Chapter 5
Applying Newton’s Laws
Week 7 homework – due Friday 10/28
• Read 5.1-5.3
• Questions: 5.1, 2, 4, 7, 8, 11, 14, 18, 19,
24, 29, 30, 32, 36, 38, 44
Mastering physics homework #7 –
due 10/21
objective
• Use Newton's laws to solve problems
involving particle in equilibrium
• Finish lab?
Homework: 5.1, 5.3, 5.5
Mastering physics – chapter 4, week 2
5.1 using Newton's 1st law: particles
in equilibrium
• A body in in equilibrium when it is at rest or moving
with constant velocity in an inertial frame of
reference.
Example 5.1: tension in a massless rope
•
A gymnast with mass mG = 50.0 kg suspends herself
from the lower end of a hanging rope. The upper end of
the rope is attached to the gymnasium ceiling.
490 N
a. What is the gymnast’s weight?
b. What force (magnitude and direction doe the rope exert
on her? 490 N
c. What is the tension at the top of the rope? Assume that
the mass of the rope itself is negligible. 490 N
a.
b.
c.
Example 5.2: tension in a rope with
mass
• Suppose that in example 5.1, the weight of
the rope is not negligible but is 120 N. find
the tension at each end of the rope.
• Lower end of the rope:
• Upper end of the rope:
Example 5.3: two-dimensional
equilibrium
• A car engine with weight w hangs from a chain that is
linked at ring O to two other chains, one fastened to the
ceiling and the other to the wall. Find the tension in each
of the three chains in terms of w, the weights of the ring
and chains are negligible.
Engine: (T1)
Ring: (T1, T2 and T3)
Example 5.4: an inclined plane
• A car of weight w rests on a slanted ramp leading to a
car-transporter trailer. Only a cable running from the
trailer to the car prevents the car from rolling backward
off the ramp. Find the tension in the cable and the force
that the tracks exert on the car’s tires.
Example 5.5 – tension over a frictionless pulley
• Blocks of granite are to be hauled up a 15o slope out of a
quarry and dirt is to be dumped into the quarry to fill up old
holes to simplify the process, you design a system in which
a granite block on a cart with steel wheels (weight w1,
including both block and cart) is pulled uphill on steel rails
by a dirt-filled bucket (weight w2, including both dirt and
bucket) dropping vertically into the quarry. How must the
weights w1 and w2 be related in order for the system to
move constant speed? Ignore friction in the pulley and
wheels and the weight of the cable.
• Dirt filled bucket
• Block and cart
test your understanding 5.1
•
a.
b.
c.
d.
e.
A traffic light of weight w hangs from two
lightweight cables, one on each side of the
light. Each cable hands at a 45o angle from the
horizontal. What is the tension in each cable?
w/2
T
T
w/√2
o
45o
45
w
w√2
2w
w
5.2 using Newton's 2nd law:
dynamics of particles
Caution: ma doesn’t belong in free-body diagrams.
Apparent weight and apparent weightlessness
• When a passenger with mass m rides in an elevator with yacceleration ay, a scale shows the passenger’s apparent
weight to be
n = m∙(g + ay)
• When the elevator is accelerating upward, ay is positive and
n is greater than the passenger’s weight w = mg. when the
elevator is accelerating downward, ay is negative and n is
less than the weight.
• When ay = g, the elevator is in free fall, n = 0 and the
passenger seems to be weightless.
• Similarly, an astronaut orbiting the earth in a spacecraft
experiences apparent weightlessness.
• In each case, the person is not truly weightless because there
is still a gravitational force acting. But the person's sensation in
this free-fall condition are exactly the same as though the
person were in outer space with no gravitational force at all.
Caution: common free-body
diagram errors
11/1 do now
Fnorm
FB on A
A
Fgrav
Fnorm
Fapp
B
Fgrav
FA on B
ii) The same as in Example 5.12
Week 9 homework – due Friday 11/4
• Read 3.3, 5.3-5.5
• Homework: 29, 30, 32, 36, 38, 44, 50, 52, 55,
56, 57, 3.28, 30, 32, 33, 50, 75 – solutions are
posted on the school website.
Mastering physics #9 – due Tuesday
11/8 – 11:00pm
• Friction force is a contact force, it usually
exists when a body slides on a surface.
• The direction of friction force is opposite of
the direction of motion.
• Sometimes we want to decrease friction
such as the oil is used in a car engine to
minimize friction between moving parts,
but without friction between the tires and
the road we couldn’t drive or run the car.
• The kind of friction that acts when a body slides
over a surface is called kinetic friction force fk.
The magnitude of the kinetic friction force can be
represented by the equation:
Where μk is a constant called the coefficient of kinetic
friction. μk depends on the types of surface. Because it is
a ratio of two force magnitudes, μk is a pure number,
without units.
