Chapter 7

Forces in Two Dimensions

equilibrium, projectile and circular motion

Two problems*

*answers on Thursday

Table lift demo: “Force

applied at an angle to the

direction of motion yields

less of an effect on the

object’s motion than force

applied in a straight line

with the motion.”

Draw a vector diagram to

show the forces and explain

the experience of the test

subject.

Derek left his physics

book on top of a drafting

table. The table is inclined

at a 35° angle. Find the

net external force acting

on the book, and

determine whether the

book will remain at rest in

this position. (given: Fg = 22N. Ff = 11N, FN

= 18N)

the Energy and force story

a quick review

what you already know

An 1850 kg car is moving right at a constant

velocity of 1.44 m/s. What is the net force on the

car?

0 as long as speed is constant

review problem

review problem .

What net external force is required to give a

25 kg suitcase an acceleration of 2.2 m/s2 to

the right?

a = F/m

2.2 = F/25 55 N

review problem

A boat moves through the water with two

forces acting on it. One is a 2.10 X 103 N

forward push by the motor, and the other is a

1.80 X 103 N resistive force due to the water.

What is the acceleration of the 1200 kg boat?

Fnet = ma

2.10 X 103 - 1.80 X 103 = (1200)(a)

.3 X 103 /1200 = a 0.25 m/s2

review problem

A dog pulls on a pillow with a force of 5 N at an

angle of 37° above the horizon. Find the x and y

components of this force.

37°

5N

x = 5cos37 3.9 N

y = 5sin37 3 N

Sec. 7.1 forces in 2 dimensions

An object is in equilibrium

when the net force on it is 0.

an object is motionless

When in equilibrium

an object moves with

constant velocity

Equilibrium – Occurs when the resultant of three or

more forces equals a net force of zero.

A

B

C

If A+B+C=0

Then A, B & C

are at

equilibrium

A

B

C

If an object is in equilibrium,

all of the forces acting on it

are balanced and the net force

is zero.

If the forces act in two

dimensions, then all of the

forces in the x-direction and y-

direction balance separately.

The 2 forces acting in the

same direction are additive.

It is much more difficult for

the gymnast to hold his arms

out at a 45-degree angle.

each arm must still support

350 N vertically to balance

the force of gravity.

Use the vertical force vector

(y-component) to find the

total force in each of the

gymnast’s arms (x

component).

350N each arm

vertically

This is the same as in our table

demonstration from yesterday.

Using what you know about

forces and equilibrium, draw a

vector diagram to show the

forces and explain the

experience of the test subject.

The ball carrier and

tackler collide and are

still moving forward.

How can a 2nd tackler

stop the forward

motion (produce

equilibrium)?

When 2 forces are

acting on an object

and the sum is not 0

– a 3rd force can be

added that produces

equilibrium.

*a force that produces equilibrium

finding the equilibrant:

15

To find this force, first find the sum of the two forces already being exerted on the object.

This single force that produces the same effect as the two individual forces added together, is the resultant force.

finding the equilibrant of A and B

16

equilibrant in the

opposite direction

The equilibrant force is one with the same magnitude as the resultant force, but in the opposite direction.

Forces are balanced

and the net force is 0

Review: Components of vectors

Going backwards to resolve a resultant vector into

the vectors that produced it (vector resolution)

R x

+x

+y

R y

q

17

To find the x component, Rx: cos θ = a/h

cos 40 = Rx/5.0km Rx = 5.0km (cos 40)

Rx = 3.8 km

To find the y component, Ry: sin θ = o/h

sin 40 = Ry/5.0km Ry = 5.0km (sin 40)

Ry = 3.2 km

R x

+y

+x

R y

q = 40o

18

3.8 km

3.2 km

A 168N sign is hanging motionless by two ropes

that make 22.5° angles with the horizontal. What

is the tension in the ropes?

22.5 22.5

First identify all the forces involved.

Example problem pg. 151

Equilibrium Problem Solving

Draw all forces on the object

Break forces into (x,y) components

Combine forces that are in the same direction

The sum of the forces must equal zero

Fa Fb

Fw

Fa Fb

Fw

Break forces into horizontal

and vertical components.

