Transcript Force and motion 1

Force and motion
Objectives
When you have competed it, you should
* understands Newton’s first law of motion
* know some different types of forces
* know and be able to apply Newton’s second law to simple
examples of objects moving in a straight line
* understands the idea of equilibrium.
Force and motion
Newton’s first law
A particle will remain at rest or will continue to move with
constant velocity in a straight line unless the forces act on it
to change that state.
Example
When you are riding a bicycle on a level path and start to free-wheel,
you can keep up an almost constant velocity force some time.
But eventually you will slow down, partly because of air
resistance.
direction of motion
air resistance.
Force and motion
Newton’s first law
Example
When a stone is sent sliding across the frozen surface of a lake.
However smooth the ice appears, the stone eventually slows down
because of friction between the two surfaces
Force and motion
Newton’ second law
The force F applied to a article is proportional to the mass m of the
particle and the acceleration produced.
Force = mass  acceleration
SI units:
F = ma
Mass  kilogram (kg)
Acceleration  metres per second per second (m s-2)
Force 
Newton (N)
1 N is the force needed to give 1 kg an acceleration of 1 ms-2.
Types of forces
Pushing
or Thrust
Tension
T
Driving force
DN
Force and motion
Example
An ice-yacht of mass 400 kg has an acceleration of 1.2 ms-2.
What force is needed to produce this ?
Solution
1.2 ms-2
400 kg
F
F=ma
F = 400  1.2 = 480 N
Force and motion
Example
Find the magnitude of the acceleration produced when a particle
of mass 5 kg is acted upon by a force of magnitude 22 N.
Solution
a ms-2
5 kg
F = ma
F = 22 N
22 = 5  a
a=
4.4 ms-2
Force and motion
Example
A car of mass 1000 kg is pushed with a force of 200 N. Calculate
the acceleration of the ca; and find how long it will take to reach a
speed of 3 ms-1 form rest.
Solution
F=ma:
Data: u = 0
200 = 1000 
v=3
a
: a = 0.2 ms-2
a = 0.2
t?
Required equation v = u + at
3  0  0.2t
 t  15 sec s
Force and motion
Example
A curling stone of mass 18 kg is launched across ice with a speed
of 2 ms-1, and goes a distance of 25 metres before coming to rest.
Calculate the deceleration, and find frictional ore between the
stone and the ice.
Solution
Data:
u= 2
Use the equation : v2 = u2 + 2as
gives: a =
F = ma :
v= 0
s = 25
: 0 2 = 22 + 2 
- 0.08 ms-2
F = 18  ( - 0.06) =
- 1.44 N
a  25