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```PHYSICS 231
Lecture 8: Forces, forces & examples
Remco Zegers
Walk-in hour: Monday 11:30-13:30
Helproom
PHY 231
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Newton’s Laws
 First Law: If the net force exerted on an object
is zero the object continues in its original state of
motion; if it was at rest, it remains at rest. If it
was moving with a certain velocity, it will keep on
moving with the same velocity.
 Second Law: The acceleration of an object is
proportional to the net force acting on it, and
inversely proportional to its mass: F=ma
 If two objects interact, the force exerted by the
first object on the second is equal but opposite in
direction to the force exerted by the second
object on the first: F12=-F21
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Forces seen in the previous lecture
• Gravity: Force between massive objects
• Normal force: Elasticity force from supporting surface
n=-FgL
FgL=mgcos 
Fg//=mgsin
Fg=mg 
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Gravity, mass and weights.
Weight=mass times gravitational acceleration
Fg(N)=M(kg) g(m/s2)
Newton’s law of universal gravitation:
Fgravitation=Gm1m2/r2
G=6.67·10-11 Nm2/kg2
For objects on the surface of the earth:
•m1=mearth=fixed
•The earth is not a point object relative to m2
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Measuring mass and weight.
Given that gearth=9.81 m/s2, gsun=274 m/s2, gmoon=1.67 m/s2,
what is the mass of a person on the sun and moon if his
mass on earth is 70 kg? And what is his weight on each of
the three surfaces?
• The mass is the same on each of the surfaces
• On Earth: w=686.7 N
• On the Moon: w=116.7 N
• On the Sun: 19180 N
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Jumping!
The pelvis has a mass of 30.0 kg. What
is its acceleration?
Decompose all forces in x and y directions
Force
x (N)
y (N)
300 N
-122
-274
690 N
236
648
Weight
0
-249
Resultant 114 N
80.3 N
Total Force: F=(1142+80.32)=139 N
Direction: =tan-1(Fy/Fx)=35.2o
Acceleration: a=F/m=139/30.0=4.65 m/s2
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Tension
T
The magnitude of the force T
acting on the crate, is the same
as the tension in the rope.
Spring-scale
You could measure the tension by inserting
a spring-scale...
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Newton’s second law and tension
m1
n
T
Fg
No friction.
T What is the acceleration of
the objects?
m2
Fg
Object 1:
Object 2:
F=m1a, so
F=m2a, so
T=m1a
Fg-T=m2a
m2g-T=m2a
Combine 1&2 (Tension is the same): a=m2g/(m1+m2)
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Problem
What is the tension in the string and
what will be the acceleration of the
two masses?
Draw the forces: what is positive &
negative???
T
T
For 3.00 kg mass: F=ma
T-9.813.00=3.00a
For 5.00 kg mass: F=ma
9.815.00-T=5.00a
T=36.8 N
a=2.45 m/s2
Fg
Fg
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Friction
Friction are the forces acting on an object due to interaction
with the surroundings (air-friction, ground-friction etc).
Two variants:
• Static Friction: as long as an external force (F) trying to
make an object move is smaller than fs,max, the static
friction fs equals F but is pointing in the opposite direction:
no movement!
fs,max=sn s=coefficient of static friction
• Kinetic Friction: After F has surpassed fs,max, the object
starts moving but there is still friction. However, the
friction will be less than fs,max!
fk=kn k=coefficient of kinetic friction
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Problem
n=-FgL
A)If s=1.0, what is
the angle  for which the
block just starts to slide?
F
=mgsin
g//
FgL=mgcos 
B)The block starts moving.
Given that k=0.5, what is the
acceleration of the block?
Fg=mg 
Fs,k
A) Parallel direction: mgsin-sn=0 (F=ma)
Perpendicular direction: mgcos-n=0 so n=mgcos
Combine: mgsin-smgcos=0
s=sin/cos=tan=1 so =45o
B) Parallel direction: mgsin(45o)- smgcos(45o)=ma (F=ma)
g(½2-¼2)=a so a=g¼2
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All the forces come together...
n
T
T
If a=3.30 m/s2 (the 12kg block
is moving downward), what is
the value of k?
Fg
For the 7 kg block parallel to
the slope:
T-mgsin-kmgcos=ma
For the 12 kg block: Mg-T=Ma
Fk
Fg
Solve for k
 M ( g  a)  m g sin   m a
k 
 0.25
 m g cos
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General strategy
• If not given, make a drawing of the problem.
• Put all the relevant forces in the drawing, object by
object.
• Decompose the forces in direction parallel to the
motion and perpendicular to it.
• Write down Newton’s first law for forces in the
parallel direction and perpendicular direction.
• Solve for the unknowns.