Fax

Fay Fby

Fbx

Fa

Fax

Fay

Fb

Fby

Fbx

Fay=Fasin22.5

Fax=Facos22.5

Fby=Fbsin22.5

Fbx=Fbcos22.5

Fa Fb

Fw

Fax

Fay Fby

Fbx

Resolve Forces By Direction

Fa Fb

Fw

Fax

Fay Fby

Fbx

Fnetx=0

Fnetx= -Fax+Fbx

0=-Facos22.5+Fbcos22.5

Facos22.5=Fbcos22.5

Fa=Fb

Fnety=0

Fnety= 0 = Fay + Fby - Fw

0 = 2Fasin22.5 - Fw

2Fasin22.5 = Fw

Fa = 168N/2sin22.5

Fa = 220N = Fb

Practice problems 1-4, pg. 151/152

problem 2, pg 151:

An 8.0 N weight has one horizontal rope exerting a

force of 6.0 N on it.

a. What are the magnitude and direction of the

resultant force on the weight?

b. What force (magnitude and direction) is

needed to put the weight into equilibrium?

Draw a vector diagram

(Draw all forces on the object)

Break forces into (x,y) components

Combine forces that are in the same direction

Determine the equilibrant

6.0 N

An 8.0 N weight has one horizontal rope exerting a force of 6.0 N on it.

a. What are the magnitude and direction of the resultant force on the weight?

8.0 N

FR ?

ϴ?

FX2 + FY

2 = FR2

FR = √(6.0 N)2 + (8.0 N)2 = 10 N

ϴR = 270⁰ + ϴy

ϴy = tan-1 (6.0/8.0) = 37⁰ ϴR = 270⁰ + 37⁰ = 307⁰

ϴy

FR = 10 N at 310⁰

b. What force

(magnitude and

direction) is needed

to put the weight into

equilibrium?

FE = 10 N at (307⁰ - 180⁰) = 10 N at 127⁰

Forces on an Incline

remember:

• Define your coordinates

• Draw a FBD

• Identify all forces.

• Resolve into x- and y- components.

How do you determine if a body is in equilibrium?

• Assign a coordinate system

• Resolve all forces into their x- and y-components

• When ΣFx = 0 AND ΣFy = 0, then the object is in

equilibrium. (velocity will not change)

θ?

Fg = 22 N

Ff = 11N

FN = 18N

Remember: normal forces are always

directed perpendicular to the surface

that the object is on.

As shown in the diagram, there are always at least two

forces acting upon any object on an inclined plane - the

force of gravity and the normal force. The force of gravity

(weight) acts in a downward direction; yet the normal

force acts in a direction perpendicular to the surface

Analyzing the forces acting upon objects on inclined planes

involves resolving the weight vector (Fgrav) into two

perpendicular components - one directed parallel to the

inclined surface and the other directed perpendicular to the

inclined surface.

Together the two components replace the affect of the

force of gravity.

The perpendicular component of the force of gravity is directed

opposite the normal force (balances the normal force).

The parallel component of the force of gravity is not balanced by any

other force.

An object will accelerate down the inclined plane due to the presence

of an unbalanced force. It is the parallel component of the force of

gravity that causes this acceleration. The parallel component of the

force of gravity is the net force.

inclined plane problems can be simplified through a

useful trick known as "tilting the head."

Once the force of gravity has been resolved into its two components and the inclined plane has been tilted, merely ignore the force of gravity (since it has been replaced by its two components) and solve for the net force and acceleration.

The task of determining the

magnitude of the two

components of the force of

gravity is a matter of resolving

the (resultant) gravity vector into

its two components, x and y

example problem pg. 152

A trunk weighing 562 N is resting on a plane inclined

at 30°. Find the components of the weight force

parallel and perpendicular to the plane.

