Transcript Free Body Diagrams

Section 4-7
Solving Problems with Newton’s Laws; Free Body Diagrams
“It sounds like an implosion!”
• Forces are VECTORS!!
• Newton’s 2nd Law:
∑F = ma
∑F = VECTOR SUM of all forces on mass m
 VECTOR addition
is needed add forces in the 2nd Law!
– Forces add according to
the rules of VECTOR ADDITION! (Ch. 3)
Problem Solving Procedures
1. Draw a sketch. For each object separately, sketch a
free-body diagram, showing all the forces acting on
that object. Make the magnitudes & directions as
accurate as you can. Label each force.
2. Resolve vectors into components.
3. Apply Newton’s 2nd Law separately to each object &
for each vector component.
4. Solve for the unknowns.
Note that this often requires algebra,
like solving 2 linear equations in 2 unknowns!
Example
Find the resultant force FR
Example
Find the resultant force FR
FR = [(F1)2 + (F2)2](½) = 141 N
tanθ = (F2/F1) = 1, θ = 45º
Example 4-9
Find the resultant force FR
If the boat moves with
acceleration a,
∑F = FR = ma
FRx = max, FRy = may
Example 4-9
• Illustrates the procedures for Newton’s 2nd Law problems:
STEP 1: Sketch the situation!!
– Sketch a “Free Body” diagram for EACH body in problem
& draw ALL forces acting on it.
• Part of your grade on exam & quiz problems!
STEP 2: Resolve the forces on each body into components
– Use a convenient choice of x,y axes
• Use the rules for finding vector components from Ch. 3.
STEP 3: Apply N’s 2nd Law to EACH OBJECT SEPARATELY
∑F = ma, Note: This is the LAST step, NOT the first!
We NEED A SEPARATE equation like this for each object!
Resolved into components: ∑Fx = max, ∑Fy = may
Conceptual Example 4-10
Moving at CONSTANT v, with NO friction,
which free body diagram is correct?
Example 4-11
A box of mass m = 10 kg is pulled by an attached cord along a horizontal
smooth (frictionless!) surface of a table. The force exerted is FP = 40.0 N
at a 30.0° angle as shown. Calculate:
a. The acceleration of the box.
b. The magnitude of the upward normal force FN exerted by the table
on the box.
Free Body
Diagram
The normal force, FN is NOT
always equal & opposite to the weight!!
Example 4-12
Two boxes are connected by a lightweight (massless!) cord & are resting on
a smooth (frictionless!) table. The masses are mA = 10 kg & mB = 12 kg. A
horizontal force FP = 40 N is applied to mA. Calculate:
a. The acceleration of the boxes. b. The tension in the cord connecting the boxes.
Free Body
Diagrams
Example 4-13 (“Atwood’s Machine”)
Two masses suspended over a (massless frictionless) pulley by a flexible
(massless) cable is an “Atwood’s machine”. Example: elevator &
counterweight. Figure: Counterweight mC = 1000 kg. Elevator mE = 1150 kg.
Calculate
a. The elevator’s acceleration. b. The tension in the cable.
a


