Glencoe Physics Chapter 7 Forces and Motion in Two Dimensions.

Glencoe Physics Chapter 7Forces and Motion in Two Dimensions

Objectives 7.1

Determine the force that produces equilibrium when three forces act on an object

Analyze the motion of an object on an inclined plane with and without friction

Objectives 7.2

Recognize that the vertical and horizontal motions of a projectile are independent

Relate the height, time in air (hang time), and initial vertical velocity of a projectile using its vertical motion , then determine the range

Explain how the shape of the trajectory of a moving object depends upon the frame of reference from which it is observed

Objectives 7.3

Explain the acceleration of an object moving in a circle at a constant speed

Describe how centripetal acceleration depends upon the object’s speed and the radius of the circle

Recognize the direction of the force that causes centripetal acceleration

Explain how the rate of circular motion is changed by exerting torque on it

7.1 Forces in Two Dimensions

We have already learned about how to find the resultant force of two forces acting on an object, as well as 3 or more forces.

What we want to consider now are cases where Net Force acting on an object is zero, and the object is in Static equilibrium.

Static Equilibrium

Condition of an object when net forces equal zero

Object is motionless

mg

1

T1

2

T2Since the sign is not accelerating in any direction, it’s in equilibrium. Since it’s not moving either, we call it Static Equilibrium.

Thus, red + green + black = 0.

Hanging sign f.b.d.Free Body Diagram

Equilibrant If you remove one force from a static condition,

the system is no longer in static equilibrium The removed force is called the

EQUILIBRANT. The equilibrant is the force, that if added to

other forces, will bring the other forces into static equilibrium.

Equilibrant is always equal to, but opposite, the sum of all the other force vectors

What force represents the equilibrant of the tensions of the strings?

FT1 FT2

mg

Example problem

What is the tension in the ropes holding the mass in place?

Θ = 22.5°m= 168 N

22.5°

168 N

Solution

Fnet =0 (system is in equilibrium)

Fax = -Fbx

Fay + Fby – mg = 0

FaSin 22.5° + FbSin 22.5° = 168 N

2Fa Sin 22.5 = 168 N

Fa = 220 N

Weight of the Picture?

Weight is equal to F1y + F2y

So…. 25N + 25 N = 50 N

What is the weight of this picture?

T2

mg

T1

1 2

T1 cos 1

(x component)

T2 cos 2

(X component)

T1 sin 1

(y component)

T2 sin 2

(Y component)

T1 cos 1 = T2 cos 2

Horizontal:

Vertical:T1 sin 1 + T2 sin 2 = mg

Components & Scalar Equations

If in Equilibrium……..the following would be

true

Sample ProblemA mother and daughter are outside playing on the

swings. The mother pulls the daughter and swing (total mass 55.0 kg) back so that the swing makes an angle of 40.0° with the vertical (50.0 ° from horizontal)

A. What is the tension in each chain holding the swing seat and the daughter?

B. How hard did the mother have to pull to hold the daughter at that position?

A. 703N

B. 452N

How to budge a stubborn mule

Big Force

Little Force

It would be pretty difficult to budge this mule by pulling directly on his collar. But it would be relatively easy to budge him by tying the rope to a tree and then pulling up (or pushing sideways) in the middle. Why would this work?????

The tension in the rope has two components, one horizontal and one “vertical”

FT

FTx

Sample Problem…

Is this a case of equilibrium?

Calculate the magnitude of the net force

45°40 °

150N160N

75N

Equilibrium or Motion… along an Inclined Plane

Is the “sled” on the inclined plane in equilibrium?What are the forces acting on the sled?Draw a Free Body Diagram of the forces

Inclined Plane

parallel component Fgx

m

mg

perpendicular component Fgy

Note: red + blue = black

Inclined Plane )

m

mg

mg Sin

mg Cos

Fgx=

Fgy=

m

mg

mg Sin

mg Cos

Be careful what you use in your calculation…..remember you must use the angle, as measured from (+X) (East)

This calculation will look different than what is in your book (p. 152)…but is easier to use if you remember the above caution!!!!!

Fn on block

Fgx

of increasing the angle.