Fg,x = mg x sinθ (562)(sin 30°) = 281 N

Fg,y = mg x cosθ (562)(cos 30°) = 487 N

Fg

FN

30°

remember: sinθ = Fg,x/Fg

Fg = W = mg

remember: cosθ = Fg,y/Fg

Fg = W = mg

X

Y

the short cut equation

Once you understand how to derive the components –

use the “shortcut” equations

The equations for the parallel and

perpendicular components are:

F = mg x sin θ

F = mg x cos θ

X+

y+

In the absence of friction and other forces the acceleration

of an object on an incline is the value of the parallel (x)

component (mg x sin θ) divided by the mass (m). This yields

the equation:

a = g x sin θ

F = ma

mg x sin θ = ma

mg x sin θ/m = a

g x sin θ = a

a 1000-kg roller coaster on

the first drop of a roller

coaster ride is at an angle

of 45°.

Determine the net force on

the riders and acceleration

of the roller coaster car.

(Assume a negligible affect of

friction and air resistance.)

try an example…

Fgrav = m • g = (1000 kg) • (9.8 m/s2) =

9800 N

The parallel and perpendicular components of the gravity force can be determined from their respective equations:

Fll(x) = mg • sin 45° = 6930 N Fperpendicular = mg • cos 45° = 6930 N

There are no other forces to counteract the parallel component of gravity (x). the net force = the Fll value.

The forces directed perpendicular to the incline balance each other.

Fnet = 6930 N, down the incline (the force on the riders)

The acceleration: a = Fnet /m: = 6930 N/1000 kg

a = 6.93 m/s2, down the incline

when other forces are involved…

The perpendicular component of force balances the normal force. Now the frictional force must also be considered when determining the net force of the X (parallel) component. The two parallel forces must be added to determine the net force. The parallel component and the friction force add together to yield 5 N. The net force is 5 N, directed along the incline towards the floor.

In the presence of friction (or other forces), the situation is slightly more complicated.

will the physics book slide?

Known: unknown:

Fg = 22 N ΣFx = 0

Ff = 11N ΣFy = 0

FN = 18N

θ = 35°

need to determine the x and y components for the Fg

35°

Fg = 22 N

FN = 18N

35°

Fg = 22 N

FN = 18N Ff = 11N Fx = mg x sin θ

Fy = mg x cos θ

Fy = 22 x cos 35

= 18 N

ΣFN – Fy = 0

The parallel forces determine if the book will slide

Fx= 22 x sin 35

= 12.6 N know Ff = 11N

ΣFx – Ff = 12.6 – 11 = Fnet of 1.6 N downward – the book will slide

try another problem

a 100-kg crate is

sliding down an

inclined plane at an

angle of 30°.

The coefficient of

friction between the

crate and the incline is

0.3.

Determine the net force and acceleration of the crate.

quick review – coefficient of friction

Ffriction = μFN Lee Mealone is sledding with his

friends when he becomes

disgruntled by one of his friend’s

comments. He exerts a rightward

force of 9.13N on his 4.68 kg sled

to accelerate it across the snow. If

the acceleration of the sled is

0.815 m/s2, what is the coefficient

of friction between the sled and

the snow?

known: Fapplied = 9.13N,right m = 4.68 kg a = 0.815

m/s2 unknown: μ between sled and snow

FN = Fg

9.13N

Ffriction = μFN

Fg = ma

= (4.68)(9.8)

= 45.86 N

FN

Fg

Fnet = 0 because the sled has an acceleration

Fnet = (4.68)(0.815) = 3.81 N, right so Fapp > Ff

Ff = Fnet – Fapp = 3.81N – 9.13N = 5.32N, left

μ = Ff/FN = 5.32/45.86 = 0.116

one more A 405N rightward force is used to drag a

large box across the floor with a constant

velocity of 0.678 m/s. The coefficient of

friction between the box and the floor is

0.795. Determine the mass of the box.

known:

Fapp = 405N,right μ = 0.795 a = 0

unknown: m of the box

405 N

Velocity is constant so all forces are balanced,

therefore: Fapp = Ffriction = 405 N AND FN = Fg

FN = Ff /μ = 405/0.795 = 509.43N

FN = Fg so 509.43N = m(9.8) = 51.98 kg

back to a problem on an incline……

a 100-kg crate is sliding

down an inclined plane

at an angle of 30°.

The coefficient of friction

between the crate and

the incline is 0.3.

Determine the net force and acceleration of the crate.

First: find the force of gravity acting on the crate and the components of this force parallel and perpendicular to the incline.

