Transcript f - Michigan State University

PHYSICS 231
Lecture 8: Forces, forces & examples
Next week:
Walk-in hour Monday 4-5 pm
Tuesday: homework due
Wednesday: Review (Prof. Lynch)
Friday Exam
I am absent Tue-Fri.
Remco Zegers
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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
•r=“radius” of earth=fixed
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QUIZ! (extra credit)
In the absence of friction, to keep an object that is
traveling with a certain velocity moving with exactly the
same velocity over a level floor, one:
a) does not have to apply any force on the object
b) has to apply a constant force on the object
c) has to apply an increasingly strong force on the object
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Jumping!
The pelvis has a mass of 30.0 kg. What
is its acceleration and direction of
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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question
ceiling
A block of mass M is hanging
from a string. What is the
tension in the string?
M
a)
b)
c)
d)
e)
T=0
T=Mg with g=9.81 m/s2
T=0 near the ceiling and T=Mg near the block
T=M
one cannot tell
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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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friction
N pull
20 kg
Fs
Fg
A person wants to drag a crate
of 20 kg over the floor. If he pulls
the crate with a force of 98 N,
the crate starts to move. What
is s?
Answer: Fs= sN=sMg=20x9.8xs=196s
Just start to move: Fs=Fpull 98=196s
so: s=0.5
After the crate starts moving, the person continues to
pull with the same force. Given k=0.4, what is the
acceleration of the crate?
F=ma F=98-0.4Mg=98-0.4x20x9.8=98-78.4=19.6 N
19.6=20a a=0.98 m/s2
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20
100
A person is pushing an ice-sledge of 50kg over a frozen lake
with a force of 100N to the east. A strong wind is pushing
from the south-west and produces a force of20N on the
sledge. a) What is the acceleration of the sledge (no friction)?
b) if the coefficient of kinetic friction is 0.05, what is the
acceleration?
a) force
Horizontal vertical
person 100
0
wind
20cos(45)
20sin(45)
sum
100+14.1=114.1 14.1
Total force: (114.12+14.12)=115 N
F=ma 115=50a a=2.3 m/s2
b) Ffriction=n=0.05*m*g=0.05*50*9.8=24.5N
F=ma 115-24.5=50a a=1.8 m/s2
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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.
• Think about the axis
• Think about the signs
• 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.
• use in further equations if necessary
• Check whether your answer makes sense.
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