Download Topic 3: Newton`s Laws

Topic 3: Newton’s Laws
Newton’s first law
Every body remains at rest or moves in a straight line with constant velocity unless some external
force compels it to act otherwise.
Example 1
8 m/s
7N
x
The diagram shows a body moving in the horizontal plane under the influence of two horizontal
forces. State the magnitude of the force x.
Solution 1
This body is moving at a constant speed so there must be no net horizontal force. It follows
therefore that the two forces should balance.
Therefore
X = 7N
Example 2
30N X
4 m/s
20 N
Y
70N
The diagram shows a body moving in the horizontal plane under the influence of a system of forces.
Given that the body is moving at a constant speed of 4 m/s in the direction shown find X and Y.
Solution 2
Since there is no acceleration in the horizontal and vertical planes there must be no net force in
either of these directions.
Horizontal
20N = Y
Vertical
30 + X = 70
X = 70 – 30
X = 40 N
Newton’s second law
The acceleration of a body is proportional to the force impressed on the body and acts along the
same straight line as the force.
Newton’s second law can be written as the equation
F = ma
Where F is the force acting on the body in newtons (N)
m is the mass of the body in kg
a is the acceleration of the body in m/s²
One common example of Newton’s second law is the weight of a body. Weight is the force which
the earth exerts on a body and calculated using the above equation as follows:
F = ma
Weight = mg
(where g is acceleration due to gravity- 9.8 m/s²)
Note: The units of weight are newtons (N)
Example 3
72N
10 kg
2 m/s²
Y
The diagram illustrates a body with mass 10 kg moving in the horizontal plane with acceleration 2
m/s² under the action of 2 forces find Y.
Solution 3
The net force in the direction of acceleration is 72 – Y. Using Newton’s second law;
F = ma
72 – Y = 10 x 2
72 – 20 = Y
Y = 52 N
Example 4
The diagram below shows a body of mass 4 kg moving in the horizontal plane under the action of a
system of forces. Given that the body accelerates at 3 m/s² in the direction of the 58N force,
find X and Y.
20N
80N
X
4 kg
Y
3 m/s²
58N
Solution 4
Since the body experiences no acceleration in the direction of the Y force;
Y = 80N
The body experiences an acceleration of 3 m/s² in the plane of the force x.
Here the net force is 58 – 20 – X
Using
F = ma
58 – 20 – X = 4 x 3
38 – X = 12
38 – 12 = X
26 = X
X = 26N
Example 5
A body of mass 4 kg travels vertically upwards under the action of a force of 80N. Calculate its
upwards acceleration.
Solution 5
80N
4 kg
a m/s²
weight
Weight = mass x g
= 4 x 9.8
= 39.2 N
Therefore the net force in the direction of acceleration is (80 – 39.2) N.
Using
F = ma
80 – 39.2 = 4a
40.8 = 4a
40.8 ÷ 4 = a
10. 2 = a
a = 10.2 m/s²
Note: You can only use F=ma if there is an acceleration in the plane you are dealing with.
Newton’s third law
For every action there is an equal and opposite reaction.
If an object is sitting on a desk its weight is acting downwards on the desk. The desk is exerting
an equal and opposite force on the object, this force is called the normal reaction.