The force of gravity (W = mg) is 980 N The components of this force are Fll(x) = 490 N (980N • sin30°)

and Fy = 849 N (980 N • cos30°).

Now the normal force can be determined to be 849 N (it must

balance the perpendicular component of the weight vector).

100 kg

Determine the net force and acceleration of the crate.

The net force is the vector sum of all the forces.

The forces directed perpendicular to the incline balance;

The forces directed parallel to the incline do not balance.

Fnet = 490 N - 255 N = 235 N

acceleration is Fnet/m = 235 N/100 kg) = 2.35 m/s2.

Determine the force of friction Ffrict = μFN = 0.3 x 849 N = 255 N

Try another (example pg. 153)

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. How fast is the skier

going 5.0 s after starting from rest?

37°

Fg = mg = 62(9.8) = 607.6

Fx = sin 37° x mg = 365.7N

Fy = FN = cos 37° x mg = 485.2N

Ffriction = μFN

= (.15)(485.2) = 72.78N

Fnet = Fx – Ff = 365.7 – 72.78 = 292.9N

Fnet = Fx – Ff = 365.7 – 72.78 = 292.9N

a = Fnet/m

= 292.9/62

= 4.72 m/s2

Vf = V0 + at

= (4.72)(5s)

= 23.6 m/s

Practice problems 5-8, pg. 154

Study guide problems

Worksheet problems

Projectile Motion Sec. 7.2

The most common example of an object moving in two dimensions

is a projectile.

After a projectile has been given initial thrust, it moves only by the force of gravity (ignoring air resistance).

If you know the initial thrust on a projectile your can figure out its trajectory**.

** its path

types of projectiles

A projectile is any object that once projected continues in

motion by its own inertia and is influenced only by the

downward force of gravity.

Remember: A force is not required to keep an object in motion. A force is only required to maintain an acceleration.

Gravity is the downward force on a projectile that influences

its vertical motion and causes the parabolic trajectory that is

characteristic of projectiles.

Recap:

• A projectile is an object upon which the only force is

gravity.

• Gravity acts to influence the vertical motion of the

projectile, thus causing a vertical acceleration.

• The horizontal motion of the projectile is the result of

the tendency of any object in motion to remain in motion

at constant velocity.

• Due to the absence of horizontal forces, a projectile

remains in motion with a constant velocity. Horizontal

forces are not required to keep a projectile moving

horizontally.

The only force acting upon a projectile is gravity!

There are the two components of the projectile's motion - horizontal and vertical motion (they are independent of each other).

Horizontal motion: Law of inertia (absence of gravity)

Vertical motion: acceleration of gravity

Only a vertical force acts on the projectile.

the projectile travels with a constant horizontal velocity and

a downward vertical acceleration.

Horizontal motion

The ball’s horizontal velocity remains constant while it falls because gravity does not exert any horizontal force.

Since there is no force, the horizontal acceleration is zero (ax = 0).

The ball will keep moving to the right at 5 m/sec.

Vertical motion

The vertical speed (vy) of the

ball will increase by 9.8 m/sec

after each second.

After one second has passed,

vy of the ball will be 9.8 m/sec.

After 2 seconds have passed,

vy will be 19.6 m/sec and so on.

Remember for “cliff” problems:

Horizontal and vertical motion act independently

vtdx = 2

2

1atd y =

vfx = vix = constant vfy = viy - gt

Projectile motion

is a combination

of horizontal and

vertical motion

but neither

affects the other

ANALYSIS OF MOTION

ASSUMPTIONS:

• x-direction (horizontal): uniform motion

• y-direction (vertical): accelerated motion

• no air resistance

QUESTIONS:

• What is the trajectory?

• What is the total time of the motion?

• What is the horizontal range?

• What is the final velocity?

Solving projectile problems: pg 156

Horizontal motion - same as constant velocity problems (1D)

a = 0 v = d/t dx = vt

Vertical motion – solve like a free fall problem

v = at a = v/t a = -g dy = -1/2at2

Time*

*solve for either -- same for both

How fast is the ball thrown?

