Transcript action - mrsmartinmath

Forces
Don't forget
mass in kg
distance in meters
time in seconds
Equilibrium (FR = 0)
all the x components have to add up to 0
all the y components add up to 0
x
components
F3
Totalx
y
components
F3
Totaly
The object is in equilibrium when Totalx and Totaly
both add up to 0.
Force will be 153 N
Angle will be 180o - 22.50 = 157.5o
What happens to FN ?
Mass of the box = 10 kg
FT = 30N at 45o
76.8
Fg = -98
FN =
+98
Fg
Newton's Laws of Physics
Newton's first law
in the absence of an external force....
A body at rest will stay at rest,
A body in motion will stay in motion
Example:
When we are in the car at a fast speed and the driver applies the brakes quickly...
Without a seat belt we are likely to hit the dash board.
Our bodies want to keep moving forward.
this is called inertia
- kick snow off our boots.
- salad spinner
-when a magician pulls a tablecloth out from under dishes and the
dishes stay put.
Why do objects in motion stop?
Friction.
Force of Friction
Ff(static) ≤ μsFN
means that Ff will be equal to the opposing force
up until its maximum value at which point the object will move
Ff(kinetic) = μkFN
the moving object will experience this force of friction
Newton's Second Law
F = ma
Force = mass (kg) x acceleration (m/s2)
"the fundamental principle of dynamics"
The force exerted on a body is equal to the product of the
mass (m) and its acceleration (a)
The acceleration of an object is directly proportional to the net force acting on it and is inversely proportional to its
mass.
The direction of the acceleration is in the direction of the applied net force
F = ma
Example:
Estimate the net force needed to
accelerate a 1500 kg race car
at 5.0 m/s2
F = ma = 1500kg x 5.0m/s2 = 7500N
= 7.5 x 103 N (sig figs)
What net force is required to bring a 1500 kg car to rest from a
speed of 100 km/h within a distance of 55m?
First determine the acceleration a. The car is
acting on the horizontal.
Vi = 100 km/h = 28m/s
Vf = 0 m/s
Vf2 = Vi2 + 2 a(Δx)
0 = 282 + 2a(55m)
a = (-28m/s)2/2(55m) = -7.1m/s2
Th
e net force required is then
F = ma = (1500kg) (-7.1m/s2)
F = -1.1 x 104 N
The elevator problem....
If a 75 kg person rides the elevator
which accelerates at 1.5 m/s2 What is the
total force acting on him or her?
F = ma = -(75kg (-9.8m/s2)) + 75kg (1.5m/s2) = 847.5N
The elevator problem....
Going down...
If a 75 kg person rides the elevator
DOWN
which accelerates down at 1.5 m/s2
What is the
Normal force acting on him or her?
FN = - Fg = (75 (-9.8) )= 735
Total F = FN + other forces =
75kg (9.8m/s2) - 75kg (1.5m/s2) =
622.5N
The spider problem
When the spider is
in the elevator
there is tension on
the thread.
p. 314 # 3
A 10.2 g spider is suspended from an elevator
ceiling by a thread.
The spider problem
a) What is the tension in the thread if the elevator
rises at a constant velocity?
It is in equilibrium therefor the only force on the string if
Fg
Fg = 0.0102kg x -9.8 m/s2 = -0.100N
0.0102 kg
b) The elevator accelerates at a rate of 2.0 m/s2 going up.
What is the tension?
FT = Fg + ma = 0.100N + (0.0102kg x 2.0 m/s2)
= 0. 120N
c) decelerates at the same rate:
FT = Fg + ma = 0.100N + (0.0102kg x -2.0 m/s2)
= 0.080N
Newton's third law
For every action there is an equal and
opposite reaction
FA⇒B = -FB⇒A
action - reaction
the foot exerts a force on the ball and at the same time
the ball exerts a force on the foot
In other words, things are never one sided...
a hammer on a nail exerts a force and the
reaction of the nail on the hammer is an
equivalent force in the opposite direction.
When we walk, our foot pushes on the ground, The ground exerts an equal and opposite force
back on the person. This force moves him or her forward.
At the skating rink Jessica pushes Jordan and while Jordan
accelerates away from Jessica, Jessica also
accelerates away from Jordan
Action - Reaction