Transcript Chapter 4, Part III

Applications & Examples of Newton’s 2nd Law
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. Acceleration of the boxes. b. 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
= 300 N
Newton’s 3rd Law 
FBR = -FRB, FCR = -FRC
FRBx = -FRBcosθ
FRBy = -FRBsinθ
FRCx = FRCcosθ
FRCy = -FRCsinθ
Example: 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
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!
Problem 25
 FT1
a
m1g   FT2
Take up as positive.
FT2 
a
 m2g
Newton’s 2nd Law
∑F = ma (y direction)
for EACH bucket separately!!!
m1 = m2 = 3.2 kg
m1g = m2g = 31.4 N
Bucket 1: FT1 - FT2 - m1g = m1a
Bucket 2: FT2 - m2g = m2a
Problem 29
Take up as positive.
Newton’s 2nd Law: ∑F = ma
(y direction) on woman + bucket!
FT 
m = 65 kg, mg = 637 N
 FT
 FP
a
a
 mg
FT + FT - mg = ma
2FT - mg = ma
Also, Newton’s 3rd Law:

FP = - FT