Vertical dy = -1/2at2

5 = -.5(9.8)t2

t= 1 second

► Horizontal d = vt

► v = d/t

► v= 20m/1s

► v= 20 m/s

The ball is thrown at 20 m/s

example pg. 157

A stone is thrown horizontally at 15 m/s from the

top of a cliff 44 m high.

a. How far from the cliff does it hit the ground?

b. How fast is it moving when it hits?

Know:

vx = 15 m/s

a = -g

vy = 0

Want to know:

x when y= - 44m

v at that time

a. Solve for t in the air t = √-2d/a t = √ -2(44 m) = 3 s dy = -1/2at2 -9.8 m/s/s

b. How fast is it moving?

t = 3s

dx = vt = (15 m/s)(3.0 s) = 45 m

vy = -gt = (-9.8 m/s/s)(3.0s) = -29 m/s

v2net = v2x + v2y

v = √ (15 m/s)2 + (-29 m/s)2 = 33 m/s

vy

vx

one more example (#11, pg 158):

A steel ball rolls with constant velocity across a tabletop

0.95 m high. It rolls off and hits the ground 0.352 m from

the edge of the table. How fast was it rolling?

vy = 0 y = -1/2gt2 t = √-2y/g

t = √ -2(-.95 m) = 0.44 s 9.8 m/s/s

vx = d/t v = 0.352 m/ 0.44s = 0.800 m/s

0.95 m

another example problem

Wile E. Coyote accidentally runs off a cliff horizontally at

8.0 m/s. The cliff is 64 m high. How far from the base of

the cliff should the road runner find him?

d = vxt = (8 m/s)(3.6s) = 28.8 m

Formulas: vx = d/t y = -1/2gt2

rearrange: d = vxt t = √2y/g

solve for time: t = √2(64) / 9.8 = 3.6 s

Use time to solve for dx

Example problem

1. You know the initial speed and the time.

2. Use relationships: y = – ½ gt2 and x = vox t

3. The car goes off the cliff horizontally, so voy = 0. Solve:

y = – (1/2)(9.8 m/s2)(2 s)2 y = –19.6 m. (negative means the

car is below its starting point)

Use x = voxt, to find the horizontal distance: x = (20 m/s)(2 s)

x = 40 m.

A stunt driver steers a car off a cliff at a

speed of 20 meters per second. He

lands in the lake below two seconds

later. Find the height of the cliff and

the horizontal distance the car

travels.

A cow walks off a bridge at a speed of 2 m/s. It takes 9

seconds for the cow to land.

a) How far away from the base of the

bridge will the cow land?

V = d / t

2 m/s = d / 9 s

d = 18 m

Recall that the cow walked at a speed of 2 m/s and

took 9 seconds to land.

b) How tall is the bridge?

d = ½ g x t2

d = ½ (10m/s2) x 9 s2

d = 405 m

While driving Eleanor his freshly boosted 1967 Shelby

Mustang, Randall “Memphis” Reines decides to

escape from Det. Castlebeck by jumping over a car

accident using the flat bed tow truck. The bed of the

tow truck is 4 m high and Eleanor is traveling at 44

m/s.

How long does it take for Eleanor to reach the ground on

the other side of the car accident?

How far horizontally from the

end of the tow truck bed did

Eleanor fly?

Gone in 60 seconds

How long does it take for Eleanor to hit the ground on the

other side of the accident?

d = ½ g x t2

t = √2 d / g

t = √ 2 (4 m)/(10 m/s2)

t = 0.9 s

How far horizontally from the end

of the tow truck bed did Eleanor fly?

v = d / t . d = v x t

. d = 44 m/s x 0.9 s

. d = 39.6 m

Be sure to have all these formulas:

Horizontal:

Vx = Vxo

x = Vxt

Vertical:

y = 1/2 gt2 (on the cliff problems only)

Vy = Vyo - gt

y = Vyot - 1/2 gt2

Vy2 = Vyo

2 - 2gy

non-horizontally launched projectiles

(7.2 Projectile Motion at an angle)

How can you predict the range of a launched marble?

Projectile Motion and the Velocity Vector

The path a projectile

follows is called its

trajectory.

The distance the

projectile travels is its

range.

Projectiles launched at an angle

A soccer ball kicked off the ground is a projectile, but it starts with an initial velocity that has both vertical and horizontal components.

