Forces and Motion
What You Will Learn
- Newton’s Laws of Motion
- Motion Terminology
- Sample Problems
Newton’s Three Laws of Motion
Newton’s First Law of Motion – An object at rest will remain at rest and an object in
motion will continue in motion at a constant velocity in a straight line unless it is
acted upon by a net force.
This is often called the Law of Inertia because it implies that any mass offers resistance to a
change in its velocity.
What are other implications?
-
No applied force to an object means no change in its velocity. Or, if velocity changes,
then there has to be a net force that is making that change.
-
If there is an acceleration, then there has to be a velocity change and a net force present.
Newton’s Three Laws of Motion
Newton’s Second Law of Motion – An acceleration is directly proportional to the applied
force and inversely proportional to the mass of the object. For us; F = ma.
What are the implications?
- You can not have a force with out an acceleration, since mass of an object does not
change.
- Force is a vector (has magnitude and direction).
Newton’s Three Laws of Motion
Newton’s Third Law of Motion – States that all forces come in pairs that are equal in
magnitude but opposite in direction. To paraphrase; For every action there is an equal
but opposite reaction.
What are the implications?
- You can not have a force with out an acceleration, since mass of an object does not
change.
Types of Forces
-
System – An object of interest.
-
Force – A push or a pull.
Contact Force – Physically touching.
Field Force – Exert without contact.
Agent – It is a contact or field force and is specifically affects a given system.
-
Free Body Diagram – Physical representation that shows the forces acting on a
system.
-
Net Force – Is the vector sum of the forces acting on a system. Symbolized by Fnet.
-
Equilibrium – A system in which the velocity remains constant. Note: an object at
rest is just a special case of velocity, namely v = 0.
-
Weight – The gravitational force experience by an object. Fg = mg.
-
Apparent Weight – A weight that is not true weight, but it is one you experience due
to your current conditions.
Types of Forces
-
Weightlessness – The absents of apparent weight.
-
Drag Force – The force exerted by a fluid on an object opposing motion through the
fluid. Note: air is a fluid.
-
Interaction Pair – Newton’s Third Law. All forces come in pairs.
-
Tension – Force exerted by a rope, string, or cable. The rope, string, or cable is
generally considered massless.
Tension
y
Wall or attachment point
T
Tx= T cos 
Ty= T sin 

x
Fa
 Fy = 0 = Ty + Fg
Fg
 Fx = 0 = Tx + Fa
0 = T sin 120o + Fg
0 = T cos 120o + Fa
– T sin 120o = Fg
– T cos120o = Fa
T=
F𝑔
– sin 120o
T=
F𝑎
– cos 120o
Tension
y
Wall or attachment point
T
Tx= T cos 
Ty= T sin 

x
Fa
What happens to tension as
 gets bigger?
Fg
Tension
Fg = –100 N Fax is the applied force
o
T1 90
T2
60o
T4
o
T3 45
Fg
Fg
What is the tension in T1, T2, T3, and T4?
Team Work
Fa4
Fa3
Fa2
Fg
30o
Fg
Tension
Fg = –100 N Fax is the applied force
y
y
o
T1 90
T2
60o
T4
o
T3 45
x
x
Fg
y
y
Fg
x
Fa2
30o
x
Fa3
Fg
Fa4
Fg
 F = 0 = T1 + Fg
 Fy = 0 = T2y + Fg
 Fy = 0 = T3y + Fg
 Fy = 0 = T4y + Fg
–T1 = Fg
0 = T2 sin 120o + Fg
0 = T3 sin 135o + Fg
0 = T4 sin 150o + Fg
– T2 sin 120o = Fg
– T3 sin 135o = Fg
– T4 sin 150o = Fg
(−100 𝑁)
T2 =
− sin 120𝑜
(−100 𝑁)
T3 =
− sin 135𝑜
(−100)
T4 =
− sin 150𝑜
T2 = 115 N
T3 = 141 N
T4 = 200 N
T1 = – Fg
T1 = – (100 N)
T1 = 100 N
Mouse or Enter to Continue
Tension
Fg = –100 N Fax is the applied force
o
T1 90
T2
60o
T4
o
T3 45
Fg
Fa4
Fa3
Fa2
Fg
30o
Fg
Fg
T1 = 100 N
T2 = 115 N
T3 = 141 N
T4 = 200 N
Tension
m = 1.00  103 kg.
Find T, no acceleration
 Fy = 0 = T + Fg
T
–T = Fg
pulley
𝐓 = 1000 𝑘𝑔 9.8
T = – (– 9800 N)
T = 9.8 103 N
Fg
𝑚
𝑠2
sin 270o
Tension
m = 1.00  103 kg.
𝑚
Find T, anet = 3.25 𝑠2
 Fy = Fnet = T + Fg
T
manet = T + Fg
pulley
1000 𝑘𝑔 3.25
𝑚
𝑚
=
𝐓
+
1000
kg
9.8
sin 270o
2
2
𝑠
𝑠
3250 N = 𝐓 − 9800 N
T = 13050 N
Fg
T = 1.3 104 N
What You Have Hopefully Learned
- Newton’s Three Laws of Motion
- Motion Terminology
- Sample Tension Problems
END OF LINE
And Now, The Crane Problem.
Two cranes are holding a load that has a mass that will be given to you by the teacher.
Find the value of T1 and T2. Show the force diagram and the vector equation needed to
solve this problem.
30o
T1
100o
T2
mass = 2000 kg
y
Simplified Force
Diagram
o
T1
T2y = T2 sin 30o
T1y = T1 sin 130o
100o
T1x = T1 cos 130o
T2x = T2 cos 30o
 Fy = 0 = T1y + T2y + Fg
0 = T1 sin 130o + T2 sin 30o – 19600 N
19600 N = T1 sin
19600 N = 𝐓1 sin
−𝐓1 cos 130o
+ cos 30o
sin 30o
− cos 130o
o
130
sin 30o
cos 30o
19600 N
𝐓1 =
= 17235.95 N
o
−
cos
130
sin 130o
sin 30o
cos 30o
T1 = 17000 N = 1.7  104 N
x
Fg = mg = 2000 kg 9.8
Fg = –19600 N
𝑚
𝑠2
sin 270o
 Fx = 0 = T1x + T2x
19600 N = T1 sin 130o + T2 sin 30o
130o
T2
0 = T1 cos 130o + T2 cos 30o
– T1 cos 130o = T2 cos 30o
−𝐓1 cos 130o
= 𝐓2
cos 30o
−17000 cos 130o
= 𝐓2
cos 30o
12617.86 N = 1.3  104 N = T2