Chapter 5 - Applying Newtons Laws - Online Version

8/11/2019 Chapter 5 - Applying Newtons Laws - Online Version

1/54

Chapter 5 -Applying Newtons Laws

Learning Goals

How to use Newtons first law to solve problems involvingthe forces that act on a body in equilibrium.

How to use Newtons second law to solve problems

involving the forces that act on an accelerating body.

The nature of the different types of friction forcesstaticfriction, kinetic friction, rolling friction, and fluid resistance

and how to solve problems that involve these forces.

How to solve problems involving the forces that act on abody moving along a circular path

The key properties of the four fundamental forces of

nature.


2/54

5.1 using Newton's 1stlaw: particles

in equilibrium

A body in in equilibrium when it is at rest ormoving with constant velocity in an inertial frame

of reference.


3/54

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 ofthe rope is attached to the gymnasium ceiling.

a. What is the gymnasts weight?b. What force (magnitude and direction doe the rope exert

on her?

c. What is the tension at the top of the rope? Assume thatthe mass of the rope itself is negligible.

490 N

490 N

490 N


4/54

Example 5.2: tension in a rope with

mass

Suppose that in example 5.1, the weight ofthe rope is not negligible but is 120 N. find

the tension at each end of the rope.


5/54

Example 5.3: two-dimensional

equilibrium

A car engine with weight whangs from a chain that islinked at ring Oto two other chains, one fastened to theceiling 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.


6/54

Example 5.4: an inclined plane

A car of weight w rests on a slanted ramp leading to acar-transporter trailer. Only a cable running from thetrailer 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 cars tires.


7/54

Example 5.5 tension over a frictionless pulley

Blocks of granite are to be hauled up a 15oslope out of aquarry 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 theweights w1and w2be related in order for the system to

move constant speed? Ignore friction in the pulley and

wheels and the weight of the cable.


8/54

test your understanding 5.1

A traffic light of weight w hangs from twolightweight cables, one on each side of the

light. Each cable hands at a 45oangle from the

horizontal. What is the tension in each cable?

a. w/2

b. w/2

c. w

d. w2e. 2w w

TT

45o45o


9/54

5.2 using Newton's 2ndlaw:

dynamics of particles

Caution: madoesnt belong in free-body diagrams.


10/54


11/54


12/54


13/54


14/54


15/54

Apparent weight and apparent weightlessness

When a passenger with mass mrides in an elevator with y-acceleration ay, a scale shows the passengers apparentweight to be n = m(g + ay)

When the elevator is accelerating upward, ayis positive andnis greater than the passengers weight w = mg.when theelevator is accelerating downward, ayis negative and nis

less than the weight.

When ay= g, the elevator is in free fall, n = 0and thepassenger seems to be weightless.

Similarly, an astronaut orbiting the earth in a spacecraftexperiences apparent weightlessness.

In each case, the person is not truly weightless because thereis 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.


16/54


17/54

Caution: common free-body

diagram errors


18/54


19/54


20/54

ii) The same as in Example 5.12


21/54

Friction force is a contact force, it usuallyexists when a body slides on a surface.

The direction of friction force is opposite of

the direction of motion.

Sometimes we want to decrease frictionsuch as the oil is used in a car engine to

minimize friction between moving parts,but without friction between the tires andthe road we couldnt drive or run the car.


22/54

The kind of friction that acts when a body slidesover a surface is called kinetic friction forcefk.

The magnitude of the kinetic friction force can be

represented by the equation:

Where kis a constant called the coefficient of kinetic

friction. kdepends on the types of surface. Because it isa ratio of two force magnitudes, kis a pure number,without units.


23/54

The equation is only anapproximate representation of acomplex phenomenon. As a boxslides over the floor, bonds between

the two surfaces form and break,and the total number of such bondsvaries; hence the kinetic frictionforce is not perfectly constant.


24/54

Static friction

Friction forces may also act when there isno relative motion.

If you try to slide a box across the floor,

the box may not move at all because thefloor exerts an equal and opposite friction

force on the box.

This is called a static friction force fs.


25/54


26/54

The equation is a relationship between magnitudes, not avector relationship.

The equality sign holds only when the applied force Thas

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).


27/54


28/54


29/54


30/54


31/54


32/54

Rolling friction

Its a lot easier to move a loaded filingcabinet across a horizontal floor using a

cart with wheels than to slide it.

We can define a coefficient of rollingfriction rthe tractive resistance. Typicalvalues of rare 0.002 to 0.003 for steel

wheels on steel rails and 0.01 to 0.02 forrubber tires on concrete.


33/54


34/54

Fluid Resistance and Terminal speedFluid resistance is the force that a fluid (a gas or liquid) exerts

on a body moving through. The direction of the fluid resistanceforce acting on body is always opposite the direction of the

bodys 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 fof the fluid resistance

force can be described using the following equation

Where kis a proportionality constant that depends on the

shape and size of the body and the properties of the fluid.

The unit of k is Ns/m or kg/s


35/54

Air drag

In motion through air at the speed of a tossed tennis ballor faster, the resisting force is approximately proportionalto v2rather 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 greaterthan rolling friction at highway speeds.

The value of Ddepends on the shape and size of the

body and on the density of the air. The units of the

constant Dare Ns2/m2or kg/m


36/54

Because of the effects of fluid resistance, an object falling in afluid does not have a constant acceleration.

To find acceleration at a point of time, we need to use Newtons

second law.


37/54

Terminal velocityslow moving object

When the rock first starts to move, vy= 0, the resistingforce 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 thistime mg kvy= 0, the acceleration becomes zero, and

there is no further increase in speed. The final speed vt,

called the terminal speedis given by mgkvt= 0 or


38/54

Terminal velocityfast moving object

When the rock first starts to move, vy= 0, the resistingforce 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 thistime mg Dvy2= 0, the acceleration becomes zero, and

there is no further increase in speed. The final speed vt,

called the terminal speedis given by mg Dvy2= 0, or

Fy= mg + (-Dvy2) = may


39/54

Graphs of the motion of a body falling without fluid


40/54

Graphs of the motion of a body falling without fluid

resistance and with fluid resistance proportional to the

speed.


41/54


42/54


43/54


44/54

Uniform circular motion is governed by Newtons 2ndLaw. To


45/54

U o c cu a ot o s go e ed by e to smake a particle accelerate toward the center of the circle, the

net force Fon the particle must always be directed toward

the center. The magnitude of the acceleration is constant, so

the magnitude Fnetmust also be constant.

Uniform circular motion can result

from any combination of forces,

just so the net force F is alwaysdirected toward the center of the

circle and has a constant

magnitude.


46/54

T = (60 s) / 5 = 12 s


47/54

In terms of m, , g, L


48/54


49/54


50/54


51/54

Banked curves and the flight of airplanes

The example 5.23 alsoapply to an airplanewhen it makes a turn in

level flight.

Lcos= mg Lsin=mv2/R

tan= v2/gR


52/54

Motion in a vertical circle


53/54

iii


54/54

Documents

Chapter 5 - Applying Newtons Laws - Online Version