*The launch angle determines how the initial velocity divides

between vertical (y) and horizontal (x) directions.

Calculating the components

of a velocity vector

1. Draw a free-body diagram

2. Use vx = v cos θ and vy = v sin θ.

3. Solve:

vx = (10 m/s)(cos 30o) = (10 m/s)(0.87) = 8.7 m/s

vy = (10 m/s)(sin 30o) = (10 m/s)(0.5) = 5 m/s

A soccer ball is kicked at a speed of 10

m/s and an angle of 30 degrees. Find

the horizontal and vertical components

of the ball’s initial velocity.

practice problem:

finding vector components

Find the component velocities

for a helicopter traveling 95

km/hr at an angle of 35° to the

ground.

vx = v(cos θ) (95 km/hr)(cos 35) = 78 km/hr

vy = v(sin θ) (95 km/hr)(sin 35) = 54 km/hr

θ

95 km/hr

Calculating acceleration on an incline

Calculate the y component of the submarine’s ascent:

vy = (10 m/s)(sin 35o) = 5.7 m/s Calculate the time to ascend 250 m

?s/5.7 m X 250 m = 43.8 s

A submarine is moving at 10 m/s

toward the surface at an angle of

35° from the ocean floor. How long

will it take to surface if it is 250 m

deep?

practice problem:

The horizontal distance a projectile moves can be

calculated according to the formula:

The vertical component of velocity alone is responsible for

giving the projectile its vertical height AND "hang" time.

y= -1/2gt2

"Hang Time"

You can easily calculate your own hang time.

Run toward a doorway and jump as high as you can, touching the wall

or door frame.

Have someone watch to see exactly how high you reach.

Measure this distance with a meter stick.

The vertical distance formula can be rearranged to solve for time:

Calculating range and velocity

punkin chunkin trailer

world record air cannon

George Fox

Univ.

Human powered

Upwardly Launched Projectiles What’s missing?

What’s

staying the

same?

What does gravity do to its horizontal speed?

NOTHING!!!!

But its vertical speed when it lands will equal the

speed at which at was launched.

Example of a Cannon

Which cat will go further?

A B C

Which cat will go higher?

A B C

Which cat will be in the air longer?

A B C

What determines how far the projectile

will go?

The initial angle of the projectile and its

horizontal velocity determine its range.

Projectiles launched at the same speed but

different angles.

What do you notice about the diagram?

A ball launched at a

steep angle will have a

large vertical velocity

component and a small

horizontal velocity.

A ball launched at a

low angle will have a

large horizontal

velocity component

and a small vertical

one.

Which angle produces the largest range for the

projectile?

45o

Which angle will produce

the longest hang time

for the projectile?

90o

The range of a projectile is calculated from the

horizontal velocity AND the time of flight.

If a ball is kicked at the same speed, angles that add up

to 90 degrees have the same range.

In those cases in which the vertical displacement is zero,

the horizontal displacement (also known as range) can be

expressed in terms of initial launch speed and angle:

Shot Put - 45º is not the ideal angle, why?

You are already off the ground 1-2 meters.

boy, have we got problems!!!

take some of these on

Be sure to have all these formulas:

Horizontal:

Vx = Vxo

x = Vxt hang time

Vertical:

y = - 1/2 gt2 (on the cliff problems only)

Vy = Vyo - gt

y = Vyot - 1/2 gt2

Vy2 = Vyo

2 - 2gy range

An Australian football is kicked at an angle of 37.0° with a velocity of 20.0 m/s. Find: (a) the max height, (b) the time in air before striking the ground, (c) the horizontal distance traveled, (d) the velocity vector at the max height, and (e) the acceleration vector at max height.

Horizontal Vertical

Displacement ? ?

Time ? ?

Initial Velocity Vo cos 37.0o Vo sin 37.0o

Average

Velocity ? ?

Final Velocity ? ?

Acceleration 0 9.8 m/s/s

Here’s what we know:

Where do you begin?

