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Room Frequency BA. Clicker Question. A glider (Yoda) is gliding along an air track at constant speed. There is no friction and no air resistance. What can you say about the net force (total force) on the glider? A) The net force is zero. - PowerPoint PPT Presentation
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A glider (Yoda) is gliding along an air track at constant speed. There is no friction and no air resistance.
What can you say about the net force (total force) on the glider?
A) The net force is zero.
B) The net force is non-zero and is in the direction of motion.
C) The net force is non-zero and is in the direction opposite the motion.
D) The net force is non-zero and is perpendicular to the motion.
Velocity is constant, thus acceleration is zero. F = ma = 0. So, net force is zero.
Clicker Question Room Frequency BA
Announcements
• CAPA Set #5 due Friday at 10 pm
• This week in Section Assignment #3: Forces, Newton’s Laws, Free-Body Diagrams Print out and bring the assignment to your section…
• Reading in Chapter 4.1-4.6 (Forces and Force diagrams)
• Exam #1 grades posted on CULearn Solutions for version #1 posted on CULearn More details at “Exam Information” link
I. “Law of Inertia”: Every object continues in its state of rest, or uniform motion (constant velocity in a straight line), as long as no net force acts on it.
II. Acceleration is proportional to force and inversely proportional to mass. Famous equation.
III.Whenever one object exerts a force on a second object, the second exerts an equal force in the opposite direction on the first.
Newton’s Laws (as stated) hold in an “Inertial Reference Frame” (non-rotating / non-accelerating):
Newton’s Three Laws
F
net ma
F
AB F
BA
Newton’s Second Law
Acceleration is proportional to force andinversely proportional to mass.
F
net ma
Knowing the forces acting on an object of mass (m), we can calculate the accelerations.
Then we can use these accelerations to calculate the kinematics (motion).
Note that the direction of the acceleration is the same as the direction of the net force vector.
The unit of force in the SI system is the Newton (N).
Note that the pound is a unit of force, not of mass, and can
therefore be equated to Newtons but not to kilograms.
1 Newton ~ 0.22 pounds.
Force is a vector, so Newton’s 2nd Law is a vector equation:
F
ma
Fx max
Fy may
All Kinds of ForcesPulling / Pushing
Springs
Gravity
Air Resistance / Drag
Friction
Lots of others too (electric, magnetic, etc.)
Forces add like vectors, not like scalars.
Consider two forces, , both of magnitude 1 N. One points to the right and other straight up. What’s the magnitude and direction of the net force?
A) 2 N, straight up.B) √2 N, straight up.C) 0 N.D) √2 N, up and right.E) -√2 N, left.
1 N
1 N
√2 N
F
1 and F
2
F
1
F
2
Fnet
F
1 F
2
Clicker Question Room Frequency BA
Newton’s Third LawWhenever one object exerts a force on a
second object, the second object exerts an equal and opposite direction force
on the first.
F
AB F
BA
A moving van collides with a sports car in a high-speed head-on collision. Crash!
During the impact, the truck exerts a force with magnitude Ftruck on
the car and the car exerts a force with magnitude Fcar on the truck.
Which of the following statements about these forces is true:
A) The magnitude of the force exerted by the truck on the car is the same size as the magnitude of the force exerted by the car on the truck: Ftruck
= Fcar
B) Ftruck > Fcar
C) Ftruck < Fcar
F
truck F
carM m
Guaranteed by Newton’s Third Law
Clicker Question Room Frequency BA
A moving van collides with a sports car in a high-speed head-on collision. Crash!
F
truck F
car
During the collision, the imposed forces cause the truck and the car to undergo accelerations with magnitudes atruck and acar. What is the relationship between atruck and acar?A) atruck > acar
B) acar > atruck C) atruck = acar
D) Indeterminate from information given.
M m
Clicker Question Room Frequency BA
Newton’s Second Law: F = ma
Draw all forces for each object (truck and car). Consider only horizontal direction.
m
M
F
truck
F
car Fcar = M atruck
Ftruck = m acar
Free-Body Diagram
CAR
TRUCK
|||| cartruck FF
By Newton’s Third Law (and with opposite directions)
Force of truck acting on the car
Force of car acting on the truck
truckcar Mama truckcar amMa
Demonstration with Skateboard
Consider an Everyday OccurrenceAssume you are in a tie with your sumo-wrestler friend.Draw the free-body diagrams (i.e. forces) acting on you.
F (tension in rope)
F (pulling on the rope)?
F (friction with floor)
Consider an Everyday OccurrenceDraw the free-body diagrams (i.e. forces) acting on you.
F (tension in rope)
F (friction with floor)
Any other forces acting on you?F (gravity)
F (Normal)
Clicker Question Room Frequency BA
F (tension in rope)
F (friction with floor)
F (gravity)
F (Normal)
What is the direction of the friction force acting on the sumo wrestler’s feet? A) Left B) Down
C) Right D) Up
F
friction F
friction
We often draw free-body diagrams schematically
N
Fgr
mg
= “normal force”
= force of gravity
N
Fgr
M g
mM
you)on pullingtension (F
Sumo)on pullingtension (F
Be very careful to indicate on which object the force is acting (not who does the acting).
N
F
mg
mF
friction
Since you are not moving….
No acceleration in x or y directions
Net force in x must be zero.
Net force in y must be zero.
x
y
Point-Like ObjectsWe often treat objects as point-like with all forces acting on that point.Reasonable for rigid (non-deformable) objects and ignoring rotations.
tension)(F
Label Coordinate Axes
0tension)( friction)( FF
0Normal)( gravity)( FF
Slip-n-SlidePart I – going down the slide
Physics approximation….1) You are a small rectangular box (that we will
treat as a point-like object anyway).2) The slope of the slide is constant (angle = q)
Step 1: Draw a free-body diagram
Fg = mg
How large a net force will you feel down the slope?What is the acceleration down the slope?
How fast will you be going after traveling down the slope for some time t?
Note that “Normal Force” is always Perpendicular
(i.e. normal) to the surface.
In this case, it is not in the opposite direction to the
gravitational force.