1. FORCES AND NEWTON’S LAWS OF MOTION

2. Newton’s First Law
An object at rest will remain at rest and an object in motion will remain in motion unless acted upon by an outside force
Net force is the vector sum of the forces acting on an object
If Net Forces = Zero. Object is in Equilibrium

3. Inertia
The natural tendency of an object to remain at rest or in motion at a constant speed along a straight line
The mass of an object is a quantitative measure of inertia
SI unit of Inertia and mass: kilogram (kg)

4. Newton’s Second Law of Motion
If a net force is acting upon an object, the sum of the forces acting on the object is equal to the mass of the object times its acceleration
ΣFnet = F1 + F2 + F3…= ma
Fnet = ma

5. Example 1:
A 900 kg car accelerates from rest to 25 m/s in 6 s. What is the net force acting on the car?
SI unit of force = 1 Newton = 1kg∙m/s2

6. Find the magnitude and Direction of the block’s accel.
ΣF = ma ΣFx = max ΣFy = may
Example 2: wooden block
Find the magnitude and Direction of the block’s accel.
F1 = 80N
30o
20 kg
F2 = 60N
Force
x comp
y comp F1
-80N 0N F2
(cos 30o)(60N)
(sin 30o)(60N)
ΣFx= N
ΣFy = -30N
-28.04N = 20kg*ax
-30N = 20kg*ay

7. Example 3:
F1 = 28 N
25o
F2 = 10 N
Force
x comp
y comp F1 F2

8. Example 4: one force acting on an object
a = 4.0 m/s2 straight down
What are the x and y components of the second force?
F1 = 147N
a = 4.0 m/s2
Force
x comp
y comp F1
147N 0N F2
F2x
F2y
ΣFx = 147N + F2x
ΣFy = 0 + F2y = may
ax = 0
ΣFy = F2y = may
ΣFx = 0 = 147N + F2x
ΣFy = 63kg ∙ -4.0 m/s2
F2x = -147N
F2y = -252N

9. Newton’s Third Law
If Object A exerts a force on Object B, then Object B exerts a force on Object A. These forces are equal in magnitude but opposite in direction.
FAB
FBA A B
FAB = -FBA

10. Example 5: a skater pushing against a wall
A key to the correct
application of the third
law is that the forces
are exerted on different
objects. Make sure you
don’t use them as if
they were acting on the
same object.

11. Identifying Forces Non-contact Forces Gravity Magnetism electric
Push
Pull

12. Apparent Weight ΣFy = +FN – mg = ma
The force that the object exerts on the scale with which it is in contact
Two forces acting
W=mg
Normal force (FN)
ΣFy = +FN – mg = ma

13. Apparent Weight If acceleration is 0 m/s2, Fy = +FN – mg = ma
If acceleration is upward, apparent weight is greater than true weight
FN – mg = ma
Apparent
weight
True

14. Apparent Weight
If acceleration is downward, apparent weight is less than true weight
Fy = +FN – mg = ma
FN = -ma + mg
If acceleration is equal to gravity (free fall), apparent weight is 0
FN – mg = -ma
FN = 0

15. Example 11: 735N man in an elevator (75kg)
Acceleration = 0 ΣFy = 0
ΣFy = FN – mg = ma
735N – (75kg)(9.8m/s2) = 0
735N = 735N
Acceleration = 3.0 m/s2 upward
ΣFy = FN – mg = (75kg)(3.0m/s2)
FN -735N = 225N
FN = 960N

16. FN – (75kg)(9.8m/s2) = (75kg)(-2.5m/s2)
Acceleration = 2.5m/s2 downward
ΣFy = FN – mg = ma
FN – (75kg)(9.8m/s2) = (75kg)(-2.5m/s2)
FN – 735N = N
FN = 547.5N
Acceleration = 9.8m/s2
FN – 735N = (75kg)(-9.8m/s2)
FN = 0

18. Example 12: a neck in tractionF
Both act on 1st vertebrae
Dr. wants 34N to be F
What is m1? m1
ΣFx = F – T
F = T
T = mg
F/g = m
34N / 9.8m/s2 = 3.5 kg

19. Example 13: a foot in traction
ΣFy = T1 sin35 – T2 sin35 = 0
ΣFx = T1 cos35 + T2 cos35 - F = 0
T1 = T2 = T solve for F
T1 + T2 = R
R = 2Tcos35o
R is determined 2.2 kg mass
F = 2T cos35o = 2mg cos35o
2(2.2kg)(9.8m/s2) cos 35o = 35N T1
35o
35o T2 F
2.2 kg

20. Example 14: pulling a boat
Find the sum of the two vectors
FAx = FA cos45o = (40.0N)(0.707) = 28.3N
FAy = FA sin45o = (40.0N)(0.707) = 28.3N
FBx = FB cos37o = (30.0N)(0.799) =24.0N
FBy = FB sin37o = -(30.0N)(0.602) = -18.1N
FRx = FAx + FBx = 28.3N N = 52.3N
FRy = FAy + FBy = 28.3N -18.1N = 10.2N
Then use Pythagorean theorem and tan-1