a
aE = - a
Free Body
Diagrams
aC = a
Conceptual Example 4-14
Advantage of a Pulley
A mover is trying to lift a piano
(slowly) up to a second-story
apartment. He uses a rope
looped over 2 pulleys.
What force must he exert on the
rope to slowly lift the piano’s
mg = 2000 N weight?
mg = 2000 N
Free Body Diagram
Example 4-15: Accelerometer
A small mass m hangs from a thin
string & can swing like a pendulum.
You attach it above the window of
your car as shown. What angle does
the string make
a. When the car accelerates at a
constant a = 1.20 m/s2?
b. When the car moves at constant
velocity, v = 90 km/h?
Free Body Diagram
Inclined Plane Problems
The tilted coordinate
System is convenient,
but not necessary.
Engineers & scientists
MUST understand these!
a
Understand ∑F = ma & how to resolve it into x,y
components in the tilted coordinate system!!
Example 4-1b: Sliding Down An Incline
A box of mass m is placed on a smooth (frictionless!) incline that
makes an angle θ with the horizontal. Calculate:
a. The normal force on the box. b. The box’s acceleration.
c. Evaluate both for m = 10 kg & θ = 30º
Free Body
Diagram
Example
= 300 N
Free Body
Diagram
FT1x = -FTcosθ
FT1y = -FTsinθ
FT2x = FTcosθ
FT2y = -FTsinθ
Problem 32
Take up as positive!
Newton’s
m = 65 kg
2nd Law
mg = 637 N
FT   FT
 FP
a
 mg
FT + FT - mg = ma
2FT -mg = ma
FP = - F T
Free
Body
Diagram
Newton’s
3rd Law!!
∑F = ma (y direction) on the woman + the bucket!
Problem 32 Solution
Take up as positive!
Newton’s
m = 65 kg
2nd Law
mg = 637 N
FT   FT
 FP
a
 mg
FT + FT - mg = ma
2FT -mg = ma
FP = - F T
Free
Body
Diagram
Newton’s
3rd Law!!
∑F = ma (y direction) on the woman + the bucket!
Problem 32 Solution
FT
The window washer pulls down on the rope with a tension force so
the rope pulls up on her hands with a tension force The tension in
The rope is also applied at the other end of the rope, where it
FT attaches to the bucket. So there is another force pulling up on the
bucket. The bucket-washer combination has a net force upwards.
The free body diagram shows only forces on the bucket-washer,
not forces exerted by them (the pull down on the rope by the
person) or internal forces (normal force of bucket on person).
(a) Write Newton’s second law in the vertical direction, with up as
positive. The net force must be zero if the bucket and washer have
constant speed.  F  FT  FT  mg  0  2 FT  mg 
mg
FT  12 mg 
1
2
 72 kg   9.80 m

s 2  352.8 N  350 N
(b) Now the force is increased by 15%, so again write Newton’s
second law, but with a non-zero acceleration.
F  F
T
a
 FT  mg  ma 
2 FT  mg
m


2  405.72 N    72 kg  9.80 m s 2
72 kg
  1.47 m s
2
 1.5 m s 2
Problem 33
 FT1
a
We draw free-body diagrams for each bucket.
a. Since the buckets are at rest, their acceleration is 0.
Write Newton’s 2nd Law for each bucket, calling UP
the positive direction.
 F1  FT1  mg  0 


FT1  mg   3.2 kg  9.80 m s 2  31N
m1 g 
F
2
 FT2
 FT2
a
 FT2  FT1  mg  0 


FT2  FT1  mg  2mg  2  3.2 kg  9.80 m s 2  63 N
b. Now repeat the analysis, but with a non-zero
acceleration. The free-body diagrams are unchanged.
F  F
1
T1
 mg  ma 


FT1  mg  ma   3.2 kg  9.80 m s2  1.25m s2  35.36 N  35 N
F
2
 m2g
 FT2  FT1  mg  ma  FT2  FT1  mg  ma  2 FT1  71N
General Approach to Problem Solving
1. Read the problem carefully; then read it again.
2. Draw a sketch, then a free-body diagram.
3. Choose a convenient coordinate system.
4. List the known & unknown quantities; find relationships
between the knowns & the unknowns.
5. Estimate the answer.
6. Solve the problem without putting in any numbers
(algebraically); once you are satisfied, put the numbers in.
7. Keep track of dimensions.
8. Make sure your answer is REASONABLE!
Chapter 4 Summary
• Newton’s 1st Law: If the net force on an object is zero, it will
remain either at rest or moving in a straight line at constant speed.
• Newton’s 2nd Law:
• Newton’s 3rd Law:
• Weight is the gravitational force on an object.
• Free-body diagrams are essential for problem-solving. Do
one object at a time, make sure you have all the forces, pick a
coordinate system & find the force components, & apply
Newton’s 2nd Law along each axis.