Look at the formulas again, they’re taunting you aren't they?

set-up the table and we can see we can find initial velocity for both the horizontal and vertical:

Vxo = Vo cos 37.0o = (20.0 m/s)(0.799) = 16.0 m/s Vyo = Vo sin 37.0o = (20.0 m/s)(0.602) = 12.0 m/s

(a)The max height is attained where Vy = 0, this occurs when t = Vyo/a, ergo (12.0 m/s)/(9.80 m/s2) =

1.22 s

fill in the table as we go along

Time for another formula: y = Vyot -

1/2 gt2

= (12.0 m/s)(1.22 s) - 1/2(9.80 m/s2)(1.22 s)2 = 7.35 m.

(b) To find the time it takes for the ball to return to the ground, we use the following equation and set y = 0 (for ground level).

y = Vyot - 1/2 gt2

0 = (12.0 m/s)t - 1/2(9.80 m/s2)t2

thus, t = 2(12. m/s)/(9.80 m/s

2 ) = 2.45 s

you might have found t = 0, this is a solution, congratulations, but it is wrong, it is 0 when the initial point y is zero.

(c) The total distance traveled horizontally is found by applying the equation x = Vxot, remember: a = 0,

(remember from your table: Vxo = 16.0 m/s)

x = (16.0 m/s)(2.45 s) = 39.2 m

(d) The velocity at max height At the max height there is no vertical component to the velocity, only horizontal.

so v = Vxo cos 37.0o = 16.0 m/s.

(e) The acceleration vector at max height The acceleration vector is always 9.80 m/s2 downward.

are you ready??

Test – Wednesday!!!

Calculating acceleration on an incline

1. You know the mass, friction force, and angle.

2. Use relationships: a = Fnet ÷ m and Fx = mg sinθ.

3. Calculate the x component of the skier’s weight:

Fx = (50 kg)(9.8 m/s2) × (sin 20o) = 167.6 N

Calculate the force: Fnet = 167.6 N – 30 N = 137.6 N

Calculate the acceleration: a = 137.6 N ÷ 50 kg = 2.75 m/s2

A skier with a mass of 50 kg is on a

hill making an angle of 20 degrees.

The friction force is 30 N. What is

the skier’s acceleration?

Tuba Tony gets into a fight with his girlfriend Charity the

Cheer Queen at the last home football game of the year.

While Tuba Tony is on the field during the half time show,

Charity throws Tony’s class ring off the top of T-bird

Stadium. .

The stadium is 10 m high and the ring was thrown at a

velocity of 10 m/s. Tony, distraught over the loss of his

class ring decides to search for it. Let’s use physics to

determine where he should look!

How long does it take for Tuba Tony’s ring to hit the

ground?

How far horizontally from the base of the stadium does

Tuba Tony’s class ring land?

How long does it take for Tuba Tony’s ring to hit

the ground?

y = ½ gt2

t = √ 2y/g

t = √2(10 m) / (10 m/s2)

t = 1.41 s

How far horizontally from the

base of the stadium does Tuba Tony’s

class ring land?

v = d / t d = v x t

. d = 10 m/s x 1.41 s

. d = 14.1 m

Pirate monkeys on a boat are positioned at rest

100 m from the base of a 40 m high cliff. A

cannon sits on the cliff and fires horizontally at

the boat with a velocity of 35 m/s.

Will the cannonball hit the boat?

100 m

40 m

35 m/s

How long will it take the cannonball to hit the water?

d = ½ g t2

t = 2d / g

t = 2(40m) / (10m/s2)

t = 2.83 s

How far will it travel horizontally

from the base of the cliff?

v = d / t . d = v x t

d = 35 m/s x 2.83 s

=

99.05 m

A mountain climber encounters a crevasse in an ice field.

The opposite side of the crevasse is 2.75 m lower and is

separated horizontally by a distance of 12 m. To cross the

crevasse, the climber gets a running start and takes off

horizontally. What speed will the climber have to take off

to make it safely cross?

Choose a formula:

y = – 1/2gt2

Solve for time:

-2.75 = -1/2gt2

t = 0.75 s

Calculate the horizontal velocity needed:

x = vxt v = x/t

= 12/.75 = 16 m/s

A hunter aims his arrow directly at a target (on the same

level) 100 m away. If the arrow leaves the bow at a

speed of 75 m/s, by how much will it miss the target?

What should be the launch angle to hit the target?

1.3 s