Objectives: The students will be able to: Use the methods of vector algebra to determine the net force acting on an object. Draw an accurate free body.
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Objectives: The students will be able to:
• Use the methods of vector algebra to determine the net force acting on an object.
• Draw an accurate free body diagram locating each of the forces acting on an object or a system of objects.
• Use free body diagrams and Newton's laws of motion to solve word problems.
Reviewhttp://www.youtube.com/watch?v=NYVMlmL0BPQ
Newton’s First Law:
Objects in motion tend to stay in motion and objects at rest tend to stay at rest unless acted upon by an unbalanced force.
Newton’s Second Law:
Force equals mass times acceleration (F = ma).`
Newton’s Third Law:
For every action there is an equal and opposite reaction.
1stlaw: Homer is large and has much mass, therefore he has much inertia. Friction and gravity oppose his motion.
2nd law: Homer’s mass x 9.8 m/s/s equals his weight, which is a force.
3rd law: Homer pushes against the ground and it pushes back.
Forces
• Newton’s Laws of Motion
• Weight
• Free fall
• Force and motion problems in 1-D
• Normal force
• Tension
• Free body diagrams
• Atwood device
• Static and kinetic friction
• Coefficients of friction
• Air resistance
• Terminal velocity
Examples of Forces
• A force is just a push or pull. Examples:– an object’s weight
– tension in a rope
– friction
– attraction between an electron and proton
• Bodies don’t have to be in contact to exert forces on each other, e.g., gravity.
Fundamental Forces of Nature• Gravity
– Attraction between any two bodies w/ mass– Weakest but most dominant
• Electromagnetic – Forces between any two bodies w/ charge– Attractive or repulsive
• Weak nuclear force – responsible for radioactive decay
• Strong nuclear force – holds quarks together (constituents of protons and neutrons)
4-6 Weight – the Force of Gravity; and the Normal Force
Weight is the force exerted on an object by gravity. Close to the surface of the Earth, where the gravitational force is nearly constant, the weight is:
Sect. 4-6:Weight & Normal Force
Weight The force of gravity on an object.• Write as FG W.
• Consider an object in free fall.
Newton’s 2nd Law is:
∑F = ma• If no other forces are acting, only FG ( W) acts (in the
vertical direction). ∑Fy = may
Or: (down, of course)
• SI Units: Newtons (just like any force!).
g = 9.8 m/s2 If m = 1 kg, W = 9.8 N
“Normal” Force• Suppose an object is at rest on a table.
No motion, but does the force
of gravity stop? OF COURSE NOT!• But, the object does not move:
2nd Law ∑F = ma = 0
There must be some other force acting
besides gravity (weight) to have ∑F = 0.
• That force The Normal Force FN (= N)“Normal” is a math term for perpendicular ()
FN is to the surface & opposite to the weight
(in this simple case only!) Caution!!!
FN isn’t always = & opposite to the weight, as we’ll see!
Normal Force
• Where does the normal force come from?
• From the other object!!!
• Is the normal force ALWAYS equal & opposite to the weight?
Normal Force
• Where does the normal force come from?
• From the other object!!!
• Is the normal force ALWAYS equal & opposite to the weight?
NO!!!
An object at rest must have no net force on it. If it is sitting on a table, the force of gravity is still there; what other force is there? The force exerted perpendicular to a surface is called the Normal Force FN. It is
exactly as large as needed to balance the force from the object. (If the required force gets too big, something breaks!)
∑F = ma = 0 or Newton’s 2nd Law for Lincoln: FN – FG = 0 or FN = FG = mg
Note! FN & FG AREN’T action-reaction pairs from N’s 3rd Law! They’re
equal & opposite because of N’s 2nd Law! FN & FN ARE the action-reaction pairs!!
“Free Body Diagrams” for Lincoln. Showall forces in proper directions.
m = 10 kgThe normal
force is NOTalways equal & opposite to the weight!!
Example 4-6
Find: Normal force on box from table for Figs. a., b., c. Always use g N’s 2nd Law to
m CALCULATE FN!
Example 4-7
m = 10 kg∑F = ma
FP – mg = ma
What happens when a person pulls upward on the box in the previous example with a force greater than the box’s weight, say 100.0 N? The box will accelerate I upward because FP > mg!!
Note: The normal force is zero here because the mass isn’t in contact with a surface!
m = 10 kg, ∑F = maFP – mg = ma100 – 98 = 10a
a = 0.2 m/s2
Example 4-8: Apparent “weight loss”A 65-kg woman descends in an elevator that accelerates at 0.20g downward. She stands on a scale that reads in kg.
(a) During this acceleration, what is her weight & what does the scale read?
