1. Newton’s 1st and 2nd laws of Motion

2. 4 Forces of the Universe
Nuclear : Strong forces holding the nucleus of an atom together (gluon)
Weak : Forces as a result of some atoms that decay the nucleus ( radioactive )
Electromagnetic : a force based on electricity, magnetism and light properties
Gravitational : The weakest force caused by the 2 factors, MASS and DISTANCE

3. Mechanical Forces
Inertia: Newton’s First Law, net or unbalanced force is needed to change the state of motion
Friction: the contact force that acts to oppose sliding motion between surfaces
Normal : the contact force exerted perpendicular to the surface contact
Tension : the pull exerted by a cable when attached to a body
Compression : the push exerted to the object
Weight : force of attraction due to distance from the center of the earth and mass of the object
Units are Measured in NEWTONS ( N )

4. Newton's First Law of Motion ( inertia )
An object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

5. Stationary and Moving objects

6. Balanced and Unbalanced Forces
The book is said to be at equilibrium. There is no unbalanced force acting upon the book and thus the book maintains its state of motion

7. Balanced and Unbalanced Forces
The book is not at equilibrium and subsequently accelerates

8. Newton’s Second law of motion
The acceleration of a body is directly proportional to the net force on it and inversely proportional to the mass. ( F = ma )
Unit of Force = kg-m or N (Newton) s2
Calculation:
mass of an apple is 100g or .1 kg
F = ma = (.1kg) (10 m/s2) = 1 kg-m/s2 = 1 N

9. Sample Force Problem #1
Using Newton’s Second law of Motion ( the acceleration of a body is directly proportional to the net force on it and inversely proportional to the mass, F = ma).
What net force is
required to accelerate
a 1500 kg car
at 3.00 m/s2?

10. Solution Problem #1 F = m a F = 1500 kg x 3.00 m/s2 F = 4500 kg-m/s2or
F = 4500 N
Given:
m = 1500 Kg
a = 3.00 m/s2
Find:
F = ?

11. Static Balance (free body) Problem #2
Forces of 10 N and 15N are arranged according to diagram. What is the net force on this object. What is the rate of acceleration of the 12 Kg object? 10 15
12 Kg

12. Solution #2 Given: Force 10 N and 15 N in opposite directions
Mass = 12 Kg
Find :
Net Force
Acceleration
Net Force = F1 - F2 = 15N – 10N = 5 N
F = ma
5N = 12 Kg (a)
5N = a
12 Kg
0.417 m/s2 = a

13. Sample Question #3
The figure shows a block that is being pulled along the floor. According to the figure, what is the acceleration of the block?

14. Solution Problem #3 F = m a 40N = 20 kg (a) 2 m/s2 = a Find: a = ?
Given:
m = 20 Kg
Ffriction= 10 N
Fpull = 50 N
Find:
a = ?

15. Sample problem #4
A 50-kg child on a skateboard experiences a 75-N force as shown.
What is the expected acceleration of the child?

16. Solution Problem #4 F = m a 75 N = 50 kg (a) 1.5 m/s2 = a Find: a = ?
Given:
m = 50 Kg
Fpush= 75 N
Find:
a = ?

17. Uniform Circular Motion
Circular motion at a constant
speed HAS an ACCELERATION!
Keeping the magnitude of the velocity constant, but changing the direction will create a change in acceleration
Using vector addition, a centripetal = v2 r

18. Sample Problem #3
During a hammer throw, an athlete rotates a 7.00 kg hammer tied to the end of a 1.3 m chain. The hammer moves at the rate of 3 m/s in a circle.
What is the centripetal acceleration of the hammer
What is the tension in the chain?
If the chain breaks at the point shown which direction will the hammer fly?

19. Solution Problem #3 m = 7.00 kg v c = 3.0 m/s r = 1.3 m a c = v2 / r
a c = (30 m/s)2 / 1.3 m
a c = 900 m2/s2 / 1.3 m
a c = m/s2
F c = m a c
F c = 1.3 kg ( m/s2)
F c = 900 the tangent (direction 2)

20. Summary
Forces are invisible measurements that explain the states of motion
Newton derived laws of motion that explained the way things move.
Forces can be balanced (equilibrium) or unbalanced (acceleration).
Uniform circular motion is accelerating because of the constant change in direction