ground pushes back and causes you to
go up to a height of, say 1 meter.
However, your feet pushing against the ground also
cause the Earth to move in the opposite direction
from you.
Because the Earth is a lot bigger and heavier than
you, it only moves a tiny bit. ( about
0.00000000000000000000001 meter)
The Newton
• One Newton is the force that causes a
mass of one kilogram to have an
acceleration of one metre pre second
every second
Relativity
• Einstein’s theory of special relativity says
that as an object starts travelling very fast
(near to the speed of light), it’s mass
increases. Therefore we can’t use F=ma
for these objects
Summary Q
1. A car of mass 800 kg accelerates uniformly along a straight line
from rest to a speed of 12 ms-1 in 50 seconds. Calculate:
a. The acceleration of the car
b. The force on the car that produced this acceleration
c. The ratio of the accelerating force to the weight of the car.
a. The acceleration of the car ‘a’
a = (v-u)/t
= (12 – 0)/50
= 0.24 ms-2
b. The force on the car that produced this acceleration
‘F’
F = ma
F = 800 x 0.24 = 192 N
c. The ratio of the accelerating force to the weight of the car.
F/mg
= 192/(800x9.8)
= 0.024
Summary Q
2. An aeroplane of mass 5000 kg lands on a runway at a speed of
60ms-1 and stops 25 seconds later. Calculate:
a. The deceleration of the aeroplane
b. The braking force on it
a. The acceleration of the aeroplane ‘a’
a = (v-u)/t
= (0 –60)/25
= -2.4 ms-2
b. The braking force ‘F’
F = ma
F = 5000 x -2.4 = -12000N
The negative sign indicates the direction of the opposing force.
3. A vehicle of mass 1200 kg on a level road accelerates from
rest to a speed of 6.0 ms-1 in 20 s, without change of
direction.
a. Calculate the force that accelerates the car.
b. The vehicle now is fitted with a trailer of mass 200 kg.
Calculate the time taken to reach a speed of 6.0ms-1 from
rest for the same force in part a.
a. Acceleration = (v-u)/t
= (6 – 0)/20
F = ma = 1200 x 0.3 = 360 N
b. Total mass = 1400 kg
Force = 360 N
a = F/m = 360/1400 = 0.26 ms-2
a = (v-u)/t
t = (v-u)/a
= (6 – 0)/0.26
=
23 s
= 0.3ms-2
4. A bullet of mass 0.002 kg travelling at a speed of 120 ms-1 hit a
tree and penetrated a distance of 55 mm into the tree.
Calculate:
a. The deceleration of the bullet
b. The impact force of the bullet on the tree.
a. The deceleration of the bullet
v2 = u2 + 2as
0 = (120)2 + 2 x a x 55x10-3
-(120)2 = 2 x a x 55x10-3
a = -(120)2 / (2 x 55x10-3)
= -130909 ms-2
b. The impact force of the bullet on the tree.
F = ma = 0.002 x -130909 = -261.8 N
The negative sign indicates that the direction of the force is in opposite
direction of the movement of the bullet.
A car of mass 1200 kg has an engine which provides an
engine force of 600N. Calculate:
a. Its initial acceleration
b. Its acceleration when the drag force is 400N
a. Acceleration = Force/mass
a = 600/1200 = 0.5ms-2
b. When the drag force is 400N, the resultant force
is equal to the weight – drag
resultant force = 600 – 400 = 200N
a = 200/1200 = 0.16kg
1. A rocket of mass 550 kg blasts vertically from the launch pad at
an acceleration of 4.2 ms-2. Calculate:
a. the weight of the rocket
b. The thrust of the rocket engines
Thrust T
a. the weight of the rocket ‘W’
W = mg
W = 550 x 9.8 = 5390 N
b. The thrust of the rocket engines ‘T’
The resultant force F = ma
F = 550 x 4.2
= 2310 N
F=T–W
T=F+W
= 2310 + 5390 = 7700 N
Weight W
A car of mass 1400 kg pulling a trailer of mass 400 kg accelerates
from rest to a speed of 9.0 ms-1 in a time of 60 s of a level road.
Calculate
a. The tension in the two bar
b. The engine force
a. The tension in the two bar
Acceleration a = (v-u)/t
F = ma
= (9-0)/60 = 0.15 ms-2
= 400 x 0.15 = 60N
b. The engine force = the force to accelerate the car + the force to
accelerate the trailer
The engine force = (1400 x 0.15) + (400 x 0.15) = 270 N
A lift and its occupants have a total mass of 1200 kg. Calculate
the tension in the lift cable when the lift is:
a. Stationary
b. Ascending at a constant speed
c. Ascending at a constant acceleration of 0.4 ms-2
d. Descending at a constant deceleration of 0.4 ms-2
a. Stationary: T = weight of the lift = 1200 x 9.8 = 11760N
b. Ascending at a constant speed: same as when stationary
c. Ascending at a constant acceleration of 0.4 ms-2:
Force causing the acceleration = ma = 1200 x 0.4 = 480N
Total tension in the lift = 11760 + 480
= 12240 N
d. Descending at a constant deceleration of 0.4 ms-2
Force causing the deceleration = ma = 1200 x 0.4 = 480N
Total tension in the lift = 11760 + 480
= 12240 N
A brick of mass 3.2 kg on a sloping flat roof, at 30° to the
horizontal slides at constant acceleration 2.0 m down the
roof in 2.0 seconds from rest. Calculate:
a. The acceleration of the brick
b. The frictional force on the brick due to the roof
a. The acceleration of the brick
N
s = ut + ½ a t2
F
2 = 0 + ½ a 22
a = 1 ms-2
b. The frictional force on the brick due
to the roof:
mg cos30°
30°
mg
Since the brick is accelerating there must be a resultant force on it in
the direction of the roof.
Resultant force = ma = 3.2 x 1 = 3.2 N
Resultant force = mgsin30 - Friction
3.2
= (3.2x9.8x0.5) - Friction
Friction = 12.48 N