(b) What does the scale read when the elevator descends at a constant speed of 2.0 m/s?
• Note: To use Newton’s 2nd Law for her, ONLY the forces acting on her are included. By Newton’s 3rd Law, the
normal force FN acting upward on her is equal & opposite to the scale reading. So, the numerical value of FN is equal to the “weight” she reads on the scale! Obviously, FN here
is NOT equal & opposite to her true weight mg!! How do we find FN? As always
We apply Newton’s 2nd Law to her!!
Normal force• When an object lies on a table or on the
ground, the table or ground must exert an upward force on it, otherwise gravity would accelerate it down.
• This force is called the normal force.
mg
N
In this particular case, N = mg.
So, Fnet = 0; hence a = 0.
m
Normal forces aren’t always up“Normal” means perpendicular. A normal force is always perpendicular to the contact surface.
For example, if a flower pot is setting on an incline, N is not vertical; it’s at a right angle to the incline. Also, in this case, mg > N.
N
mg
Normal force directions• Up
– You’re standing on level ground.
– You’re at the bottom of a circle while flying a loop-the-loop in a plane.
• Sideways– A ladder leans up against a wall.
– You’re against the wall on the “Round Up” ride when the floor drops out.
• At an angle– A race car takes a turn on a banked track.
• Down– You’re in a roller coaster at the top of a loop.
Cases in which N mg
N
mgmg
N
mg
N FA
a
1. Mass on incline
2. Applied force acting on the mass
3. Nonzero acceleration, as in an elevator or launching space shuttle
When does N = mg ?
If the following conditions are satisfied, then N = mg:
• The object is on a level surface.
• There’s nothing pushing it down or pulling it up.
• The object is not accelerating vertically.
N and mg are NOT an Action-Reaction Pair!
Earth
FE
m
mg
N
Fg
The dot represents the man.
mg, his weight, is the force on the man due to the Earth.
FE is the force on the Earth due to the man.
N, the normal force, is the force on the man due to the ground.
Fg is the force on the ground due to the man.
“Switch the nouns to find the reaction partner.”
The red vectors are an action-reaction pair. So are the blue vectors. Action-reaction pairs always act on two different bodies!
Box / Tension Problem
8 kg 5 kg 6 kgT1
frictionless floor
T238 N
A force is applied to a box that is connected to other boxes by ropes. The whole system is accelerating to the left.
The problem is to find the tensions in the ropes.
We can apply the 2nd Law to each box individually as well as to the whole system.
Box / Tension Analysis
T1 pulls on the 8-kg box to the right just as hard as it pulls on the middle box to the left.
T1 must be < 38 N, or the 8-kg box couldn’t accelerate.
T2 pulls on the middle box to the right just as hard as it pulls on the 6-kg box to the left.
T1 must be > T2 or the middle box couldn’t accelerate.
8 kg 5 kg 6 kgT1
frictionless floor
T238 N
Free Body Diagram – system
19 kg38 N
mg
N
The total mass of all three
boxes is 19 kg.
N and mg cancel out.
Fnet = m a implies
a = 2.0 m/s2
Since the ropes don’t
stretch, a will be 2.0 m/s2
for all three boxes.
For convenience, we’ll choose left to be the positive direction.
Free Body Diagram – right box
6 kgT2
N
mg
N and mg cancel out.
For this particular box, Fnet = m a implies:
T2 = 6a = 6(2) = 12 N.
(Remember, a = 2 m/s2 for all three boxes.)
8 kg 5 kg 6 kgT1
frictionless floor
T238 N
Free Body Diagram – middle box
5 kgT2 = 12 NT1
mg
NN and mg cancel out again.
Fnet = m a implies:
T1 – T2 = 5a. So, T1 – 12 = 5(2), and
T1 = 22 N
8 kg 5 kg 6 kgT1
frictionless floor
T238 N
Free Body Diagram – left box
8 kgT1 = 22 N38 N
mg
NLet’s check our work using the left box.
N and mg cancel out here too.
Fnet = ma implies:
38 - 22 = ma = 8(2). 16 = 16.
8 kg 5 kg 6 kgT1
T238 N
Practice ProblemsPractice #1: A fisherman yanks a fish out of the water with an acceleration of 3.5 m/s² using very light fishing line that has a breaking strength of 28 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish?
Practice #2: A 15.0-kg bucket is lowered by a rope in which there is 163 N of tension. What is the acceleration of the bucket? Is it up or down?
Practice #3: An elevator (mass 6850 kg) is to be designed so that the maximum acceleration is 0.0780g. What are the maximum and minimum forces the motor should exert on the supporting cable?
Practice #4: The cable supporting a 2001-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking?
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