1. Forces and the Laws of Motion
Physics ChAPter 4
Forces and the Laws of Motion

2. Ch 4-1 Changes in Motion

3. What is a force? The cause of an acceleration.
Force is measured in Newtons!
1N= 1kg·m/s2
F = ma

4. Field- No contact is required. Can you think of any field forces?
Types of Forces
Contact- a force that arises because of contact between 2 or more objects.
Field- No contact is required. Can you think of any field forces?
Gravity
Electromagnetism

5. Force is a vector
Force diagrams are used to analyze situations and determine resultant forces.
YAY!!

6. Force is a vector, it has magnitude and direction

7. Newton’s First Law
An object at rest stays at rest and an object in motion stays in motion at constant velocity unless acted upon by an outside force.
AKA Law of Inertia

8. Inertia- The natural tendency of an object at rest to stay at rest or an object in motion to stay in motion. Mass is a measure of Inertia.
The larger the mass the greater the inertia.

9. Newton’s Second Law of Motion- The acceleration of an object is directly proportional to net external force acting on the object and inversely proportional to the object’s mass. ΣF = ma

10. Calculating net force and accceleration
Convert both forces A and B into component vectors. Find the x coordinate of the force.
Find the y coordinate of the force.
Repeat for vector B.
Use a coordinate system for all vectors and then add the x and y coordinates, (this is your vector addition).
Convert the force vector into a magnitude angle form.
Find the angle.
Find the magnitude of the force.

11. If necessary: Convert the force into an acceleration using F=ma
If necessary: Convert the force into an acceleration using F=ma. Don’t forget your direction. If an object is at equilibrium then there is no net force acting upon it.

12. Assume that two forces are acting on a 0
Assume that two forces are acting on a 0.1 kg hockey puck as follows: A N at 53° B N at 19 °. What force is being applied to the hockey puck?
What is the acceleration of the hockey puck?

13. Working with Inclined Planes Breaking up ramps into vectors
The force on the cart is the force due to gravity, BUT the vectors are resolved using the RAMP, not the x and y axis.
The force along the plane of the ramp provides an acceleration, but the force perpendicular to the ramp does not provide an acceleration.
To resolve Fg along the ramp, start by figuring out the angle between Fg and the ramp.
Because you know that the angle’s in a triangle add up to 180 °, and the angle between Fg and the ground is 90 °, and the angle between the ground and the ramp is Θ, then the angle between the ramp and Fg (the vertical) is (90° - Θ).
So….. Since sin Θ = cos(90- Θ)
Fg along the ramp = Fg · cos (90° - Θ) = Fg sin Θ.
This makes sense because as Θ goes to 0°, the force along the ramp also goes to zero

14. Fn perpendicular to the ramp
Fn (perpendicular to the ramp) = Fg sin (90 – Θ)
since cos Θ = sin (90 - Θ ) then:
Fn perpendicular to the ramp = Fg cos Θ = mg cos Θ
for an object at an angle
Try this one: A cart with a mass of 1.0 kg is sitting on a ramp at 30°. What are the forces normal to the ramp and Fg along the ramp?

15. NEWTON’S SECOND AND THIRD LAWS
Section 4-3
NEWTON’S SECOND AND THIRD LAWS

16. Force is proportional to mass and acceleration.
ΣF = ma
NEWTON’S SECOND LAW
The acceleration of an object is directly proportional to the net external force acting on the object and inversely proportional to the objects mass.

17. NEWTON’S THIRD LAW
When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.

18. Frictional forces
The NORMAL Force- the force pushing two surfaces together
Fn = mg = weight when perpendicular to the ground.
On an inclined plane Fn = mg cos Θ
Fn is always perpendicular to the surface providing the friction.
Coefficient of Friction (μ mu) depends on the two surfaces and the contact between them.
Ff = μFn

19. Coefficients of Static and Kinetic friction
Static friction is stronger than kinetic friction because on a microscopic level, the two surfaces have had a chance to fully interlock.
So… the coefficient of static friction will be larger than the coefficient of kinetic friction.
Ffs = μs Fn = μs mg on a flat surface
Ffk = μk Fn = μk mg on a flat surface

20. Uphill friction Here come components again…YEA!
I have a 100 kg refrigerator to move up a 3 m ramp at 30° with a static coefficient of friction of 0.2 and a kinetic coefficient of friction of 0.15.
The force of gravity acting along the ramp = mg sin Θ
But I still need to add the frictional force to the force along the ramp to determine how much force must be applied.
Fapplied = mg sin Θ Ff = mg sin Θ + μs Fn
Find the normal force perpendicular to the ramp:
Fn = mg cos Θ.
So Fapplied = mg sin Θ + μs mg cos Θ

21. How far will the refrigerator slide if we let go of it?
The force acting along the ramp = mg sin Θ
Fn = mg cos Θ
Ffk = μk mg cos Θ
Fnet = mg sin Θ - μk mg cos Θ
If we need to find the speed of the fridge:
Vf2 = 2a (distance)

22. Q: Two cats are on a roof. Which slides off first?
A:  The one with the smaller mew (mu).