• The equation is only an
approximate representation of a
complex phenomenon. As a box
slides over the floor, bonds between
the two surfaces form and break,
and the total number of such bonds
varies; hence the kinetic friction
force is not perfectly constant.
Static friction
• Friction forces may also act when there is
no relative motion.
• If you try to slide a box across the floor,
the box may not move at all because the
floor exerts an equal and opposite friction
force on the box.
• This is called a static friction force fs.
The equation is a relationship between magnitudes, not a
vector relationship.
The equality sign holds only when the applied force T has
reached the critical value at which motion is about to start.
When T is less than this value, the inequality sign holds. In
that case we have to use the equilibrium conditions (∑F = 0)
to find fs. If there is no applied force (T = 0), then there is
not static friction force either (fs = 0).
Rolling friction
• It’s a lot easier to move a loaded filing
cabinet across a horizontal floor using a
cart with wheels than to slide it.
• We can define a coefficient of rolling
friction μr the tractive resistance. Typical
values of μr are 0.002 to 0.003 for steel
wheels on steel rails and 0.01 to 0.02 for
rubber tires on concrete.
Happy belated birthday
Yasmin
• Homework: 5.35, 5.37, 5.39
11/4 do now
Fnorm
• Draw a free body
diagram for the
two boxes (as one
object)
v
Fapp
Ffric
Fgrav
objective
• Fluid friction
• Homework: 5.47, 5.48
Fluid Resistance and Terminal speed
Fluid resistance is the force that a fluid (a gas or liquid) exerts
on a body moving through. The direction of the fluid resistance
force acting on body is always opposite the direction of the
body’s velocity relative to the fluid.
The magnitude of the fluid resistance force usually increase
with speed of the body through the fluid.
For very low speed, the magnitude f of the fluid resistance
force can be described using the following equation
Where k is a proportionality constant that depends on the
shape and size of the body and the properties of the fluid.
The unit of k is N∙s/m or kg/s
Air drag
• In motion through air at the speed of a tossed tennis ball
or faster, the resisting force is approximately proportional
to v2 rather than v. It is then called air drag or simply
drag. Airplanes, falling raindrops, and bicyclists all
experience air drag. The air drag on a typical car is
negligible at low speeds but comparable to or greater
than rolling friction at highway speeds.
The value of D depends on the shape and size of the
body and on the density of the air. The units of the
constant D are N∙s2/m2 or kg/m
• Because of the effects of fluid resistance, an object falling in a
fluid does not have a constant acceleration.
• To find acceleration at a point of time, we need to use Newton’s
second law.
Terminal velocity – slow moving object
• When the rock first starts to move, vy = 0, the resisting
force is zero, and the initial acceleration is ay = g. As the
speed increases, the air resistance force also increases,
until finally it is equal in magnitude to the weight. At this
time mg – k∙vy = 0, the acceleration becomes zero, and
there is no further increase in speed. The final speed vt,
called the terminal speed is given by mg – k∙vt = 0 or
Terminal velocity – fast moving object
∑Fy = mg + (-Dvy2) = may
• When the rock first starts to move, vy = 0, the resisting
force is zero, and the initial acceleration is ay = g. As the
speed increases, the air resistance force also increases,
until finally it is equal in magnitude to the weight. At this
time mg – D∙vy2 = 0, the acceleration becomes zero, and
there is no further increase in speed. The final speed vt,
called the terminal speed is given by mg – D∙vy2 = 0, or
Graphs of the motion of a body falling without fluid
resistance and with fluid resistance proportional to the
speed.
a. i, iii
b. ii, iv
c. v
objectives
• Dynamics of circular motion
• Homework – mastering physics – due
Sunday night – no exceptions
Chapter 5 review
• Class work and homework 5.83, 5.91,5.95,
5.109
10/28 do now (2004 exam)
• Three blocks of masses 3m, 2m, and m are connected to
strings A, B, and C as shown. The blocks are pulled along
a rough surface by a force of magnitude F exerted by
string C. The coefficient of friction between each block and
the surface is the same. Which string must be the
strongest in order not to break?
1. A
3m
A
2m
B
C
m
2. B
3. C
68%
4. They must all be the same strength.
5. It is impossible to determine without knowing the
coefficient of friction.
F
10/29 do now (2004 exam)
Force (N)
6
5
4
3
2
1
0
Time t (s)
1 2
3
A block of mass 3 kg, initial at rest, is pulled along
a frictionless, horizontal surface with a force show
as a function of time t b the graph
A. what is the acceleration of the block at t = 2 s?
88%
B. what is the speed of the block at t = 2 s?
33%