SECTION 5: THREE NEWTON’S LAWS (2nd)
Isaac Newton (1643-1727)
Write the formula of Newton’s second law and equation words.
Answer:
1. Vocabulary and term
Pull (v) [pul] :
Push (v) [pu∫]:
Interact (v) /,intзr`ækt/:
Action (n) [`æk∫n]:
Reaction(n) [ri:`æk∫n]
Weight (n) [`weit]:
Gravitation/gravity (n) [`græviti]:
Friction (n) : [`frik∫n]
Contact force :
opposite in direction:
equal in magnitude:
1. Vocabulary and terms
Pull (v) [pul] : Kéo
Push (v) [pu∫]: Đẩy
Interact (v) /,intзr`ækt/: tác động qua lại.
Action (n) [`æk∫n]: Lực tác dụng
Reaction(n) [ri:`æk∫n] Phản lực.
Weight (n) [`weit]: Trọng lượng.
Gravitation/gravity (n) [`græviti]: Trọng lực
Friction (n) : [`frik∫n] Lực ma sát.
Contact force : Lực tiếp xúc.
opposite in direction: Ngược hướng
equal in magnitude: Cùng độ lớn.
2. Identifying forces
+ Pushes and pulls. You can make an object accelerate by pushing and pulling it. Your force is shown by an arrow pushing (or pulling) the object.
+ Weight. This is the force of gravity acting on the object. It is usually shown by an arrow pointing vertically downwards from the object’s centre of gravity.
+ Friction. This is the force which arises when two surfaces rub over one another. If an object is sliding along the ground, friction acts in the opposite direction to its motion.
Contact force.
Upthrust.
Tension.
3. Newton’s third law of motion
When two bodies interact, the forces they exert on each other are equal in magnitude and opposite in direction.
Action and reaction, have the following characteristics:
+ They act on different objects.
+ They are equal in magnitude.
+ They are opposite in direction.
+ They are forces of the same type.
4. Test yourself
1 . Calculate the force needed to give a car of mass 800 kg anacceleration of 2.0 m s-2.
Solution:
Subtitude the values a=2.0 m.s and m=800kg
We have the formula:
2. A motorcyclist of mass 60 kg rides a bike of mass 40 kg. As she sets off from the lights, the forward force on the bike is 200 N.
Assuming the net force on the bike remains constant, calculate the bike’s velocity after 5.0 s.
Solution:
Applying the Newton’s second law.
Substituting these values gives:
To calculate the force, we use:
Substituting these values
3. Draw force diagram ‘Newton’s third law pair’ of forces involved in the following situations. In each case, state the object that each force acts on,
the type of force and the direction of the force.
a You step on someone’s toe.
b A car hits a brick wall and comes to rest.
c A car slows down by applying the brakes.
d You throw a ball upwards into the air.
Summary:
+ Newton’s second law of motion.
For a body of constant mass, the acceleration is directly proportional to the net force applied to it. Net force F, mass m and acceleration a are related by the equation:
net force = mass × acceleration



+ The acceleration produced by a force is in the same direction as the force. Where there are two or more
forces, we must determine the resultant force.
+ Newton’s third law of motion.
When two bodies interact, the forces they exert on each other are equal in magnitude and opposite in
direction.
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