Download forces and newton`s laws of motion

FORCES AND NEWTON’S LAWS OF MOTION
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
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)
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
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
ΣF = ma
ΣFx = max
ΣFy = may
Example 2: wooden block
F1 = 80N
30o
20 kg
F2 = 60N
Find the magnitude and Direction of the block’s
accel. Force
x comp
y comp
F1
-80N
0N
F2
(cos 30o)(60N)
(sin 30o)(60N)
ΣFx= -28.04N
ΣFy = -30N
-28.04N = 20kg*ax
-30N = 20kg*ay
Example 3:
F1 = 28 N
25o
F2 = 10 N
Force
F1
F2
x comp
y comp
Example 4: one force acting on an object
a = 4.0 m/s2 straight down
F1 = 147N
a = 4.0 m/s2
Force
F1
F2
What are the x and y
components of the second
force?
x comp
147N
F2x
y comp
0N
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
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
A
B
FAB = -FBA
FBA
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.
Identifying Forces
Non-contact Forces
Gravity
Magnetism
electric
Contact Forces
Push
Pull
Apparent Weight
• 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
Apparent Weight
• If acceleration is 0 m/s2,
Fy = +FN – mg = ma
FN = mg + ma
Apparent
True
weight
weight
• If acceleration is upward, apparent weight
is greater than true weight
Fy = +FN – mg = ma
FN – mg = ma
FN = mg + ma
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
Fy = +FN – mg = ma
FN – mg = -ma
FN = 0
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
Acceleration = 2.5m/s2 downward
ΣFy = FN – mg = ma
FN – (75kg)(9.8m/s2) = (75kg)(-2.5m/s2)
FN – 735N = -187.5N
FN = 547.5N
Acceleration = 9.8m/s2
ΣFy = FN – mg = ma
FN – 735N = (75kg)(-9.8m/s2)
FN = 0
Example 12: a neck in traction
T
F
m1
Both act on 1st vertebrae
Dr. wants 34N to be F
What is m1?
ΣFx = F – T
F=T
T = mg
F/g = m
34N / 9.8m/s2 = 3.5 kg
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
35o
35o
T2
F
T1 + T2 = R
R = 2Tcos35o
R is determined 2.2 kg mass
F = 2T cos35o = 2mg cos35o
2.2 kg
2(2.2kg)(9.8m/s2) cos 35o = 35N
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 + 24.0N = 52.3N
FRy = FAy + FBy = 28.3N -18.1N = 10.2N
Then use Pythagorean theorem and tan-1