The next 3 slides show the effect

CD 5.3 p 21CD 5.3 p-22

Fg on block

Fn on block

F gx

CD 5.3 p 21CD 5.3 p-22

Fg on block

Fn on block

F gx

CD 5.3 p 22

Fe on block

A

B

C

net force greatest at:acceleration greatest at:Speed increases, acceleration:AA

decreases

Example p. 153 of your book…. A 62 kg person on skis is going down a hill sloped at

37°. The coefficient of kinetic friction between the skis and the snow is 0.15.

A) What is the Horizontal component of the skier’s weight?

B) What is the Vertical component of the skier’s weight?

C) What is the Normal force acting on the skier?D) What is the Frictional force acting on the skier?E) What is the Net Force acting on the skier?F) What is the acceleration of the skier?G) How fast is the skier traveling after 5.0s, starting

from

7.2 Projectile Motion in Two

Dimensions:

Fountain at ¡Explora! Science Museum, Albuquerque, NM

What is Projectile Motion?

Any object moving horizontally, with only the force of gravity acting on it, will exhibit “projectile motion”. (not moving straight up or down)

An object displaying projectile motion will follow a “trajectory”, or a curved, parabolic-shaped path.

Projectile Motion

At a given location on the earth and in the absence of air resistance, all objects fall with the same uniform acceleration. Thus, two objects of different masses, dropped from the same height, will hit the ground at the same time.

An object’s horizontal and vertical motion are independent. Object projected horizontally will reach the ground in the same time as an object dropped vertically. No matter how large the horizontal velocity is, the downward pull of gravity is always the same.

Why does a projectile follow a parabolic path?

What happens to the horizontal displacement during each successive time period?What happens to the vertical displacement during each successive time period?

Horizontal motion….

…+ Vertical = parabolic path

Horizontal Vs. Vertical Velocities We can see from the diagrams that the

horizontal velocity of a projectile remains constant (with no air resistance)

We can also see that that vertical velocity is constantly changing, at a rate of 9.8 m/s2.

Horizontal Vs. Vertical Velocities The horizontal and vertical velocities

are also independent of each other, that is, they have no effect on each other.

The only thing the horizontal and vertical motions have in common is that each motion takes exactly the same time to occur.

Equations of Projectile Motion

Working Projectile motion problems… Separate the problem into two problems…

vertical motion and horizontal motion Vertical motion is exactly that of an object

dropped or thrown straight up or down. Use free-fall equations (or accelerated motion equations)

Horizontal motion is constant velocity (Newton’s 1st Law), work this part using constant velocity equations.

Both Vertical and horizontal motion are tied together by TIME. (Tv=Th)

Sample Problem…. A rock is thrown horizontally off a tall cliff at

20.m/s. If the cliff is 70.m high… A) How far horizontally from the bottom of the cliff

does the rock land? B) What is the magnitude of the rock’s total

velocity at the instant before landing?

Another example in book on page 157

Monkey Hunter

A young native hunter in the Amazon is hunting monkeys with a bow and arrow. The hunter see a monkey sitting on a branch, and since the bow does not have a sight, the hunter pulls back the bow and points the arrow directly at the monkey. As the hunter releases the arrow the monkey notices the movement and drops off the branch. What happens next?

Throw at monkey with no gravity…

Throw above monkey with gravity…

Throw at monkey fast with gravity…

Horizontal Range…What is the horizontal range of the projectile?

dx = vxt

dx = (v0 cos Θ)t

25.0°

V=30.0 m/s

dx = (v0Cos Θ) (V0 Sin Θ/4.9)

dx = (30.0 m/s Cos 25.0°)(30.0 m/s Sin 25.0°)/4.9

dx = 70.4 m

First…find time using vertical motion..

ΔVy =at

t= Δvy

a

=- V0Sin25.0 ° -V0Sin25.0 ° -9.8m/s2

=2.58s

Then, use that time to find range…!

Notice nice equation to use to find vertical time in air!!!

Alternate Range equation…same thing….

Range (dx) =V02 Sin 2 Θ

g

Max height & hang time depends only on initial vertical velocity

Each initial velocity vector below has the a different magnitude (speed) and the projectiles will have different ranges (green the greatest), but each object will spend the same time in the air and reach the same max height. This is because each vector has the same vertical component.

Max Range at 45Over level ground at a constant launch speed, what angle maximizes the range, R ? First consider some extremes: When = 0, R = 0, since as soon as the object is launched it’s back on the ground. When = 90, the object goes straight up and lands right on the launch site. The best angle is 45, smack dab between the extremes.

45

All launch speeds are the same; only the angle varies.

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